y=xTo solve the system of equations, follow the steps below.
Step 01: Substitute the value of y from equation 2 in equation 1.
In the second equation:
[tex]y=x^2-x-42[/tex]In the first equation:
[tex]13x-y=90[/tex]So, let's substitute y by x² - x - 42.
[tex]\begin{gathered} 13x-y=90 \\ 13x-(x^2-x-42)=90 \\ 13x-x^2+x+42=90 \end{gathered}[/tex]Adding the like terms:
[tex]-x^2+14x+42=90[/tex]Subtracting 90 from both sides:
[tex]\begin{gathered} -x^2+14x+42-90=90-90 \\ -x^2+14x-48=0 \end{gathered}[/tex]Step 02: Use the quadratic formula to solve the equation.
For a quadratic equation ax² + bx + c = 0, the quadratic formula is:
[tex]\begin{gathered} x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ \end{gathered}[/tex]In this question, the equation is -1x² + 14x + -48 = 0, then, teh coeffitients are:
a = -1
b = 14
c = -48
Substituting the values and solving the equation:
[tex]\begin{gathered} x=\frac{-14\pm\sqrt{14^2-4*(-1)*(-48)}}{2*(-1)} \\ x=\frac{-14\pm\sqrt{196-192}}{-2} \\ x=\frac{-14\pm\sqrt{4}}{-2}=\frac{-14\pm2}{-2} \\ x_1=\frac{-14-2}{-2}=\frac{-16}{-2}=8 \\ x_2=\frac{-14+2}{-2}=\frac{-12}{-2}=6 \end{gathered}[/tex]Step 03: Substitute the values of x in one equation and find y.
Knowing that:
[tex]y=x^2-x-42[/tex]Let's substitute x by 6 and 8 and find the ordered pairs that are the solution of the system.
First, for x = 8:
[tex]\begin{gathered} y=8^2-8-42 \\ y=64-8-42 \\ y=14 \end{gathered}[/tex]Second, for x = 6:
[tex]\begin{gathered} y=6^2-6-42 \\ y=36-48 \\ y=-12 \end{gathered}[/tex]So, the solutions for the system of equations are (8, 14) and (6, -12).
Answer: (8, 14) and (6, -12).
