In order to solve the system of equations algebraically you have to follow those steps
we have
[tex]y=x^{2}-x-3[/tex] ------> equation A
[tex]y=-3x+5[/tex] -------> equation B
Step 1
Equate the equation A and equation B
[tex]x^{2}-x-3=-3x+5[/tex]
[tex]x^{2}-x-3+3x-5=0[/tex]
[tex]x^{2}+2x-8=0[/tex]
Step 2
Convert the quadratic equation in factored form
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex]x^{2}+2x=8[/tex]
Complete the square. Remember to balance the equation by adding the same constants to each side.
[tex]x^{2}+2x+1=8+1[/tex]
[tex]x^{2}+2x+1=9[/tex]
Rewrite as perfect squares
[tex](x+1)^{2}=9[/tex]
Square root both sides
[tex]x+1=(+/-)3[/tex]
[tex]x=-1(+/-)3[/tex]
[tex]x1=-1+3=2[/tex]
[tex]x2=-1-3=-4[/tex]
Step 3
Find the values of y
Substitute the value of x in the equation B
For [tex]x=2[/tex]
[tex]y=-3*2+5=-1[/tex]
For [tex]x=-4[/tex]
[tex]y=-3*(-4)+5=17[/tex]
therefore
the answer is
[tex](2,-1)[/tex] and [tex](-4,17)[/tex]
