SOLVING SYSTEMS OF LINEAR AND QUADRATIC EQUATIONS ALGEBRAICALLY

There are two methods to solve a system of linear and quadratic equations.

1. Substitution

2. Elimination

Substitution

Step 1 :

Isolate one of the two variables in one of the equations.

Step 2 :

Substitute the expression that is equal to the isolated variable from Step 1 into the other equation.

Step 3 :

Solve the resulting quadratic equation to find the x-values.

Step 4 :

Substitute the x-values into the equation from step 1.

Elimination

Step 1 :

Add or subtract to eleiminate y.

(To eleiminate y, some times, you may have to multiply one of the equations or both equations by some constant depending on the coefficient of y in both the equations.)

Step 2 :

Solve the resulting equation for x.

Step 3 :

Substitute the x-values from step 2 into one of the two equations and solve for y.

Example 1 :

Solve the system of equations by substitution and verify your solution.

4x - y + 3 = 0

2x2 + 8x - y + 3 = 0

Solution :

4x - y + 3 = 0 ----(1)

2x2 + 8x - y + 3 = 0 ----(2)

Isolate y in (1).

4x - y + 3 = 0

4x + 3 = y

or

y = 4x + 3 ----(3)

Substitute '4x + 3' for y in (2).

2x2 + 8x - (4x + 3) + 3 = 0

2x2 + 8x - 4x - 3 + 3 = 0

2x2 + 4x = 0

2x(x + 2) = 0

2x = 0  or  x + 2 = 0

x = 0  or  x = -2

 Substitute x = 0 in (3).y = 4(0) + 3y = 0 + 3y = 3 Substitute x = -2 in (3).y = 4(-2) + 3y = -8 + 3y = -5

The two solutions are (0, 3) and (-2, -5).

Verification :

Verify the solution (0, 3).

Substitu x = 0 and y = 3 into the original equations.

4x - y + 3 = 0 :

4(0) - 3 + 3 = 0

0 = 0

2x2 + 8x - y + 3 = 0 :

2(0)2 + 8(0) - 3 + 3 = 0

2(0) + 0 - 3 + 3 = 0

0 = 0

Verify the solution (-2, -5).

Substitu x = -2 and y = -5 into the original equations.

4x - y + 3 = 0 :

4(-2) - (-5) + 3 = 0

-8 + 5 + 3 = 0

-8 + 8 = 0

0 = 0

2x2 + 8x - y + 3 = 0 :

2(-2)2 + 8(-2) - (-5) + 3 = 0

2(4) - 16 + 5 + 3 = 0

8 - 16 + 5 + 3 = 0

0 = 0

Both solutions are correct.

Example 2 :

Solve the system of equations by elimination and verify your solution.

x - y + 1 = 0

x2 - 6x + y + 3 = 0

Solution :

x - y + 1 = 0 ----(1)

x2 - 6x + y + 3 = 0 ----(2)

In the above system of two equations, y has different signs.

By simply adding the two equations, y can be eliminated.

(1) + (2) :

(x - y + 1) + (x2 - 6x + y + 3) = 0

x - y + 1 + x2 - 6x + y + 3 = 0

Combine the like terms.

x2 - 5x + 4 = 0

Factor and solve.

x2 - x - 4x + 4 = 0

x(x - 1) - 4(x - 1) = 0

(x - 1)(x - 4) = 0

x - 1 = 0  or  x - 4 = 0

x = 1  or  x = 4

 Substitute x = 1 in (1).1 - y + 1 = 0-y + 2 = 0-y = -2y = 2 Substitute x = 4 in (1).4 - y + 1 = 0-y + 5 = 0-y = -5y = 5

The two solutions are (1, 2) and (4, 5).

Verification :

Verify the solution (1, 2).

Substitu x = 1 and y = 2 into the original equations.

x - y + 1 = 0 :

1 - 2 + 1 = 0

0 = 0

x2 - 6x + y + 3 = 0 :

12 - 6(1) + 2 + 3 = 0

1 - 6 + 2 + 3 = 0

0 = 0

Verify the solution (4, 5).

Substitu x = 4 and y = 5 into the original equations.

x - y + 1 = 0 :

4 - 5 + 1 = 0

0 = 0

x2 - 6x + y + 3 = 0 :

42 - 6(4) + 5 + 3 = 0

16 - 24 + 5 + 3 = 0

0 = 0

Both solutions are correct.

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