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