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Generally, a linear congruence is a problem of finding an integer x that satisfies the equation ax = b (mod m). Thus, a linear congruence is a congruence in the form of ax = b (mod m), where x is an unknown integer. In a linear congruence where x0 is the solution, all the integers x1 are x1 = x0 (mod m).

You can use several methods to solve linear congruences. The most commonly used methods are the Euclidean Algorithm Method and the Euler's Method.

**Example:** Solve the linear congruence ax = b (mod m)

**Solution:** ax = b (mod m) _____ (1)

a, b, and m are integers such that m > 0 and c = (a, m).

If c cannot divide b, the linear congruence ax = b (mod m) lacks a solution.

If c can divide b, the congruences ax = b(mod M) has an incongruent solution for modulo m.

As mentioned, ax = b (mod m) is equal to ax - my = b. If we apply Theorem 19, you realize that c cannot divide b; thus, it is impossible to get a solution for the equation ax - my = b.

Alternatively, if c can divide b, it implies that there are several solutions whose variable is x denoted by x = x0 + (m/c)t _____ (2)

Thus, the values of x become the solution for the congruence ax = b (mod m). You can now easily find the number of all the incongruent solutions.

Assuming the two solutions are incongruent such that:

x0 + (m/c)t1 = x0 + (m/c)t2 (mod m) _____ (3)

(m/c)t1 = (m/c)t2 (mod m) _____ (4)

You now realize that (m, m/c) = m/c

Thus t1 = t2 (mod c) _____ (5)

You now have a set of incongruent solutions given by X = X0 + (M/C)T, where T is given as modulo C.

Mathematically, if C = (A, M) = 1, there is always a definite unique solution for modulo M in the linear congruence AX = B(mod M).

The Euclidean Algorithm Method is one of the simplest methods of solving linear congruences. The technique works so that if d is the Greatest Common Divisor of two positive integers, say a and b, the d divides the reminder (r). This remainder (r) results from dividing the smaller of a and b into the larger.

The Euclidean Algorithm Method allows you to find the middle ground pathways of prime numbers while solving linear congruences.

**Example:** Solve the linear congruence 3x = 12 (mod 6)

**Solution:**

3x = 12 (mod 6)

(3, 6) = 3 and 3/12

Thus, you get three incongruent solutions for mod 6.

You should use the Euclidean Algorithm Method to find the solution for the resultant linear diophantine equation 3x - 6y= 12 to give you x0 = 0.

The three incongruent solutions are given by:

x1 = 6 (mod 6)

x1 = 6+2 = 2 (mod 6)

x2 = 6+4 = 4 (mod 6)

The GCD of 24 and 16 is 8.

Thus; 24 = 1(16) + 8

Therefore the remainder is 8 when you divide 24 by 16, which is divisible by 8.

To better understand how to use the Euclidean Algorithm Method, we will use the equation 11x = 1 mod 23.

**Example:** Solve the Linear Congruence 11x = 1 mod 23

**Solution:** Find the Greatest Common Divisor of the algorithm.

23 = 2(11) +1

11 = 1(11) + 0

Therefore the GCD is 1.

You can see from the equation that 1 = (1)(23) + (-2)11 (mod 23)

(-2)(11) = 1 mod 23

Thus, x = -2

To write the answer as the value from 0 to 22, you realize that -2 = 21 mod 23; thus, the solution is x = 21.

**Example:** Consider the linear congruence 16x = 5 mod 29

**Solution: **

First, solve the congruence 16x = 1 mod 29

29 = 1(16) +13 (2)

16 = 1(13) + 3 (3)

13 = 4(3) + 1 (4)

Hence you get:

1 = 1(13) + (-4)(3).

Substitute in the equation 1 = 1(13) + (-4)(3) using the fact that from (3), 3 = 16 -1(13) you get:

1 = 1(13) + (-4)(16-1(13)) = 5(13) + (-4)(16) (5)

You can now use (2) with the knowledge that 13 = 29 -1(16) to substitute for 13 in (5).

1 = 5(29-1(16)) + (-4)(16) = (-9)(16) +(5)(29)

Therefore the last equation mod 29 results to:

(-9)(16) = 1 modulo 29

The value of x is thus -9, which in this case, is congruent to modulo 29 to 30.

It doesn't end here, though.

Now that you know 16(20) is congruent to 1 mod 29, multiply both sides of the equation by 5 to get 100(16), a congruent to modulo 29. And because 100 is congruent to 13 mod 29, the solution to the linear congruence 16x = 5 modulo 29 is 13.

Lastly, verify that 16(13)-5 will leave a zero remainder when you divide it by 29.

This method applies to solve a linear diophantine equation. A linear diophantine equation is any equation expressed as ax + by = c. Euler's method applies the knowledge of solving linear diophantine equations to solve linear congruences.

It is important to note that there exists a close relationship between linear diophantine equations and linear congruences.

This is so because in the equation a = b (mod n), n divides (a-b) or a-b = nt for some t, or a= b + nt.

Also, the equation a = b + nt can be converted to modulo n:

a = b + nt

a = b + 0t mod n

Hence a = b mod n

You can easily convert the linear congruence 13x = 4 mod 37 to a diophantine equation 13x = 4 + 37y.

Solving linear congruences using Euler's Method involves changing congruences to equations. You then change the equation to a congruence modulo using the smallest coefficient.

Solve the following diophantine linear equation.

23x + 49y = 102

23x + 49y = 102 _____ (1)

First, change the diophantine equation into a linear congruence. You realize from the equation that mod 23 is the smallest coefficient.

49y = 101 mod 23

Now simplify the equation to get:

3y = 10 mod 23

Now use Euler's Method to change the equation into equality.

3y +10 + 23a _____ (2)

Now reduce the equation to congruence mod 3, which is the smallest coefficient.

0 = 10 + 2a mod 3 or 0 = equiv 1+ 2a mod 3.

Simplify this further to an equation to get:

0 = 1 + 2a + 3b _____ (3)

Before we go to the next step, take note of the introduction of a new variable (b).

Next, let us take mode 2 to get:

0 = 1 +3b mod 2

Simplify further to get:

0 = 1 + b mod 2

This results in:

b −1 mod 2

Now that you have b mod 2 pick any b integer satisfying this congruence, for example, b = -1.

Go back to the x equation (2) to get:

1 + 2a + 3(−1) = 0.

1 + 2a − 3 = 0

Next, go to the y in equation (2) to get:

3y = 10 + 23a = 10 + 23 = 33

Therefore y = 11

Go back to the original equation (1) to get:

23x + 49 11 = 102

23x = 102 − 539 = 437

x = 19

Therefore the final solution is:

x = 19

y = 11

Understanding the Euclidean Algorithm and Euler's Methods makes solving linear congruences less complicated. The two methods allow us to extend modular arithmetic beyond simple addition, multiplication, and subtraction to give room for division. It has cemented a fundamental relationship between integer linear combination of numbers and their GCD.

Take your time to learn the two methods and start solving linear congruences with ease.

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If you have data with a mean *μ* and standard deviation* σ, *you can create models of this data using typical distribution. We can find the probability within this data based on that mean and standard deviation by standardizing the normal distribution.

