# How To Solve A System Of Quadratic Equations By Elimination?

## Solve the system of quadratic equations by elimination

If xy + 3y² – x + 4y = 7 and 2xy + y² – 2x – 2y = -1 then solve the system of quadratic equations using elimination

## Solution

Equations have two variables, we need a single-variable quadratic equation to apply the quadratic formula or solve using factorization. We can do this by eliminating one variable from given equations, so we can create a new equation in a single variable

Let

xy + 3y² – x + 4y = 7…………..eq(1)

2xy + y² – 2x – 2y = -1………….eq(2)

Multiply equation 1 with 2

2(xy + 3y² – x + 4y) = 2 × 7

⇒ 2xy + 6y² – 2x + 8y = 14………eq(3)

subtract equation 2 from equation 3 then

2xy + 6y² – 2x + 8y – (2xy + y² – 2x – 2y) = 14 – (-1)

⇒ 2xy + 6y² – 2x + 8y – 2xy – y² + 2x + 2y = 14 + 1

⇒ 5y² + 10y = 15

We can simplify this equation to **y² + 2y – 3 = 0**, This is a simple quadratic equation in y so we can apply factorization to find the value of y

y² + 2y – 3 = y² – y + 3y – 3

⇒ y² + 2y – 3 = y(y – 1) + 3(y – 1)

⇒ y² + 2y – 3 = (y – 1)(y + 3)

so, (y – 1)(y + 3) = 0

Thus, y = 1 or y = -3

When y = 3 from equation 1

xy + 3y² – x + 4y = 7

⇒ -3x + 3(9) – x + 4(-3) = 7

⇒ -4x = -8

Thus, x = 2

When y = 1 from equation 1

xy + 3y² – x + 4y = 7

⇒ x + 3 – x + 4 = 7

⇒ x – x = 0

which means any real value of x satisfy y = 1 or x ∈ R

So solutions of **(x, y) are (2, -3) and (1, R)**