# 6. Algebraic Solution of Systems of Equations

## Solution by Substitution

Similar to the linear case in the previous section, we can solve a system of equations by substitutiong one of the expressions given into the other expression. Our solution will be a set of `x`-`y` coordinates.

### Example 1

Solve the system of equations algebraically:

y=x+ 1

x^{2}+y^{2}= 25

Answer

We recognize that this is a straight line intersecting a circle. (See more on the circle.)

We may have:

- no intersection point
- 1 intersection point
- 2 intersection points

We can simply substitute the right hand side of the first equation into the second equation:

x^{2}+ (x+ 1)^{2}= 25

This gives:

x^{2}+x^{2}+ 2x+ 1 = 252

x^{2}+ 2x− 24 = 0

x^{2}+x− 12 = 0(

x+ 4)(x− 3) = 0

So `x = −4` or `x = 3`.

This gives our intersecting points to be: `(−4, −3)` and `(3, 4)`.

Is it correct? The graph showing the line intersecting the circle is as follows:

Graphs of `y = x+1` and `x^2+y^2=25`: Intersection line and circle

We can see from the graph that our solutions `(−4, −3)` and `(3, 4)` are correct.

## Solution by Addition or Subtraction

This method works by **eliminating** one of the variables
from the equations. We then find the value(s) of the remaining
variable.

### Example 2

Solve the system of equations by adding or subtracting

x^{2}+y= 5

x^{2}+y^{2}= 25

Answer

NOTE: This system represents a **parabola** intersecting a
**circle***.* We expect:

- no intersection point or possibly
- 1, 2, 3 or 4 intersection points

If we subtract the first line from the second, we have:

y^{2}−y= 25 − 5

y^{2}−y− 20 = 0(

y+ 4)(y− 5) = 0So

y= −4 or 5

The corresponding *x* values are going to be:

`x = +3` or ` −3`, and `x = 0`

So the solution set will be:

`(−3,−4)`, `(3,−4)` and `(0,5)`.

The sketch shows:

Graphs of `y = -x^2+5` and `x^2+y^2=25`: Intersection parabola and circle

We can see from the graph that our 3 solutions `(−3,−4)`, `(3,−4)` and `(0, 5)` are correct.

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### Exercises

1. Solve algebraically:

6

y−x= 6

x^{2}+ 3y^{2}= 36

Answer

We first solve the first line for *y* (so we can
substitute):

`y=(x+6)/6`

Substituting in the second row gives:

`x^2+3((x+6)/6)^2=36`

Expand the brackets:

`x^2+3((x^2+12x+36)/36)=36`

Cancel the 3 and 36:

`x^2+((x^2+12x+36)/12)=36`

Multiply throughout by 12:

`12x^2+x^2+12x+36=432`

`13x^2+12x-396=0`

Solving using the quadratic formula gives:

`x = 5.077`, or `x = −6`

This gives us solutions of: `(5.077, 1.85)` and `(−6,0)`.

Graphically, we have:

Graphs of `y = (x+6)/6` and `x^2+3y^2=36`: Intersection of line and ellipse

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2. Solve algebraically:

3

x^{2}−y^{2}= 4

x^{2}+ 4y^{2}= 10

Answer

The first equation is a hyperbola, while the second is an ellipse.

We multiply the first row by 4 so we can eliminate the `y^2` term:

`12x^2-4y^2=16`

`x^2+4y^2=10`

Now adding the two rows, we obtain:

`13x^2=26`

`x^2=2`

`x=+-sqrt(2)`

Substituting `+sqrt(2)` into the question's first equation gives us

`y=+-sqrt(2)`

Likewise, substituting `-sqrt(2)` into the first equation also gives us

`y=+-sqrt(2)`

This gives us the solutions

`(sqrt(2),sqrt(2)),` `(sqrt(2),-sqrt(2)),` `(-sqrt(2),sqrt(2)),` `(-sqrt(2),-sqrt(2)) `

≈ (±1.414, ±1.414), (±1.414, ∓1.414)

The graph shows the intersection of the ellipse and the hyperbola. We see 4 intersection points, with the same values that we found algebraically.