The equation for the probability of a function or an event looks something like this (x - *μ*)/ *σ *where σ is the deviation and μ is the mean. Using the standard or z-score, we can use concepts of integration to have the function below.

This might appear strange at first, but what it means is that anyone can find probabilities for any given normal distribution as long as they have the mean and the standard deviation without having to do any integration. As long as you have the standardized table with a standardized normal curve with a standard deviation (unity) and a single mean, you can calculate probability using the z-score. It is this same table that we will use to calculate probabilities in the examples below.

You can download the Standard Normal Distribution Table from websites like Maths Arizona as a pdf or excel file. Look closely at the table; you will see that it contains values from negative infinity to x. X values are from 0 to 3, and in very rare cases, 4 bringing the probability daringly close to unity or one.

This means that *P*(*X* ≤ *x*) =

Calculating P(x) may appear straightforward, but what if you want to calculate for a range of numbers, say p(X > x)? This is outside of the values on the table but *P*(*X* > *x*) = 1 – *P*(*X* ≤ *x*). In this case, we'll look for the value of *P*(*X* ≤ *x*) and subtract from one.

**1**. What is the probability that 5 is greater than x in a normally distributed data given that the mean is 6, and the standard deviation is 0.7.

P(X < 5) the first step is to find the z- score. We find that using the formula above.

z = (x – μ (mean)) / σ (standard deviation) this means that

*For *P(X < 5), z = (5 - 6)/0.7

-1/7 = - 1.42857 which is rounded up to – 1.43

Now in the table, we will look for the value of -1.4 under 3

= 0.07636

The normal return for the z-score is usually less than, and because the function is asking for the probability of x being less than 5, this will be our final answer.

**2**. What is the probability that x is greater than 4.5 in a normally distributed data given that the mean is 6, and the standard deviation is 0.7.

P(X > 4.5) => the first step is to find the z- score. We find that using the formula below

z = (x – *μ (mean)*) / *σ (standard deviation) this means that *

*For *P(X > 4.5), z = (4.5 - 6)/0.7

-1.5/0.7 = - 2.14285 which is rounded up to – 2.14

Now in the table, we will look for the value of -2.1 under 4

= 0.01618

The normalization table returns for the z-score is usually less than, but the function is asking for the probability of x being greater than 4.5; this means that the value we got is for x less than 4.5 and not greater than 4.5. To get the probability for x greater than 4.5, we will have to subtract the answer from unity.

=> 1 - 0.01618 = 0.9838

**3.** Find the probability that x is greater than 3.8 but less than 4.7 in a normally distributed data given that the mean is 4 and the standard deviation is 0.5.

This problem is a bit different from the rest. Here we are asked to find the probability for two values when x is greater than 3.8 and less than 4.7. This means it falls between 3.9 and 4.6.

We can express this as P (3.8 < x <4.7).

Here we will be finding the z-score for P (x > 3.8) and P (x < 4.7). We find that using the formula below

z = (x – μ (mean)) / σ (standard deviation) this means that

for P (X > 3.8), z = (3.8 - 4)/0.5

-0.2/0.5 = - 0.400

Now in the table, we will look for the value of -0.4 under 0

= 0.34458

*For *P (X < 4.7), z = (4.7 - 4)/0.5

0.7/0.5 = 1.40

Now in the table, we will look for the value of 1.4 under 0

= 0.91924

We are going to subtract the upper limit by the lower limit

0.91924 - 0.34458 = 0.57466

The probability that x is greater than 3.8 but less than 4.7 is 0.57466

**4.** Find the probability that x is less than 6 but greater than 4 in a normally distributed data given that the mean is 5 and the standard deviation is 0.6.

We are looking for the probability that x ranges from 4.1 to 5.9

We can express this as P (4 < x < 6).

Here we will be finding the z-score for P (x > 4) and P (x < 6). We find that using the formula below

z = (x – μ (mean)) / σ (standard deviation) this means that

*For *P (X > 4), z = (4 - 5)/0.6

-1/0.6 = - 1.67

Now in the table, we will look for the value of -1.6 under 7

= 0.04746

*For *P (X < 6), z = (6 - 5)/0.6

1/0.6 = 1.67

Now in the table, we will look for the value of 1.6 under 7

= 0.95254

We are going to subtract the upper limit by the lower limit

0.95254 - 0.04746= 0.90508

The probability that x is less than 6 but greater than 4 are 0.90508

In a normally distributed data set, you can find the probability of a particular event as long as you have the mean and standard deviation. With these, you can calculate the z-score using the formula z = (x – μ (mean)) / σ (standard deviation). With this score, you can check up the Standard Normal Distribution Tables for the probability of that z-score occurring.

No matter the value of the mean and the standard deviation, the probability of x being equal to any number is automatically zero. There is an emphasis on a normally distributed data set because if your data isn't distributed normally, you may have to consider different factors like kurtosis.

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Trigonometric identities are different ways of representing the same expression. They are used to solve a trigonometric equation when applied to the given scenario.

When a criminal is on the run, they will choose an Italian passport to assume a covert Roman identity for travel purposes. An identity is selected and applied to the expression for a solution to a trigonometric equation.

Let’s examine the fundamental identities before verifying them to solve trigonometric equations.

Identities are the enablers that simplify complicate trigonometric expressions or equations. They are vital tools in the solution of trigonometric equations. Trigonometric identities work alongside factoring, finding special formulas, and using common denominators.

Like an algebraic equation, trigonometric equations are composed of basic formulas and properties of algebra. Perfect square and difference of square simplify working with expressions and trigonometric equations. It’s common knowledge that all trigonometric functions are closely related. That’s because they are all definitions of the unit circle, and their identities can be written in several ways.

To solve trigonometric identities requires that you start with the more complicated part of the equation. You'll have to essentially rewrite the trigonometric expression until it’s transformed to become similar to the other part of the equation.

To obtain the desired results, you may have to expand or factor expressions while finding common denominators. You can also use any other algebraic strategy to transform the expression.

Consider this function*: f(x) = 2x² + x.*

*Solve f(x) = 0.*

You already know that solving the function requires simple algebra. This will work out like;

*2x² + x = 0*

*x (2x + 1) = 0*

*x = 0 or*

*x = − ½*

Given the same scenario with g (t) = sin (t) and being asked to solve g (t) = 0. You can find the solution for this function using unit circle values.

These may include* sin (t) = 0 for t = 0, π, 2π *and others.

Using similar concepts, you can now consider the following functions and their composition.

*f (g(t)) = 2(sin(t))² + (sin(t)) = 2sin²(t) + sin(t)*

The equation created is called a polynomial-trigonometric function. To solve trigonometric equations using the like functions, use identities, the quadratic formula, and algebraic techniques such as factoring.

Six or more ratios exist that can derive trigonometric elements. They are known as trigonometric functions. These are sine, cosine, secant, co-secant, tangent, and co-tangent.

