## Solving Quadratic Simultaneous Equations

Try to figure out how to solve this equation: x2 + y2 = 29 y - x = 3

## How to Find the Answer

Lets begin by rewriting our two equations, x²+y²=29…..iy-x=3……….ii Next we take our equation ii and equate y in terms of x as follows, y-x=3y=x+3… Now we substitute y term in equation i:. x²+y²=29 where there’s the value of y we replace it with x+3 x²+(x+3) (x+3) =29 we rewrite the equation in quadratic formatx²+x²+3x+3x+9=292x²+6x+9=29 2x²+6x+9-29=02x²+6x-20=0……now we have formed our quadratic equation. To solve the equation we may use **the factorization method or completing square method.** So in the case here, I prefer to use the completing square method because our equation doesn’t have perfect squares. Which is [- b±(b²_4ab)¹/²]/2a Next, we substitute the coefficients of x in our equation above, where a=2, b=6, and c=-20[-(6)±(6²-{4×2×-20})¹/²]/2×2{-6±(36+160)¹/²}/4 {-6±(196)¹/²}/4 where the root of 196=14(-6±14)/4x=(-6+14)/4=2x=(-6-14)/4=-5 Now we have the possible values of x 2&-5, then we substitute these values on our y equation y=x+3y=2+3y=5y=-5+3y=-2 Therefore our values for both x and y will be x=2 or – 5y=5 or – 2 This is because our simultaneous equation had a quadratic equation.