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Which represents the solution(s) of the graphed system of equations, y = x2 + 2x – 3 and y = x – 1?



(1, 0) and (0, –1)
(–2, –3) and (1, 0)
(0, –3) and (1, 0)
(–3, –2) and (0, 1)

Respuesta :

luisejr77

Answer:

Second option: (-2,-3) and (1,0)

Step-by-step explanation:

Given the system of equations [tex]\left \{ {{y = x^2 + 2x-3} \atop {y = x - 1}} \right.[/tex], you can rewrite them in this form:

[tex]x^2 + 2x-3= x - 1[/tex]

Simplify:

[tex]x^2 + 2x-3-x+1=0\\\\x^2+x-2=0[/tex]

Factor the quadratic equation. Choose two number whose sum be 1 and whose product be -2. These are: 2 and -1, then:

[tex](x+2)(x-1)=0\\\\x_1=-2\\\\x_2=1[/tex]

Substitute each value of "x"  into any of the original equation to find the values of "y":

[tex]y_1= (-2) - 1=-3\\\\y_2=(1)-1=0[/tex]

Then, the solutions are:

(-2,-3) and (1,0)

kudzordzifrancis

ANSWER

The solutions are (-2,-3) and (1,0).

EXPLANATION

The given system has equations:

[tex]y = {x}^{2} + 2x - 3[/tex]

and

[tex]y = x - 1[/tex]

We equate both equations:

[tex] {x}^{2} + 2x - 3 = x - 1[/tex]

[tex] {x}^{2} + 2x - x - 3 + 1 = 0[/tex]

[tex] {x}^{2} + x - 2 = 0[/tex]

[tex](x - 1)(x + 2) = 0[/tex]

This implies that,

[tex]x = - 2 \: or \: x = 1[/tex]

When x=-2 , y=-2-1=-3

When x=1, y=1-1=0

The solutions are (-2,-3) and (1,0)

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