Identities and functions of trigonometry are derived using the right-angled triangle as a reference. They appear as;

- Sin θ: Equals the Opposite Side divided by the Hypotenuse
- Cos θ: Equals the Adjacent Side divided by the Hypotenuse
- Tan θ: Equals the Opposite Side divided by the Adjacent Side
- Sec θ: Equals the Hypotenuse divided by the Adjacent Side
- Cosec θ: Equals the Hypotenuse divided by the Opposite Side
- Cot θ: Equals the Adjacent Side divided by the Opposite Side

In some cases, a single trigonometric equation will use a variety of reciprocal identities. These are given as;

- sin θ = 1 divided by cosec θ
- cot θ = 1 divided by tan θ
- tan θ = 1 divided by cot θ
- cosec θ = 1 divided by sin θ
- cos θ = 1 divided by sec θ
- sec θ = 1 divided by cos θ

Reciprocal identities arise from a right-angled triangle. When the base and height are given, you can find the sin, cos, tan, sec, cos, and cot values with trigonometric formulas.

By using trigonometric functions, reciprocal trigonometric identities can also be derived.

Also called co-function identities, the periodicity formulas are used for shifting angles by π/2, π, and 2π and so forth.

Periodicity formulas are useful in calculating complex geometry, evaluating trigonometric functions, and proving other identities. The first two are; *sin (90°−x) = cos x, and cos (90°−x) =sin *x, and are the most commonly used.

Since trigonometric identities are cyclic, they repeat themselves following the periodicity constant. For *different trigonometric identities, the radian constant for periodicity is different.*

*Tan 45° = tan 225°, and the same is applicable for cos 45° = cos 225°.*

You can also represent the co-function of periodic identities in degrees. These include *sin (90°−x) = cos x, cos (90°−x) = sin x, tan (90°−x) = cot x, cot (90°−x) = tan *x and so on.

Co-function or periodic identities relate to trigonometric function co-pairs at x and π/2. For instance, *sin (π/2 - x) = cos(x), cos (π/2 - x) = sin(x), tan (π/2 - x) = cot(x), and cot (π/2 - x) = tan(x).*

The angle addition formula can derive double angled identities. They are not principally hard to memorize and usually come in handy in a lot of trigonometric equations.

You can use the double angled identities to find the cosine and sine of 2x in terms of the cosines and sines of x following from the angle-sum formula.

The sine area formula for a triangle *is A = ½ **⋅ ab sin C*. Angle sum identities tell how to find the sine and cosine of *x + y *when the cosines and sines of x and y are given.

These may look intimidating, but they are simple to derive. The double angled formula can further generate half angled identities.

You can use the half angled identities to find the sine and cosine of x/2. This is in terms of the cosines or sines given for x following after the double angled formulas.

The negative-angle identities are based on the unit circle and are often called odd, even identities. You can find the trigonometric functions at –x when the identities of x relate to values at opposing angles –x and x.

For instance;

*sin(−t) = −sin(t)cos(−t) = cos(t)tan(−t) = −tan(t)csc(−t) = −csc(t)sec(−t) = sec(t)cot(−t) = −cot(t)*

Alongside reciprocal identities, you can use these to solve a single equation.

You are given a trigonometric equation that looks a lot like a quadratic equation.

*2sin2 (t) + sin (t) = 0*

The problem requires all solutions with 0≤t<2π. It is also called a quadratic in-sine equation because of the sin (t) instead of a quadratic variable.

Use the quadratic formula or factoring techniques as with all quadratic equations. By factoring out the common sin (t) factor, the expression factors well.

*sin (t)(2sin(t)+1) = 0*

If either factor is zero, you know that the product on the left equals zero. It is also called the zero product theorem, and it enables you to break the equation into two expressions.

*sin (t) = 0 or,*

*2sin (t) + 1 = 0*

These equations can then solved independently.

*sin (t)= 0t = 0 or,*

*t = π*

*2sin(t)+1 = 0sin(t) = −12t=7π6 or,*

*11π6*

It will give you four solutions for the *0≤t<2π: t=0, π, 7π6, 11π6 *equation. To check if the answers are reasonable, you can compare the zeros after graphing the function.

To solve trigonometric equations, you need to use identities and reference angles to memorize alongside algebra.

These equations require that you think and have a good grasp of the first quadrant trig-ratio value and the workings of the unit circle. Be ready to identify the various trigonometric functions in the first period or the relationship between degrees and radians.

One of the key concepts to take with you is that multiple representations of a trigonometric expression exist. A verified identity will illustrate how to simplify the equation by rewriting the expression.

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- Acute angles: angles that are 0-90 degrees
- Obtuse angles: angles that are 90-180 degrees

The sides that form the angles of an oblique shape are never perpendicular. In simple terms, these sides will never meet.

When someone talks about oblique shapes, they refer to a shape, either plane or space, that has either an acute or obtuse (an oblique) angle.

You can figure out if two planes are oblique or not by looking at the angle they create when they intersect. When two plane figures intersect, squares, for example, and create an oblique angle, the planes are called oblique planes.

The most straightforward example of an oblique plane figure to visualize is an oblique triangle. An oblique triangle is any triangle that is not a right triangle. This is because it has an oblique angle and does not contain an angle that is 90 degrees. You can instead call this triangle an acute or obtuse triangle.

An acute triangle has three internal angles of less than 90 degrees. Obtuse triangles have one internal angle greater than 90 degrees; the other two are less than 90. Regardless of the type of triangle, it still follows that all three internal angles will equal 180 degrees.

Finding the area of oblique figures is not that different from finding the area of right figures.

Triangles: base x height / 2 = area

An oblique space figure, like a rectangular prism, has faces that do not align with the bases. This makes the prism slanted with its base still flush. This is the case for any oblique prism, including a cylinder. The Leaning Tower of Pisa is a fun way to think about oblique space figures. If the figure leans like the tower, it’s oblique.

Note: an oblique space figure is not tipped so that the bottom is balancing on one of its corners. This is simply a tipped right space figure.

Like with plane figures, there really isn’t much of a difference in figuring out the volume of an oblique space figure.

Cylinder: base x height = volume

Pyramid: base x height (of apex when oblique) / ⅓ = volume

Essentially, all formula will remain the same. You must consider any measurement changes that could have happened when the right space figure turned into an oblique prism.

There are some shapes out there that we may not use quite often, but they are quite important to know in the realm of oblique planes. One of those figures is a rhomboid. A rhomboid is a parallelogram with oblique, internal angles. When a rhomboid has sides of equal length, it is a Rhombus.

But is a parallelogram oblique? Not always.

If you want to define a parallelogram, it is a quadrilateral with opposite sides equal in length and parallel. Technically, a rectangle can be a parallelogram. When the rectangle is not right, it becomes this oblique shape.

There is also a type of drawing called an oblique pictorial. You’ve likely drawn the most common oblique pictorial, a cube.

To create an oblique pictorial, you take a plane figure and make it 3D. You do this by extending parallel, angled lines from the figure and connecting the lines to mirror the shape. Usually, these angles are at a 35, 45, or 60-degree angle.

Can you guess why we refer to these drawings as oblique drawings? Look again at the standard angles used to make the plane figure 3D -- they are all less than 90 degrees. Therefore, we draw the extended lines at an oblique angle.

