SOLVING A SYSTEM OF QUADRATIC-QUADRATIC EQUATIONS ALGEBRAICALLY

The most efficient method to solve a system quadratic-quadratic equations algebraically is elimination.

The following steps would be useful to solve the system by elimination.

Solve the systems of quadratic-quadratic equations algebraically and verify the solutions :

Example 1 :

2x² - 16x - y = -35

2x² - 8x - y = -11

Solution :

2x² - 16x - y = -35 ----(1)

2x² - 8x - y = -11 ----(2)

In the above system of two quadratic equations, y has the same coefficent and same sign.

By simply subtracting the two equations, y can be eliminated easily.

(2) - (1) :

(2x² - 8x - y) - (2x² - 16x - y) = -11 - (-35)

2x² - 8x - y - 2x² + 16x + y = -11 + 35

Group the like terms ang combine them.

(2x² - 2x²) + (-8x + 16x) + (-y + y) = 24

0 + 8x + 0 = 24

8x = 24

Divide both sides by 8.

⁸ˣ⁄₈ = ²⁴⁄₈

x = 3

Substitute x = 3 into (1) and solve for y.

2(3)² - 16(3) - y = -35

2(9) - 48 - y = -35

18 - 48 - y = -35

-30 - y = -35

Add 30 to both sides.

-y = -5

Multiply both sides by -1.

y = 5

The solution is (3, 5).

Verification :

Verify the solution (3, 5).

Substitu x = 3 and y = 5 into the original equations.

2x² - 16x - y = -35 :

2(3)² - 16(3) - 5 = -35

2(9) - 48 - 5 = -35

18 - 48 - 5 = -35

-35 = -35 ✔

2x² - 8x - y = -11 :

2(3)² - 8(3) - 5 = -11

2(9) - 24 - 5 = -11

18 - 24 - 5 = -11

-11 = -11 ✔

The solution (3, 5) is correct.

Example 2 :

2x² + 16x + y = -26

x² + 8x - y = -19

Solution :

2x² + 16x + y = -26 ----(1)

x² + 8x - y = -19 ----(2)

In the above system of two quadratic equations, y has the same coefficient with different signs.

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

(1) + (2) :

(2x² + 16x + y) + (x² + 8x - y) = -26 + (-19)

2x² + 16x + y + x² + 8x - y = -26 - 19

Group the like terms ang combine them.

(2x² + x²) + (16x + 8x) + (y - y) = -45

3x² + 24x + 0 = -45

3x² + 24x = -45

Add 45 to both sides.

3x² + 24x + 45 = 0

3(x² + 8x + 15) = 0

Divide both sides by 3.

x² + 8x + 15 = 0

Factor and solve it.

x² + 5x + 3x + 15 = 0

x(x + 5) + 3(x + 5) = 0

(x + 5)(x + 3) = 0

x + 5 = 0  or  x + 3 = 0

x = -5  or  x = -3

The solutions are (-5, 4) and (-3, 4).

Verification :

Verify the solution (-5, 4).

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

2x² + 16x + y = -26 :

2(-5)² + 16(-5) + 4 = -26

2(25) - 80 + 4 = -26

50 - 80 + 4 = -26

-26 = -26 ✔

x² + 8x - y = -19 :

(-5)² + 8(-5) - 4 = -19

25 - 40 - 4 = -19

-19 = -19 ✔

Verify the solution (-3, 4).

Substitute x = -3 and y = 4 into the original equations.

2x² + 16x + y = -26 :

2(-3)² + 16(-3) + 4 = -26

2(9) - 48 + 4 = -26

18 - 48 + 4 = -26

-26 = -26 ✔

x² + 8x - y = -19 :

(-3)² + 8(-3) - 4 = -19

9 - 24 - 4 = -19

-19 = -19 ✔

Both solutions are correct.

Example 3 :

x² + 2x = y - 1

-4x = y + 8

The solution to the given system of equations is (x, y). What is the value of x ?

Solution :

x² + 2x = y - 1 -----(1)

-4x = y + 8 -----(2)

x² + 2x + 1 = y

y = x² + 2x + 1

y = -4x - 8

x² + 2x + 1 = -4x - 8

x² + 2x + 4x + 1 + 8 = 0

x² + 6x + 9 = 0

(x + 3)(x + 3) = 0

x = -3 and x = -3

When x = -3

y = -4(-3) - 8

= 12 - 8

y = 4

So, the required value of x is -3.

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