# Solve for x and y using the following system of equations: x^2 + y^2 = 10 2x + y = 1

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We have to solve for x and y using the equations

x^2+y^2=10...(1)

2x+y=1 ...(2)

(2)

=> y = 1 - 2x

substitute this for y in (1)

=> x^2 + ( 1- 2x)^2 = 10

=> x^2 + 1 - 4x + 4x^2 = 10

=> 5x^2 - 4x - 9 = 0

=> 5x^2 - 9x + 5x - 9 =0

=> x ( 5x - 9) +1(5x - 9) = 0

=> (x + 1) ( 5x - 9) = 0

Therefore we get x = -1 and 9/5

The corresponding values of y are 3 , -13/5

**Therefore the values of x and y are ( -1 , 3) and ( 9/5 , -13/5).**

We'll have to use substitution to solve the system.We'll solve the second equation for y and we'll substitute it into the first equation.

y = 1 - 2x

x^2 + (1 - 2x)^2=10

We'll expand the square:

x^2 + 1 - 4x + 4x^2 = 10

We'll combine like terms:

5x^2 - 4x + 1 - 10 = 0

5x^2 - 4x - 9 = 0

We'll apply the quadratic formula:

x1 = [4 + sqrt(16 + 180)]/10

x1 = (4 + 14)/10

x1 = 18/10

x1 = 9/5

x2 = (4 - 14)/10

x2 = -1

For x1 = 9/5, y1 = 1 - 18/5

y1 = -13/5

For x2 = -1, y2 = 1 + 2

y2 = 3

**The solutions of the system are: {9/5 ; -13/5} and {-1 ; 3}.**