Use the Quadratic Formula to solve the equation of the form ax^2+bx+c=0.
(-7,100) and (-10,130)
Practice makes perfect
We want to solve the given system of equations using the Substitution Method.
y=x^2+7x+100 & (I) y+10x=30 & (II)
The y-variable is isolated in Equation (I). This allows us to substitute its value x^2+7x+100 for y in Equation (II).
Notice that in Equation (II), we have a quadratic equation in terms of only the x-variable.
x^2+17x+70=0 ⇔ 1x^2+ 17x+ 70=0We can substitute a= 1, b= 17, and c= 70 into the Quadratic Formula.
We found that y=100 when x=-7. One solution of the system, which is a point of intersection of the parabola and the line, is (-7,100). To find the other solution, we will substitute -10 for x in Equation (I) again.
We found that y=130 when x=-10. Therefore, our second solution, which is the other point of intersection of the parabola and the line, is (-10,130).
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 (-7,100). We will substitute -7 and 100 for x and y, respectively, in Equation (I) and Equation (II).
