# How do you solve x² + y² = 16 and x + y = 4 using substitution?

Feb 29, 2016

(0,4) and (4,0)

#### Explanation:

first label the equations.

${x}^{2} + {y}^{2} = 16. \ldots \ldots \ldots \ldots \ldots \ldots . \left(1\right)$
x + y = 4 .......................(2)

rearrange (2) to y = 4 - x (could do x = 4 - y )

substitute y = 4 - x into (1)

hence: ${x}^{2} + {\left(4 - x\right)}^{2} = 16 \Rightarrow {x}^{2} + 16 - 8 x + {x}^{2} = 16$

and $2 {x}^{2} - 8 x + 16 - 16 = 0 \Rightarrow 2 {x}^{2} - 8 x = 0$

factor and solve : 2x(x - 4 ) = 0 $\Rightarrow x = 0 , x = 4$

substitute these values into y = 4 - x , to find corresponding values of y.

x = 0 : y = 4 - 0 = 4 → (0 , 4)

x = 4 : y = 4 - 4 = 0 → (4 , 0 )

These are the points of intersection with the line x +y = 4 and the circle ${x}^{2} + {y}^{2} = 16$