There are three types of oblique pictorial drawings, cabinet, cavalier, and general.

- Cavalier: represents the full depth
- Cabinet: represents half the depth
- General: concerned more with the idea of the pictorial rather than accuracy of depth

Outside of trying to impress your friends in middle school math class with a general oblique pictorial, they have very real-world applications. Eighteenth-century military artists in France used the cavalier pictorial method to design forts. The cabinet projection comes from its application in drawings used in the furniture business.

So when you look at building structures and even your living room furniture, look for the oblique angle that turned its face into a 3D figure.

Think of what you can call an oblique shape without having to use a protractor to measure the angles -- any imperfect tortilla chip, for one. Identifying oblique shapes is simple once you know what you’re looking for: an acute or obtuse angle! Remember, if it’s not right or a prism is leaning, it’s oblique.

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One can define gravity as a universal force that acts between two objects. It tends to pull objects towards the center of the earth. Each body in the universe possesses a particular amount of matter. This is known as mass, which is defined as the amount of matter contained in a substance. Anything that occupies space and has weight is known as matter. The SI unit for mass is the kilograms (KG), although mass can also be expressed in grams(g), milligrams (mg), and even tones (t) and is measured using a beam balance.

When gravity acts on a body, it experiences a pull effect. This pull effect of gravity is known as **weight**. Weight is measured in Newton and changes depending on the place. The SI unit for weight is Newton (N). The difference between mass and weight is brought about by the presence of gravity in the latter. Weight is a vector quantity, and it has both magnitude and direction.

An astronomer's weight on earth slightly differs from the same astronomer's weight on a different planet, but the mass remains constant. Many people think it's fine to refer to mass and weight as the same thing, but a science student needs to understand the difference between them. Weight is directly proportional to the gravitational force available; thus, mathematically, this can be expressed as:

Weight = mass * gravity

W= mg

Lifting a 1-kilogram bag requires less physical effort compared to a 2-kilogram bag. This is because a 1-kilogram bag will weigh more than a 2-kilogram bag. The 1 Kg bag has a force of about 10 Newton acting on it, and the 2 Kg bag has a force of about 20 Newton acting on it; thus, the higher the mass, the larger the force required to carry it.

A man on the moon weighs slightly lighter than he does on earth. The man's mass is constant on both the moon and earth; what brings about the difference in weight then? This is brought about by the different gravitational forces on both the earth and the moon's surface. The gravity at the surface of the moon is about 1/6th of the acceleration on earth. Thus, for a 2-kilogram bag of maize, a force of about 20 newton acts on it on earth, which translates to about 4 newton on the moon. This also happens because the moon is less heavy than the earth.

Even on the earth's surface, there are local variations in the gravitational force due to altitude, latitude, and geological structures. When a body is at rest, it experiences some acceleration due to its own weight and contents. To calculate the gravity load, one needs to get the product between the object's mass, the earth's gravitational acceleration, and the height above the ground in meters.

i.e. m*g which is (9.8 m/s2) * h

=mgh

Weight is the pull effect felt on a body due to gravity. The formulae for calculating weight as stated earlier is w = m * g ….. (i)

Where 'w' is the weight of the object, 'm' is the mass of the body measured in kilograms (Kg), and 'g' represents gravitational acceleration, which is 9.8 m/s2 when expressed in meters and 32.2 f/s2 when expressed in term of feet. Weight is a force; thus, the equation (i) above can also be written as F=mg.. .(ii)…this is expressed in Newton.

When calculating weight, the mass of the object in question should always be converted to kilograms. You shouldn't mix up units of measurement; also, during the calculation, you should only use the scientific units available and convert them accordingly if need be. The next step is to figure out the gravitational acceleration depending on where gravity is acting from as the gravitation forces vary.

For example, the gravitation acceleration on the sun is about 274.0 m/s2. This is about 28 times the acceleration on earth. This means, for instance, if a man weighs 20 kilograms, he would have a weight force of 200 Newton on earth and 560 Newton on the sun. This is proof that the wrong estimated value of gravitational acceleration could lead to finding fault results; therefore, enough research must be done concerning this before proceeding with the math.

Lastly, once the mass and gravitational force with the right units have been noted, the values are then substituted in the equation (ii), i.e., F=mg accurately and the final weight force calculated.

Example:

Let's say an object has a mass of 90 Kg; what is its weight on earth?

I) identify the mass and convert it to kilograms

In this case, the mass is already in kilograms, and therefore it remains m=90 Kg

II) Identify the gravitational acceleration

In our case above, the space in question is the earth, which ah a gravitational acceleration of 9.8 m/s2

III) Finally, after identifying the 'g' and 'm,' substitute these values to equation [ii] above

F=m*g

=9.8 * 90 =882N

For this particular object, the weight is 882 Newton

A body has a mass of 100,000grams; what is its weight on the moon?.

In this above example, we have to convert the mass from grams to kilograms by diving by 1000

The mass, in this case, is 100 Kilograms. The gravitational acceleration on the moon is 1.6 m/s2

Therefore, the body's final weight after calculation is; 160 Newton after finding the product of the mass and the gravitational acceleration.

It should be noted that the gravity force in the moon and the earth are different.

Let's take into consideration the weight of a body in a lift. At rest, the weight of the body is W=mg. During upward movement of the lift with an acceleration 'a,' then the weight becomes; W=m (a + g). and during downward movement of the lift with a deceleration '-a' the weight becomes W= m ( g – a). This explains the elevator phenomenon where one feels lighter than usual when accelerating down and heavier than usual when accelerating upwards. In the upward movement, one feels heavy because of the force of moving upwards and the gravity acting on it. For the case of a freely falling lift, the weight is zero since there is no pressure on the feet or floor .i.e there is no support force.

In weight calculations, sometimes, one may be required to determine the gravity load of different structures. This load mainly comprises the weight of the structure and its occupants. The gravity load is majorly done for objects at rest. It is like determining the potential energy of a body, and it is measured in joules. Gravity load is determined by finding the products of mass, acceleration, and height. For a particular body, let's say a cuboid-shaped structure, made out of wood, you can determine the mass of this structure from the volume and density of the wood used. With the value of the cuboid's gravitational force and height, you can determine the gravitational load and weight.

We can conclude that the calculations involving weight are not hard or tiresome; only two factors that is mass, and gravitational acceleration, are put into consideration. It is necessary, though, that one stays keen with the scientific units.

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Acceleration has both magnitude and direction, and thus it is a vector quantity. Acceleration is referred to as the change in velocity per unit time and is denoted by 'a'. The SI unit for acceleration is (m/s2) or newton per kilogram (N/Kg).

This acceleration is dependent on different factors; body mass of an object (m), the distance of the object from the center of mass(r), and the universal gravitational constant (G), which is 6.673*10-11Nm2 Kg2. These factors are related as denoted in the equation; g=GM/r2…….. (i)

, here g represents the acceleration due to gravity.

Mass is directly proportional to gravity; thus, the heavier the object, the more the gravitational pull it will experience. Gravitational force is inversely proportional to the square of the distance of separation between the bodies. Thus, the larger the distance of separation, the weaker the gravitational pull effect. As two objects are moved away from each other, the gravitational force between them decreases.

