Solve the equations in the system y=x^2-3,2x-y=3

Question

asked Nov 14, 2024 214k views

2 Answers

Katychuang · Answer 1 · 2024-11-16T11:08:11+0000

Explanation:

Isolate the last system for y

2x - y = 3

- y = - 2x + 3

y = 2x - 3

Set each equation equal to each other

${x}^(2) - 3 = 2x - 3$

${x}^(2) - 2x = 0$

x(x - 2) = 0

x = 0

x = 2

Plug in both x values to get their associated y values

${0}^(2) - 3 = - 3$

So one solution is (0,-3)

${2}^(2) - 3 = 1$

Another solution is (2,1)