# How to Solve the System of Quadratic Equations

## Algebra math problem

Solve the system of quadratic equations ** x² + y² = 25** and

**2**then find the value of

*x*+*y*= 10*,**x*+

*y*

x^2+y^2=25

2x+y=10

x+y=?

## Solution to the system of quadratic equations

Let

*x*² + *y*² = 25………………*eq*(1)

2*x* + *y* = 10………………*eq*(2)

From equation 2

2*x* + *y* = 10

⇒* y* = 10 – 2*x*……………….*eq*(3)

From equation 1 and equation 3

*x*² + *y*² = 25

⇒ *x*² + (10 – 2*x*)² = 25

⇒ *x*² + 100 – 40*x* + 4*x*² = 25

so, * *5*x*² – 40*x* + 75 = 0

⇒ *x*² – 8*x* + 15 = 0

*x*² – 8*x* + 15 = 0 is a quadratic equation so we can apply quadratic equation formula

\begin{aligned} x&=\dfrac{8 \pm \sqrt{(-8)^2 - 4 \times 1 \times 15}}{2 \times 1} \\ \\ &=\dfrac{8 \pm \sqrt{64 - 60}}{2} \\ \\ &=\dfrac{8 \pm \sqrt{4}}{2} \\ \\ &=\dfrac{8 \pm 2}{2} \\ \\ \Rightarrow x&=5 \ \ \& \ \ x= 3 \\ \\ \end{aligned}

when *x* = 5

*y* = 10 – 2*x*

⇒* y* = 10 – 2 × 5

⇒* * *y* = 0

when *x* = 3

*y* = 10 – 2*x*

⇒* y* = 10 – 2 × 3

⇒* * *y* = 4

so the solution to the system of equation is **( x, y) = (5, 0) & (3, 4)**

Now *x* + *y* = 5 + 0 = 5

*x* + *y* = 3 + 4 = 7

So* x *

**+**

*y*= 5*or*

**7**