Consider to bodies which are 2m away from the center mass and weigh 5 kilograms and 3kilograms, respectively. On substitution using the above equation, it will be seen that the acceleration due to gravity on the 5 kg mass is 1.25G m/s2 while that on the 3 kg mass is 0.75G m/s2. This proves that mass directly affects the gravitational acceleration experienced on a body, i.e, the heavier the mass, the higher the gravitational acceleration felt.

Height is also a factor that affects acceleration due to gravity. Height is inversely proportional to acceleration, which decreases with an increase in height. At an infinite distance from the earth, the value becomes zero. This can be computed in the following equation;

gh=g(1+h/R)-2……. (ii)

gh is the acceleration due to gravity at a particular height, 'h' is the height from the earth's surface and 'R' is the earth's radius.

Assume the moon has a radius of 1.64 * 106m, and has a mass of 7.24 * 1022 Kg. Substituting this, values to the equation (i) above,

I) identify the 'r' and'M' since the universal gravitational constant (G) is constant. In this case,

r = 1.64 * 106 m and M = 7.24 * 1022 Kg

G = 6.673 * 10 -11 Nm2 Kg2

II) Substituting this, values to equation (i) above

Multiply the 'G' and'M' to obtain

= 4.831 * 10 12

III) Divide the above calculated value by r2

r2 =2.689 * 10 12 m2

g = 1.796 ms-2

* Therefore, the value of acceleration due to gravity in this case is 1.796 ms-2.*

In calculating the acceleration for a body in motion on a plane, several quantities are put into consideration. These quantities are displacement, speed, and velocity. Displacement is the distance moved towards a particular direction, is denoted by 'S' and the SI unit is meter (m). Speed is the distance covered per unit time, and the SI unit is meters per second (m/s or ms-1). It is a scalar quantity.

Velocity is the change of displacement per unit time and is a vector quantity. The SI unit for velocity is (m/s).

The initial velocity is normally denoted by (u) and the final velocity denoted by (v). In calculating the acceleration of a body, the two velocities must be noted accurately, including the total time taken.

For a uniformly accelerated motion on a plane. The following thee kinematic equations are applied.

v=u+ at…… (iii)

s=ut +1/2 at2… (iv)

v2 =u2 +2as…... (v)

It is always observed in our day to day life that leaves on a tree always fall to the ground on their own. Why don't they remain floating or better still, move upwards towards the sky? Or rather, why is it that when a stone is thrown upwards, it falls back to the ground? This phenomenon is due to gravitational force, which causes a free-fall of bodies and objects.

The movement of an object towards the ground involuntarily at an accelerated speed under the sole influence of gravitational force is known as free fall. This acceleration is theoretically estimated to be 9.8 m/s2, but for calculation purposes and for the sake of accuracy, the gravitational force used is rounded off to 10m/s2.

When trying to find the acceleration due to gravity for a free fall, the above kinematic equations are applied, but in this case, acceleration 'a' is replaced with gravitational force 'since it's the force behind the increase in velocity. i.e.

V= u+gt …. (vi)

S= ut +1/2 gt 2…. (vii)

V2 =u2 +2gs…… (viii)

For vertical projections, a body goes against the force of gravity, and thus slowdown is experienced. In this case, the gravitational force is negative.

During a vertical projection, there is a maximum height that can be attained at a particular time. At this height, the final velocity is always zero as the body in motion has to come to a stop before it starts falling back to the ground. Applying the relevant equations, from equation (vi)

V=u+gt…will be V= u-gt…..since v=0 then u=gt and thus t=u/g…this shows the time taken for an object to attain the maximum height.

From equation (viii)

V2= u2 +2gs…will be v2 =u2 -2gs in the case of a vertical projection.

Since v=0...then u2=2gHmax….. i.e.'s' is replaced with the maximum height (Hmax). It's still the same thing as both of them represent displacement. Therefore, the maximum height reached can be calculated using the equation… Hmax =u2/2g

The time of flight taken by the projectile is twice the time taken to reach the maximum height, thus t=2u/g. In cases here and object returns to its original position, the total displacement equals zero. Thus equation (viii) becomes v2=u2…. therefore v=u.

Consider a stone projected vertically upwards with a velocity of 20 m/s.

I) the time taken to reach the maximum height will be;

U=20 g=10 …. thus T=20/10 the time taken to reach the maximum height is 2seconds

II) The time of flight is twice the time taken to reach the maximum height, therefore,

4seconds is the time of flight

III) The maximum height reached

Hmax =u2/2g (400/20) therefore the maximum height reached, in this case, is 20m.

IV)The velocity with which the stone lands on the ground, is equal to its initial velocity assuming that there is no air resistance. Thus, the velocity will be 20 m/s.

The projections made could also be horizontal. The path taken during a horizontal projection is known as the trajectory. The maximum horizontal distance covered at a given time I known as range (R). The acceleration is zero at the maximum horizontal distance; thus, from the equation s=ut -1/2 gt2, the range R= ut, and the vertical displacement will be h=1/2gt2.

Gravitational force is responsible for the acceleration observed in an object's movement as it moves towards the ground. A body that undergoes a free fall always accelerates at the rate of 9.8 m/s2.

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We've all heard the term “mass” in school before. But what actually is mass? And how can we calculate it if we know the force and weight of an object? Well, I’m glad you asked.

To calculate mass, you need to know the force of gravity that's acting on the object, and its weight. And then, you can calculate the mass by following this equation:

W = m x g

We will go into more detail later. First, let me explain the basics, so you have a clear understanding of how mass works.

Weight is a measure of an object’s force pressing on a surface because of gravity. The SI (Système International) of weight is Newton. SI is the international standard of measurement used by every country in the world except in the USA.

Since weight is a force, it's measured in Newtons. The weight is represented by the ‘W’ letter in the equation. It’s a common misconception that weight is often mistaken with mass, we’ll get into that later.

Mass is a measure of how much stuff or matter exists in an object. Matter, or stuff, is measured in atoms or molecules. Counting all the atoms that make up your entire body one at a time would be practically impossible, but sometimes you need that data quickly.

Instead of using ‘millions and billions of atoms’ for scale, we simply use kilograms or grams as a measure. The mass is represented by the 'm' letter in the equation.

The force that is mentioned in the equation above refers to gravity. The earth is massive, and it has a lot of mass. Mass creates gravity. Gravity pulls on every object around it, including us! Gravity is constantly pulling us down, towards the core of the earth, so that we don’t fly into outer space.

The force of gravity is measured in meters per second squared (m/s2). The gravity of the earth is represented by the ‘g’ letter in the equation. The earth’s gravity is 9.8 m/s2. It’s a constant.

So, let me repeat the equation clearly:

W = m x g

- W = Weight (Newton or kg)
- m = Mass (kg)
- g = Gravity (9.8 m/s2)

From the equation, we can conclude that weight is a direct result of mass times gravity.

