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


A(1, 0) and (0, –1)
B(–2, –3) and (1, 0)
C(0, –3) and (1, 0)
D(–3, –2) and (0, 1)

Respuesta :

irspow
For the solution(s) to exist the equations must be equal to each other.  Thus we can say y=y which means:

x^2+2x-3=x-1  subtract x from both sides

x^2+x-3=-1  add 1 to both sides

x^2+x-2=0  now factor

(x+2)(x-1)=0, so there are two solutions:

x=-2 and 1, to find the corresponding y values we can use y=x-1

y(-2)=-3 and y(1)=0 so the two solutions are the points:

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

The solutions of the graphed system of equations, y = x² + 2x – 3 and y = x – 1 are B. (–2, –3) and (1, 0)

Further explanation

Discriminant of quadratic equation ( ax² + bx + c = 0 ) could be calculated by using :

D = b² - 4 a c

From the value of Discriminant , we know how many solutions the equation has by condition :

D < 0 → No Real Roots

D = 0 → One Real Root

D > 0 → Two Real Roots

Let us now tackle the problem!

Given :

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

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

To get a solution from the two equations above, it can be done with the following substitution method:

[tex]y = y[/tex]

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

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

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

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

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

(x + 2) = 0 or (x - 1) = 0

x = -2 or x = 1

For x = - 2 :

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

[tex]y = -2 - 1[/tex]

[tex]y = -3[/tex]

∴ The solution is ( -2 , -3 )

For x = 1 :

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

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

[tex]y = 0[/tex]

∴ The solution is ( 1 , 0 )

Learn more

  • Solving Quadratic Equations by Factoring : https://brainly.com/question/12182022
  • Determine the Discriminant : https://brainly.com/question/4600943
  • Formula of Quadratic Equations : https://brainly.com/question/3776858

Answer details

Grade: High School

Subject: Mathematics

Chapter: Quadratic Equations

Keywords: Quadratic , Equation , Discriminant , Real , Number

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