We want to solve the given system of equations using the Substitution Method. First, we need to identify the possible x- values. The expression x^2-9 is in the denominator, thus is cannot be equal to 0.
x^2-9 ≠0 ⇔ x ≠± 3
Now, we have to rewrite a little our system of equations.
Notice that in Equation (I), we have a quadratic equation in terms of only the x-variable.
x^2-x-12=0 ⇔ ( 1)x^2+( - 1)x+( - 12)=0
We can substitute a= 1, b= - 1, and c= - 12 into the Quadratic Formula.
This result tells us that we have two solutions for x. One of them will use the positive sign and the other one will use the negative sign.
x=1± 7/2
x_1=1+7/2
x_2=1-7/2
x_1=8/2
x_2=- 6/2
x_1=4
x_2=- 3
However, we remember that cannot be equal to 3 or - 3, therefore we obtain obly one solution to the equation above x=4.
Now, consider Equation (II).
y=x^2-9
We can substitute x=4 and x=- 3 into the above equation to find the values for y. Let's start with x=4.
We found that y=7 when x=4. Therefore, there is one solution of the system, which is a point of intersection of the parabola and the line, and it is (4,7).