Isaac Newton is peacefully enjoying himself under the apple tree. He was about to drift asleep when suddenly *bonk* an apple fell into his head and woke him up from his half-sleep. Upon waking up, Mr. Newton immediately weighs the apple, and it weighs 250 grams.

Help Mr. Newton find his apple’s mass.

From that problem, we can conclude that:

- W = 250 grams = 0.25 kilograms = 2.4517 Newton
- g = 9.8 m/s2

So, we substitute that data into the equation, and we’ll get:

- W = m x g
- 2.4517 = m x 9.8
- m = 2.4517 / 9.8
- m = 0.25017 kg = 250.17 grams

*Voilà!* The mass of the apple is 250.17 grams.

Neil Armstrong just landed on the moon with his spacecraft. To celebrate it, Neil would like to eat an apple. Before eating it, Neil has to weigh the apple to measure his daily intake. The weight scale says that the apple weighs 41.3 grams. The gravity on the moon is 1.62 m/s2.

From that problem, we can conclude that:

- W = 0.0413 kilograms = 41.3 grams = 0.4052 Newton
- g = 1.62 m/s2

So, we substitute that data into the equation, and we’ll get:

- W = m x g
- 0.4052 = m x 1.62
- m = 0.4052 / 1.62
- m = 0.25017 kilogram = 250.17 grams

From that equation, the mass of the apple is 250.17 grams. That’s the same mass as Mr. Newton’s apples back on earth! What a coincidence.

Mass is the amount of stuff in an object. Weight is the force exerted from an object to the surface of the earth. Essentially, the mass is the volume while weight is the force. That’s why the SI of mass in kilograms and the SI of weight is Newton.

From the example above, about the apple belong to Isaac Newton and Neil Armstrong, you can deduct that mass is unchanged regardless of the location. Unless something blew up the apple and shattered it to pieces.

Meanwhile, weight is subject to gravity. Weight will change when the gravity changes. The gravity can change when the object is placed on other space bodies, like the moon or mars. You can also create an artificial gravity or increase the amount of gravity. One way to do it is to ascend tremendously quickly to the sky, like the astronauts inside a launching rocket or executing certain maneuvers on a fighter jet.

Fighter jet pilots and astronauts can experience about 6 to 8 g of force. That's 6 to 8 times of the earth's gravity. More than that, the pilot will most probably pass out. That’s because the bodyweight is also multiplied by that number. The body is buckling under pressure before eventually pass out under its own weight.

- Mass is the amount of matter that makes up an object.
- An object’s mass will always stay the same regardless of the gravity or location unless the object is undergoing an internal chemical reaction.
- Weight is a measure of an object’s force pressing on a surface because of gravity.
- Force of gravity is a measure of pull exerted by a massive object like planets towards our body.
- Gravity is measured in meter per second squared (m/s2).

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Separation of Variables, widely known as the Fourier Method, refers to any method used to solve ordinary and partial differential equations. To apply the separation of variables in solving differential equations, you must move each variable to the equation's other side.

As indicated in the introduction, Separation of Variables in Differential Equations can only be applicable when all the y terms, including dy, can be moved to one side of the equation. All the other x terms, including dx, are taken to the other side of the equation. Kindly follow these steps in Separating Variables involving Differential Equations.

Step One: Move all the y terms, including dy, to one side of the equation

Step Two: Move all the x terms, including dx, to the other side of the equation

Step Three: Integrate the y side.

Step Four: Integrate the x side. Kindly note that "+C" is the constant of the integration.

Step Four: Simply the equation.

Task: Assuming that k is constant, solve the following differential equation.

dy/dx = ky

Solution

Move all the y terms to one side of the equation and the x terms to the other side of the equation.

Thus

dy/dx = ky

dy = ky × dx ( You are simply multiplying both sides by dx)

You should then divide both sides of the equation by y.

y:dy/y = k dx

Now integrate both sides of the equation. Kindly note that the integration should be done separately.

Thus;

∫ dy/y = ∫ k dx ( note the placement of the integration sign)

Now integrate the left side of the equation:

(y) + C = ∫ k dx

Then integrate the right side of the equation:

(y) + C = kx + D

Remember that "C" is the constant of integration and that D is used for the other. Therefore it is a different constant.

You can now simplify the equation. To simplify, roll the two constants into one.

Thus;

(a=D−C): ln(y) = kx + a

e(ln(y)) = y

It would help if you now took the exponents on both sides

Thus;

y = ekx + a

And ekx + a = ekx ea

Therefore:

y = ekx ea

Assuming that you have keenly followed these steps and the example given, you can now use the formula and the procedure to solve real-life examples. It is prudent to state here that though we used y and x in our example, you are at liberty to use other variable names.

Take rhinos, for example. The more rhinos you have, the more baby rhinos you will have. When these baby rhinos grow, they automatically sire babies. Therefore there is an increase in population.

This information can translate into a differential equation.

Please let:

(N) population any time be t;

the growth rate be r and;

the rate of change in population is dNdt.

Please note that the rate of change at any given time is equal to the growth rate times the population. Simply put:dN dt = rN.

If you note keenly, the resulting equation is the same as the previous one we solved, in example (1) above. The only difference is that this time, different letters have been used. That is:

N instead of y

t instead of x

r instead of k

Therefore N = cert

Further Illustration

Solve this differential equation by separating the variables:

dy/dx = 1y

First, separate the y and x variables. Then you will have to move all the y variables to one side of the equation. Then move all the x terms to the other side of the equation.

Thus;

Multiply both sides by dx:

dy = (1/y) dx

Multiply both sides by y:

y dy = dx

It would be best if you then integrated both sides of the equation separately. Please be keen to put the integral sign (∫ ) in front of the equation.

Thus;

∫ y dy = ∫ dx

Now integrate each side:

Thus;

(y2)/2 = x + C

You have now integrated both sides of the equation in one line.

Next, simplify the equation:

Multiply both sides by 2

y2 = 2(x + C)

Find the square root of both sides:

y = ±√(2(x + C))

Note that y = ±√(2(x + C)) is not the same as y = √(2x) + C. The difference is as a result of the addition of C before finding the square root. You realize that this is common in many differential equations. C is not just added at the end of the process. You should add the C only when integrating.

Thus;

y = ±√{2(x + C)}

Task solve :dydx = 2xy1+x2

First, learn how to separate the Variables.

Multiply both sides by dx, then divide both sides by y:

Thus:

1y dy = 2x1+x2dx

Next, integrate the two sides of the equation separately.

Thus:

∫1y dy = ∫2x1+x2dx

Note that while the left side is a simple logarithm, the right side can be integrated by the substitution method.

Let u = 1 + x2, so du = 2x dx:∫1y dy = ∫1udu

You can then integrate(y) = ln(u) + C

Then make C = ln(k):ln(y) = ln(u) + ln(k)

Therefore you get:

y = uk

Lastly please put u = 1 + x2 back again.

Hence;

y = k(1 + x2)

Lastly, simplify the equation.

Therefore y = k(1 + x2)

The previous example on growth Differential Equation (dNdt = rN) is closely related overhaul Logistics Equation. The equation also referred to as the Verhult's model was introduced in 1838 and works with the assumption that growth may not last forever because the population will soon run of food.

