Solve the system of equations algebraically

Mathematics

1 answer:

lora16 [44]3 years ago

8 0

2x + 2y = 10
x + y = 5
x = 5 - y
Substitute for x in the other equation:-

(5 - y - 3)^2 + (y + 2)^2 = 16
(2 - y)^2 + (y + 2)^2 = 16
4 - 4y + y^2 + y^2 + 4y + 4 = 16
2y^2 + 8 = 16

2y^2 - 8 = 0
y^2 = 4

y = +/- 2

consider 2x + 2y = 10

when y = -2, 2x = 10 +4 giving x = 7
when y = 2, 2x = 10 - 4 giving x = 3

so solution is x = 7, y = -2 and x = 3, y=2.

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see attached

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