3. An alternating current has impedance given by *Z* = 2.00 Ω. If the
resistance *R* in the circuit is numerically equal to the
square of the reactance *X*, find
*R* and *X.*

Answer

The statement "*R* is numerically equal to the square of
the reactance *X*" simply means `R = X^2`.

Recall (from Application of Complex Numbers) that

`|Z|=sqrt(R^2+(X_L-X_C)^2`

In this case, from the definition, and to make life easier, we assume that *X*_{L} −* X*_{C} =* X*.

So

`|Z|=sqrt(R^2+X^2)=2Omega`

Now, on squaring both sides, we have

R^{2}+X^{2}= 4

But *R = X*^{2} (since *R* is equal to the
square of *X*) so

R^{2}+R^{}= 4

Then

R^{2}+R^{}− 4 = 0

Using quadratic formula gives

`R=(-1+-sqrt(1+16))/2` `=(-1+-sqrt17)/2`

Only the positive root has meaning (since we cannot have negative resistance), so

R= 1.56 Ω. and thereforeX= √1.56 = 1.25 Ω.

In this question, the 2 interssecting functions are a parabola (`X = R^2 + R - 4`) and a straight line (`X=4`). In the graph, we can see the 2 solutions we obtained, one is negative (`R=-2.56`) and the other one is positive, at (`R=1.56`).

4. Find the intersection points for the circles

(

x+ 2)^{2}+ (y− 3)^{2}= 25

and

(

x− 1)^{2}+ (y+ 4)^{2}= 16

Answer

a. Setting the right side to 0 and expanding each equation gives:

Equation [1]:

(

x+ 2)^{2}+ (y− 3)^{2}= 25(

x+ 2)^{2}+ (y− 3)^{2}− 25 = 0

x^{2}+ 4x+ 4 +y^{2}− 6y+ 9 − 25 = 0

x^{2}+ 4x+y^{2}− 6y− 12 = 0 [3]

Equation [2]:

(

x− 1)^{2}+ (y+ 4)^{2}= 16(

x− 1)^{2}+ (y+ 4)^{2}− 16 = 0

x^{2}− 2x+ 1 +y^{2}+ 8y+ 16 − 16 = 0

x^{2}− 2x+y^{2}+ 8y+ 1 = 0 [4]

Solving the above 2 results simultaneously, we subtract Row [4] from Row [3] and this gives:

6

x− 14y− 13 = 0

Solving for *y* gives:

`y=3/7x-13/14`

This means the intersection points are on the line

`y=3/7x-13/14`

b. Solve one of the circle equations for *y* using:

`y=(-b+-sqrt(b^2-4ac))/(2a)`

`y^2-6y+(x^2+4x-12)=0`

This gives:

`y=3+-sqrt(21-x^2-4x`

c. Substitute the positive case into the LHS of

`y=3/7x-13/14`, gives us:

`3+sqrt(21-x^2-4x)=3/7x-13/14`

d. Solve for *x*:

`sqrt(21-x^2-4x)` `=3/7x-13/14-3` `=3/7x-55/14`

Square both sides:

`21-x^2-4x` ` =(3/7x-55/14)^2` `=3025/196-165/49x+9/49x^2`

Re-writing for convenience:

`21-x^2-4x` `=3025/196-165/49x+9/49x^2`

Moving everything to the right hand side:

`0=9/49x^2+x^2-165/49x` `+4x` `+3025/196` `-21`

Simplifying:

`0=58/49x^2+31/49x-1091/196`

This gives us a quadratic in *x*

`58/49x^2+31/49x-1091/196=0`

or more simply

`58x^2+31x-272 3/4=0`

`x=-31/116+7/116sqrt1311` `=1.9177`

`x=-31/116-7/116sqrt1311` `=-2.4522`

e. Substitute these two *x*-values into

`y=3/7x-13/14`:

`[3/7x-13/14]_(x=1.9177)=-0.1067`

`[3/7x-13/14]_(x=-2.4522)=-1.9795`

So the points of intersection are (1.9177, −0.1067) and (−2.4522, −1.9795).

[We could have substituted the *x*-values into either circle equation and solved for *y*, but what I have done is easier.]

We can see that the circles, the line `y=3/7x-13/14` and the intersection points are all correct when we draw the graph:

As a comment, this question could be solved very quickly, and easily, using a computer graphics program. We would just need to zoom in on the intersection points until we obtained the required precision.

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