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# 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

x2 + y2 = 25

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:

x2 + (x + 1)2 = 25

This gives:

x2 + x2 + 2x + 1 = 25

2x2 + 2x − 24 = 0

x2 + 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

x2 + y = 5

x2 + y2 = 25

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:

y2y = 25 − 5

y2y − 20 = 0

(y + 4)(y − 5) = 0

So 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.

### Exercises

1. Solve algebraically:

6y x = 6

x2 + 3y2 = 36

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

2. Solve algebraically:

3x2y2 = 4

x2 + 4y2 = 10

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.

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 XL XC = X.

So

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

Now, on squaring both sides, we have

R2 + X2 = 4

But R = X2 (since R is equal to the square of X) so

R2 + R = 4

Then

R2 + R − 4 = 0

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 therefore X = √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

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

x2 + 4x + 4 + y2 − 6y + 9 − 25 = 0

x2 + 4x + y2 − 6y − 12 = 0 [3]

Equation [2]:

(x − 1)2 + (y + 4)2 = 16

(x − 1)2 + (y + 4)2 − 16 = 0

x2 − 2x + 1 + y2 + 8y + 16 − 16 = 0

x2 − 2x + y2 + 8y + 1 = 0 [4]

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

6x − 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.