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

1 Answer
Feb 29, 2016

Answer:

(0,4) and (4,0)

Explanation:

first label the equations.

#x^2 + y^2 = 16....................(1)#
x + y = 4 .......................(2)

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

substitute y = 4 - x into (1)

hence: #x^2 + (4 - x )^2 = 16rArr x^2 +16 - 8x + x^2 = 16#

and # 2x^2 -8x + 16 - 16 = 0rArr 2x^2 - 8x = 0#

factor and solve : 2x(x - 4 ) = 0 #rArr 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 #