Now solve the Verhaust's Equation using Differential Equation by Separating the Variables

dNdt = rN(1−N/k)

First, separate the two variables by multiplying both sides of the equation by dt, then divide again by N.

Thus;

dN = rN(1−N/k) dt

N(1-N/k):1N(1−N/k)dN = r dt

Next, please integrate the resulting expression.

Thus;

∫1N(1−N/k)dN = ∫ r dt

You realize that the left side of the equation seems difficult to integrate. You don't have to worry, though. With a little knowledge of Partial Fractions, you don't have to spend a whole day looking for a solution. Kindly rearrange the equation like this:

It would help if you began with:1N(1−N/k) then multiply both top and bottom by k:kN(k−N). You should then add N and -N to the top.

Thus;

N+k−NN(k−N)

You should then split the expression into two fractions.

Thus:

NN(k−N) + k−NN(k−N)

You can then simplify each fraction

1k−N + 1N.

You can now quickly solve the problem. Go ahead and separate each term separately.

Thus;

∫1k−NdN + ∫1NdN = ∫ r dt

Then, Integrate:−ln(k−N) + ln(N) = rt + C

Simplify the equation:

In(k−N) − ln(N) = −rt − C

Combine ln():ln((k−N)/N) = −rt − C

Then, take exponents on both sides:(k−N)/N = e−rt−C

Now, separate the powers of e:(k−N)/N = e−rt e−C

Remember that e−C is constant. You should therefore replace it with A:(k−N)/N = Ae−rt.

You are now close to the answer. With a little more algebra, you are assured of getting N on its own. To do this:

Separate the fraction terms:(k/N)−1 = Ae−rt

Then add 1 to both sides:k/N = 1 + Ae−rt

Next,divide both by k:1/N = (1 + Ae−rt)/k

Find the reciprocal of both sides: N = k/(1 + Ae−rt)

Therefore the solution is N = k1 + Ae−rt

Separating Variables in Differential Equations, though often understood by a few people and widely feared by many, can be a straightforward concept to grasp if you adhere to the procedures discussed in this article.

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Subtracting integers on a number line is easy as pie! All you need is a number line, some numbers, and a reason to subtract. It's really that easy. Let's start by reviewing some key terms.

Integers are countable, whole numbers. Positive and negative whole numbers are integers. Examples include 3, -3, 5, 0, etc. If you cannot write a number as a whole number, it is not an integer. That means that fractions and decimals are not integers.

You can write integers using set notation.

{ … -3, -2, -1, 0, 1, 2, 3, … }

The ellipsis means that the numbers continue in both directions and keep going and going...

A number line is a handy tool! It is a horizontal line that shows consecutive numbers in equal intervals. The center of the line is at the zero mark, and it is called the Origin of the number line. On the right of the zero mark is the positive side, while the left of the zero mark is the negative side. You write negative numbers with a minus sign in front of them, e.g., -5. All number lines have arrowheads at one or both ends of the line to show that the numbers continue.

Although a number line includes all real numbers, our number line for this article uses only integers. A number line helps you visualize and understand the concept of numbers less than zero, subtraction, and addition.

Now let's see how to find integers on a number line.

Today is your first day at your new school. Your mum gave you two biscuits for lunch. How can you locate the two biscuits on the number line?

- The first step is to draw your number line.
- Next, mark the Origin of your number line with zero. Then, write numbers 1-5 to the right of the Origin.
- You have a positive number of biscuits since they have not been eaten. (Even if they have been eaten, you can only go down to zero.) Count two places to the right on your number line, and you will end at 2.
- That is the position of your biscuits on the number line!

Now imagine again that your dad gave you and your brother some apples. But you like apples a lot, so you ate all your apples. The next day, while going to school, you had no apples while your brother still had 5 apples. So you took 2 apples from him and promised to give him 2 the next time you got apples. How do you show the debt you owe to your brother on the number line?

- Once again, draw your number line. This time, you don't have any apples, and the 2 apples you are going to eat in school are not yours. They are your brother's. This means that the 2 apples are not positive but negative. Negative numbers are less than 0.
- So, write out the integers -1 to -5 to the left of your Origin.
- Now, count 2 spaces to the left on your number line. You will land on -2!

If you want to subtract integers using a number line, you are finding the difference between two numbers. You may also think of it as finding the distance between the two numbers. Measuring this distance is subtraction. Three things to remember:

- When you subtract from a positive number, count to the left.
- When you subtract from a negative number with a positive, you count to the left. You create a bigger negative.
- When you subtract from a negative number with another negative, you count to the right. You increase the number.

Now that you can locate your numbers on the number line subtracting numbers should be a breeze!

Subtract 3 from 7 using the number line.

This means

*7 – 3*

The first step is to locate 7 on the number line. If you start at 7 and count 3 steps to the left, you will end at 4. This means that 7 – 3 = 4.

You can think of it this way. You had 7 apples, and you gave your friend 3. You will have 4 apples left.

Subtract 2 from -5.

This means

*-5 – 2*

The first step here is also to locate -5 on your number line. The 2 is positive. So since you are subtracting a positive number, you will count 2 steps to the left. This also means that you are decreasing the initial number by 2. Two steps to the left, and you will end at -7.

-5 – 2 = -7

Think of it this way. You already owe someone 5 mangoes. Then your sister is hungry, and she wants to eat two mangoes. But you don't have any mangoes! What should you do so your sister stops crying for mangoes? You borrow two more mangoes to give her. So you now owe a total of 7 mangoes.

Subtract -2 from -6.

This means

-6 – ( - 2 )

The first step is to locate -6 on the number line. The number you are subtracting now is -2, which is a negative number, so you count 2 steps to the right. You will end at -4. The double negative signs turn the -2 into a positive 2.

-6 – ( - 2 ) = ?

-6 + 2 = -4

The best way to think of this subtraction of negative numbers is to imagine that you owe your friend $6. But your friend is very kind and decides to forgive $2 of the debt. Now, you only have to pay back $4.

And that's all there is to it! With practice, you can confidently handle subtracting integers using a number line. It is an excellent tool to help you visualize the math.

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There are several systems of linear equations involving the same set of variables. A system here refers to when you have two or more equations working together.

2X + Y=6

3X - Y= 4

X-2Y +3Z=9

-X+3Y-Z=-6

X+2Y+3Z=-7

2X-3Y-5Z=9

-6X-8Y+Z=-22

A three-variable linear equation is a bit more difficult to solve compared to equations with two variables. This complexity is a result of the additional variable.

Although there are several methods for solving this type of equation, the elimination method remains the most straightforward.

Follow the procedure below to use the elimination method in solving three variable linear equations:

1. Put all the equations in standard form, avoiding decimals and fractions.

2. Select a suitable variable to eliminate.

3. Choose any two of the three equations and eliminate the selected variable.

4. Select a different set of the two equations and eliminate the same selected variable.

5. Solve the two equations in steps three and four for the two variables they contain.

6. Substitute the answers in step five into any equation that has the remaining variable.

7. Check to prove the solution with all the three original equations.

Task: Solve the following system of equations using the elimination method.

