Answer:
[tex]x=1[/tex]
[tex]y=2[/tex]
Step-by-step explanation:
[tex]y=x^2+3x-2[/tex] , [tex]y+3=5x[/tex]
Replace all occurrences of [tex]y[/tex] in [tex]y+3=5x[/tex] with [tex]x^2+3x-2.[/tex]
[tex](x^2+3x-2)+3=5x[/tex]
[tex]y=x^2+3x-2[/tex]
Add [tex]-2[/tex] and 3.
[tex]x^2+3x+1=5x[/tex]
[tex]y=x^2+3x-2[/tex]
Subtract 5x from both sides of the equation.
[tex]x^2+3x+1-5x=0[/tex]
[tex]y=x^2+3x-2[/tex]
Subtract 5x from 3x.
[tex]x^2-2x+1=0[/tex]
[tex]y=x^2+3x-2[/tex]
Rewrite 1 as [tex]1^2[/tex].
[tex]x^2-2x+1^2=0[/tex]
[tex]y=x^2+3x-2[/tex]
Check that the middle term is two times the product of the numbers being squared in the first term and third term.
[tex]2x=2[/tex] · [tex]x[/tex] · [tex]1[/tex]
[tex]y=x^2+3x-2[/tex]
Rewrite the polynomial.
[tex]x^2-2[/tex] · [tex]x[/tex] · [tex]1[/tex] [tex]+[/tex] [tex]1^2=0[/tex]
[tex]y=x^2+3x-2[/tex]
Factor using the perfect square
trinomial rule [tex]a^2-2ab+b^2=(a-b)^2,[/tex]
where a = x and b = 1.
[tex](x-1)^2=0[/tex]
[tex]y=x^2+3x-2[/tex]
Set the [tex]x-1[/tex] equal to 0.
[tex]x-1=0[/tex]
[tex]y=x^2+3x-2[/tex]
Add 1 to both sides of the equation.
[tex]x=1[/tex]
[tex]y=x^2+3x-2[/tex]
Replace all occurrences of [tex]x[/tex] in
[tex]y=x^2+3x-2[/tex] with 1.
[tex]y=(1)^2+3(1)-2[/tex]
[tex]x=1[/tex]
[tex]y=2[/tex]
