SOLVING SYSTEMS OF LINEAR AND QUADRATIC EQUATIONS WORKSHEET (Algebraically)

Problems 1-2 : Solve the given system of linear and quadratic equations by substitution and verify your solution.

Problems 1 :

5x - y - 10 = 0

x² + x - 2y = 0

Problems 2 :

3x + y + 9 = 0

4x² - x + y + 9 = 0

Problems 3-4 : Solve the given system of linear and quadratic equations by elimination and verify your solution.

Problems 3 :

x - y + 6 = 0

x² + 3x + y - 2 = 0

Problems 4 :

2x + y - 2 = 0

x² - 2x + y - 7 = 0

Answers

1. Answer :

5x - y - 10 = 0 ----(1)

x² + x - 2y = 0 ----(2)

In (1), solve for y in terms of x.

5x - y - 10 = 0

-y = -5x + 10

y = 5x - 10 ----(3)

Substitute '5x - 10' for y in (2).

x² + x - 2(5x - 10) = 0

x² + x - 10x + 20 = 0

x² - 9x + 20 = 0

Factor and solve.

x² - 4x - 5x + 20 = 0

x(x - 4) - 5(x - 4) = 0

(x - 4)(x - 5) = 0

x - 4 = 0  or  x - 5 = 0

x = 4  or  x = 5

The two solutions are (4, 10) and (5, 15).

Verification :

Verify the solution (4, 10).

Substitu x = 4 and y = 10 into the original equations.

5x - y - 10 = 0 :

5(4) - 10 - 10 = 0

20 - 10 - 10 = 0

 0 = 0 ✔

x² + x - 2y = 0 :

4² + 4 - 2(10) = 0

16 + 4 - 20 = 0

 0 = 0 ✔

Verify the solution (5, 15).

Substitu x = 5 and y = 15 into the original equations.

5x - y - 10 = 0 :

5(5) - 15 - 10 = 0

25 - 15 - 10 = 0

 0 = 0 ✔

x² + x - 2y = 0 :

5² + 5 - 2(15) = 0

25 + 5 - 30 = 0

 0 = 0 ✔

Both solutions are correct.

2. Answer :

3x + y + 9 = 0 ----(1)

4x² - x + y + 9 = 0 ----(2)

In (1), solve for y in terms of x.

3x + y + 9 = 0

y = -3x - 9 ----(3)

Substitute '-3x - 9' for y in (2).

4x² - x + (-3x - 9) + 9 = 0

4x² - x - 3x - 9 + 9 = 0

4x² - 4x = 0

Factor and solve.

4x(x - 1) = 0

4x = 0  or  x - 1 = 0

x = 0  or  x = 1

The two solutions are (0, -9) and (1, -12).

Verification :

Verify the solution (0, -9).

Substitu x = 0 and y = -9 into the original equations.

3x + y + 9 = 0 :

 3(0) + (-9) + 9 = 0

0 - 9 + 9 = 0

 0 = 0 ✔

4x² - x + y + 9 = 0 :

4(0)² - 0 + (-9) + 9 = 0

 0 + 0 - 9 + 9 = 0

 0 = 0 ✔

Verify the solution (1, -12).

Substitu x = 1 and y = -12 into the original equations.

3x + y + 9 = 0 :

 3(1) + (-12) + 9 = 0

3 - 12 + 9 = 0

 0 = 0 ✔

4x² - x + y + 9 = 0 :

4(1)² - 1 + (-12) + 9 = 0

 4 - 1 - 12 + 9 = 0

 0 = 0 ✔

Both solutions are correct.

3. Answer :

x - y + 6 = 0 ----(1)

x² + 3x + y - 2 = 0 ----(2)

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

(1) + (2) :

(x - y + 6) + (x² + 3x + y - 2) = 0

x - y + 6 + x² + 3x + y - 2 = 0

x² + 4x + 4 = 0

Factor and solve.

x² + 2x + 2x + 4 = 0

x(x + 2) + 2(x + 2) = 0

(x + 2)(x + 2) = 0

(x + 2)² = 0

Taking square root on both sides,

x + 2 = 0

x = -2

Substitute x = -2 in (1).

-2 - y + 6 = 0

-y + 4 = 0

-y = -4

y = 4

The solution is (-2, 4).

Verification :

Verify the solution (-2, 4).

Substitu x = -2 and y = 4 into the original equations.

x - y + 6 = 0

x - y + 6 = 0

-2 - 4 + 6 = 0

 0 = 0 ✔

x² + 3x + y - 2 = 0

x² + 3x + y - 2 = 0

(-2)² + 3(-2) + 4 - 2 = 0

4 - 6 + 4 - 2 = 0

0 = 0 ✔

The solution is (-2, 4) is correct.

4. Answer :

2x + y - 2 = 0 ----(1)

x² - 2x + y - 7 = 0 ----(2)

By subtracting the two equations above, y can be eliminated.

(2) - (1) :

(x² - 2x + y - 7) - (2x + y - 2) = 0

x² - 2x + y - 7 - 2x - y + 2 = 0

x² - 4x - 5 = 0

Factor and solve.

x² - 5x + x - 5 = 0

x(x - 5) + 1(x - 5) = 0

(x - 5)(x + 1) = 0

x - 5 = 0  or  x + 1 = 0

x = 5  or  x = -1

The two solutions are (5, -8) and (-1, 4).

Verification :

Verify the solution (5, -8).

Substitu x = 5 and y = -8 into the original equations.

2x + y - 2 = 0

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

10 - 8 - 2 = 0

 0 = 0 ✔

x² - 2x + y - 7 = 0

x² - 2x + y - 7 = 0

5² - 2(5) + (-8) - 7 = 0

25 - 10 - 8 - 7 = 0

0 = 0 ✔

Verify the solution (-1, 4).

Substitu x = -1 and y = 4 into the original equations.

2x + y - 2 = 0

2(-1) + 4 - 2 = 0

-2 + 4 - 2 = 0

 0 = 0 ✔

x² - 2x + y - 7 = 0

(-1)² - 2(-1) + 4 - 7 = 0

1 + 2 + 4 - 7 = 0

0 = 0 ✔

Both solutions are correct.

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