Question

Solve the system of equationsy=2x-5 y=x^2-5 PLEASE HELP FAST!!!!!!!

Answer 1

(0, - 5 ) and (2, - 1 )

Explanation:

Given the 2 equations

y = 2x - 5 → (1)

y = x² - 5 → (2)

Substitute y = x² - 5 into (1)

x² - 5 = 2x - 5 ( subtract 2x - 5 from both sides )

x² - 2x = 0 ← factor out x from each term

x(x - 2) = 0

Equate each factor to zero and solve for x

x = 0

x - 2 = 0 ⇒ x = 2

Substitute these values into (1) for corresponding values of y

x = 0 : y = 2(0) - 5 = 0 - 5 = - 5 ⇒ (0, - 5 )

x = 2 : y = 2(2) - 5 = 4 - 5 = - 1 ⇒ (2, - 1 )

Answer 2

y=2x-5 ......... 1

y = x² - 5 ......... 2

Substitute equation 1 into equation 2

You'll get

2x - 5 = x ² - 5

x² - 2x = -5 + 5

x² - 2x = 0

Factorize

x ( x - 2 ) = 0

x = 0 x = 2

Now put the values of x into any of the equation

When x = 0

y = 2(0) - 5

y = 5

When x = 2

y = 2(2) - 5

y = 4 - 5

y = -1

When x = 0 y = - 5

When x = 2 y = - 1

Hope this helps