Question

graph each equation of the system. Then solve the system to find the points of intersection.

$\left\{\begin{array}{l}y=x^2-4\\ y=6x-13\end{array}\right.$

Short answer

the point of intersection is$(3,5)$.

Step-by-step solution

Step 1. Given 

$\left\{\begin{array}{l}y=x^2-4\\ y=6x-13\end{array}\right.$

Step 2. Graphing each equation

$y=x^2-4$

$y=6x-13$

Step 3. Calculation of intersection point

$$\begin{gathered} Consider, \\ y=x^2-4.....\left(1\right) \\ y=6x-13.....\left(2\right) \\ Nowsubstitutingthevalueofyfromeq.1intoeq.2 \\ weget \\ {x}^2-4=6x-13 \\ {x}^2-6x+9=0 \\ {(x-3)}^2=0x=3 \\ Nowsubstitutingvalueofxineq.1weget \\ y={\left(3\right)}^2-4 \\ y=5 \end{gathered}$$

Step 4. Check the solution

$$\begin{gathered} Substitutetheobtainedvaluesofxandyinboththeequations \\ y=6x-13 \\ 5=6\left(3\right)-13 \\ 5=5 \\ also \\ y=x^2-4 \\ 5={\left(3\right)}^2-4 \\ 5=5 \\ holdstrue. \end{gathered}$$