X-2Y +3Z=9 ______________(1)

-X+3Y-Z=-6_______________ (2)

2X-5Y+5Z=17_____________ (3)

In this case, you realize that the system is already in standard form; therefore, choose the variable to eliminate, for instance, X. You should then select two equations with which to eliminate X. For this example, pick (1) and (2).

Thus;

X-2Y +3Z=9

-X+3Y-Z=-6

__________________

Y +2Z=3__________(4)

You should then select a different set of two equations, in this case, equations (1) and (3) to eliminate the same variable.

Thus;

X-2Y+3Z=9___________(1)

2X-5Y +5Z=17_________(3)

To eliminate X, multiply equation (1) with (-2) or any other suitable constant.

Thus;

-2X +4Y -6Z = -18

2X-5Y+5Z=17

____________________

-Y-Z=-1__________(5)

You can now solve equation (4) and (5) .

Y+2Z=3

-Y-Z=-1

__________________

Z=2

Now substitute Z=2 in any of the created 2×2 system, that is (4) or (5). In this case pick equation (5).

-Y-Z=-1

-Y-2=-1

Thus Y=-1

Use the answers obtained to substitute into any equation involving the remaining variable. In this case pick X-2Y+3Z=9___________(1)

X-2Y+3Z=9___________(1)

X-2(-1)+3(2)=9

X+2+6=9

X+8=9

X=1

Therefore the solution is X=1, Y =-1 and Z = 2.

Task: Solve the following system of equations using the elimination method.

X=3Z-5

2X+2Z=Y +16

7X-5Z= 3Y +19

Solution

First, write all the equations in standard form.

Thus;

X -3Z = -5____________(1)

2X-Y+2Z=16_________(2)

7X-3Y-5Z=19_________(3)

Next, choose a suitable variable to eliminate. In this case, select Y since it is missing in equation (1). Then, select equation (2) and (3) to eliminate Y.

2X-Y +2Z=16__________(2)

7X-3Y -5Z=19__________(3)

Please, multiply equation (2) by a suitable constant,say ( -3) to get:

-6X+3Y-6Z= 48

Then, eliminate Y.

-6X+3Y-6Z= 48

7X-3Y -5Z=19

_____________________

X-11Z=-29_________________(4)

It would be best if you then solved equation (1) and (4) since they are 2×2 system linear equation by subtracting.

Thus;

X -3Z = -5_________________(1)

X-11Z=-29_________________(4)

______________________

8Z=24

Therefore Z=3

You can then substitute Z=3 into equation (4)

X-11Z=-29_________________(4)

X-11(3)=-29

X-33=-29

X=4

Then, submit X=4 and Z=3 into equation 3 to get the value of Y.

7X-3Y -5Z=19__________(3)

7(4)-3Y-5(3)=19

28-3Y-15=19

-3Y=6

Therefore Y=-2

Here's the solution: X=4,Y=-2 and Z=3

The Gaussian Elimination Method is the best method for solving three (or more) variable equations. However, the Gaussian Elimination Method is generally for experts, as it involves a bit of set up work. You don't have to worry though, because once you have the system set up, the equation is straightforward to solve.

To use the Gaussian Elimination Method, you have to translate your equations into a matrix using placeholder zeros and ones, where necessary. You should then set up the matrix by manipulating it through basic application of the properties of matrixes. You should then add, subtract, multiply and divide the rows in the matrix to make ones and zeros in specific locations.

Solve the following equation using the Gaussian Elimination Method

system of equations.

–3x + 2y – 6z = 6

5x + 7y – 5z = 6

x + 4y – 2z =8

You realize that in this problem, no expression for a variable is solved. It means you'll have to do the multiplication and addition tasks to simplify the system.

First, do away with the leading x terms in any of the two rows. To begin, simply pick any two rows you find easy to clear out before switching the rows later to have the system in the upper triangular form.

Since there is no existing rule limiting you to pick the x term starting from the first row, pick the simpler x term in the third row which gives a simple coefficient of multiply the third row by 3 then add it to the first row. If you find this difficult, you can do the calculations on paper.

Thus;

-3X + 2Y-6Z=6

5X + 12Y-6Z=24

14Y - 12Z=30

You should now write the results in this format:

7Y- 6Z= 15

5X +7Y- 5Z= 6

X + 4Y -2Z= 8

Whereas in solving two variable expressions, you are allowed to multiply a row, there is no provision for this in three variable linear equations. It is therefore suggested that you work it out on a scratch paper. You should get this result:

14Y -12Z= 30

5X +7Y - 5Z= 6

X + 4Y - 2Z= 8

Next, multiply the first row by one- half to get smaller values for coefficients.

Thus;

7Y - 6Z = 15

5X + 7Y -5Z= 6

X +4Y - 2Z= 8

Now that you have the value of Z, divide the first row by 43. Next, arrange the rows in the upper triangular format:

Thus;

43Z = 43

Y- 7Z= -4

X +4Y- 2Z= 8

Multiply the first row by 1/43 to get:

Z= 1

Y-7Z=- 4

X+ 4Y-2Z= 8

X+ 4Y-2Z=8

Y-7Z=-4

Z=1

Please use the back solving process to get the values of Y and Z:

Y – 7(1) = –4

Y – 7 = –4

Y = 3

X + 4(3) – 2(1) = 8

X + 12 – 2 = 8

X + 10 = 8

X = –2

Thus the solution is:

X = - 2

Y= 3

Z = 1

The remaining methods of solving three variable linear equations are the substitution method and the graphical method. In substitution, the equation can quickly be written with a single variable, as the subject is identified. The identification is made by solving the equation of the variable. Next, substitution is done to the expression in which the variable appears as well as the two other equations. You will, therefore, come up with a system with fewer variables. After you have solved the smaller system, either through substitution or any other method, substitute the solutions you have found for the variables back into the first right-hand side of the expression.

Consider this example:

Solve the following three variable linear equations by the substitution method.

3X +2Y-Z= 6

-2X+2Y +Z=3

X+Y+2=4

The coefficient of Z is already Z in equation (1). Therefore, solve for Z to get:

Z= 3X+2Y-6

You can now substitute for Z in equation (2) and (3).

Thus;

-2X + 2Y +(3X+2Y-6)=3

X+Y+(3X+2Y-6)=4

You can then simplify the new system to:

X+4Y=9

4X+3Y=10

Now solve for Z in equation ( 1)

X=9-4Y

Substitute the expression X=9-4Y in equation (3).

Thus;

4(9-4Y)+3Y=10

36-16Y+3Y=10

13Y=26

Y=3

With Y=2, work back the equation X=9-4Y.

Thus,

X=9-4Y

X=1

You can now find Z's value by substituting X=1 and Y= 2 in equation (2).

Z=3X +2Y-6

Z=3(1)+2(1)-6

Z=1.

Therefore X= 1,Y=2 and Z=1.

Knowledge of three variable linear equations has actual real-life applications. In tackling life's challenges, many situations can be converted into mathematical expressions to illustrate the relationship between variables, otherwise known as linear equations.

Finally, knowledge of three variable equations is useful in supply and demand economics.

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