We want to solve the given system of equations using the Substitution Method.
y=x^2-x+12 & (I) y=2x^2+3x+7 & (II)
The y-variable is isolated in both equations. This means we can choose the equation we will use for our initial substitution. Let's substitute x^2-x+12 for y in Equation (II).
Notice that in Equation (II) we have a quadratic equation in terms of only the x-variable.
x^2+4x-5=0 ⇔ 1x^2+ 4x+( -5)=0Now, recall the Quadratic Formula.
x=- b±sqrt(b^2-4ac)/2a
We can substitute a= 1, b= 4, and c= -5 into this formula to solve the quadratic equation.
We found that y=12 when x=1. One solution of the system, which is a point of intersection of the two parabolas, is (1,12). To find the other solution, we will substitute -5 for x in Equation (I) again.
We found that y=42 when x=-5. Therefore, our second solution, which is the other point of intersection of the two parabolas, is (-5,42).
Checking Our Answer
Checking the answer
We can check our answers by substituting the points into both equations. If they produce true statements, our solutions are correct. Let's start by checking (-5,42). We will substitute -5 and 42 for x and y, respectively, in Equation (I) and Equation (II).
