Respuesta :

Cricetus

Answer:

(2,-4) And (3,-2)

Step-by-step explanation:

Here we have to solve one linear equation and a quadratic equation.

First we find the value of y in terms of x from linear equation and then substitute this value in our quadratic equation to solve it for x , Let us see how :

we have y-2x=8

y=2x-8

Now we substitute this in [tex]x^2-3x-y=2[/tex]

Hence we have

[tex]x^2-3x-(2x-8)=2\\x^2-3x-2x+8=2\\x^2-5x+8=2\\x^2-5x+8-2=0\\x^2-5x+6=0\\x^2-2x-3x+6=0\\x(x-2)-3(x-6)=0\\(x-2)(x-3)=0\\[/tex]

Thus we have

either (x-2)= 0 or (x-3)=0

or  x=2 or x=3

Now let us find the value of y by substituting them in y=2x-8 one by one.

y=2(2)-8= 4-8=-4

y+2((3)-8=6-8=-2

Hence our coordinates are

(2,-4) and (3,-2)

Answer:

(2, - 4), (3, - 2)

Step-by-step explanation:

Given the 2 equations

y - 2x = - 8 → (1)

x² - 3x - y = 2 → (2)

Rearrange (1) expressing y in terms of x

y = 2x - 8 → (3)

Substitute y = 2x - 8 in (2)

x² - 3x - 2x + 8 = 2

x² - 5x + 6 = 0 ← in standard form

(x - 2)(x - 3) = 0 ← in factored form

Equate each factor to zero and solve for x

x - 2 = 0 ⇒ x = 2

x - 3 = 0 ⇒ x = 3

Substitute these values into (3) for corresponding values of y

x = 2 : y = (2 × 2) - 8 = 4 - 8 = - 4 ⇒ (2, - 4)

x = 3 : y = (2 × 3) - 8 = 6 - 8 = - 2 ⇒ (3, - 2)