4. Solving Equations
Remember this kind of problem from primary school?
? + 5 = 7
We just needed to figure out which number should go into the box to make it a true statement. Clearly, we need to replace the question mark with "2":
2 + 5 = 7
Solving equations using algebra is really no different. Instead of using a box, we use a letter to represent a number. Our task is to find the correct number (or sometimes there may be more than one number) that makes the equation true.
Sometimes we can "see" the right answer if it is simple (maybe we can just count up with our fingers, or whatever.) But when our equations become more complicated, we need a process to follow that will eventually give us the answer.
- We are aiming to get x (or whatever letter the question uses) on the left hand side of the equals sign, by itself.
- We solve equations by balancing: whatever we do to one
side of an equation, we must do the same to the other
side. So if we add 4 to the left hand side, we must add 4 to the right hand side as well. If we multiply on the left side by 2, we multiply on the right side by 2 as well.
Solve the equation
x − 6 = 10
We need to "get rid of" the -6 on the left hand side so we are left with x only on the left hand side.
The opposite of subtracting 6 is adding 6.
If we add 6 to both sides, we will remove the -6 on the left.
x − 6 = 10
x − 6 + 6 = 10 + 6
x = 16
So the value of x needs to be 16 to make the equation true.
CHECK into the original question:
16 − 6 = 10. It checks out okay.
Solve 5x = 35
This time we are answering
5 × ? = 35
We could do this in our heads easily (right?), but if the problem is more complicated, we need to know what to do.
On the left, we are multiplying our unknown quantity by 5. We'll use "x" for this quantity.
`5x = 35`
The opposite of multiplying by 5 is dividing by 5. So we divide both sides by 5:
We obtain :
`x = 7`
CHECK: 5 × 7 = 35. It checks out okay.
[These checks seem silly with easy examples, but it is a really good idea to check your solutions for all the equation problems that you do. It means you can leave the problem feeling good that you have the right answer and also, you learn more about how the solution works.]
This time we need to do 2 steps to solve the equation. We notice there is a 4 on the bottom of the fraction.
This is equivalent to dividing by 4. The opposite to dividing by 4 is multiplying by 4. So we do that first:
`(3x)/4 xx 4=7xx4`
Cancelling the 4's on the left gives:
In the middle step we cancelled out the 4's, so we are left with no fraction.
Now we need to divide both sides by 3, since we have a "3×" on the left hand side of the equation.
Some countries (like the USA) will leave the answer as a single fraction (28/3), while the practice in other countries (like the UK and Australia) is to express the answer as a mixed numeral.
Is our answer correct?
Substituting our answer in the left hand side gives:
`(3x)/4=3/4 x=3/4 xx 28/3`
Canceling the 3's (which gives us 1) and the 28 with the 4 gives us 7:
`3/4 xx 28/3 = 7`
The right hand side in the question was 7, so we are confident our answer is correct.
Easy to understand math videos:
Solve 5 − (x + 2) = 5x
First, we expand out the bracket.
`5 - (x + 2) = 5x`
`5 - x - 2 = 5x`
`3 - x = 5x`
Now we recognise that it is easier to get all the x's on the right side, by adding x to both sides :
`3 = 6x`
Now I divide both sides by 6 and swap the sides:
`x = 0.5.`
We check our answer in both sides of the equation. If it works, it must be the right answer.
LHS = `5 - (0.5 + 2) = 2.5`
RHS = `5 xx 0.5 = 2.5` = LHS.
Solve 5x − 2(x − 5) = 4x
Expanding the bracket:
`5x - 2(x - 5) = 4x`
`5x - 2x + 10 = 4x`
`3x + 10 = 4x`
Subtracting `3x` from both sides and swapping sides gives:
`x = 10`
LHS = `5 xx 10 - 2(10 - 5) = 50 - 10 = 40`
RHS = `4 xx 10 = 40` = LHS.
If you can, solve the equation
− (7 − x) + 5 = x + 7
What do you conclude?
− (7 − x) + 5 = x + 7
Expand out the brackets:
−7 + x + 5 = x + 7
Subtract x from both sides:
Simplify the left hand side:
`-2 = 7` ????
This is not possible, so we conclude that there are no possible values for x.
[There was a hint in the question that something funny may be going on. Always be aware that an equation may not have solutions. Also, there are times when you get solutions that cannot possibly work, so you have to discount them. We find such examples later in Equations with Radicals.]