Exercise 80 Page 552

We will solve the given system of equations using the Substitution Method. y=x^2-x+12 & (I) y=2x^2+3x+7 & (II) Notice that y-variable is isolated in both equations. Since the expression equal to y in Equation (I) seems simpler, we will substitute its value 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)=0We can substitute a= 1, b= 4, and c= - 5 into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- 4±sqrt(4^2-4( 1)( - 5))/2( 1)

▼

Solve for x

CalcPow

Calculate power

x=- 4±sqrt(16-4(1)(- 5))/2(1)

IdPropMult

Identity Property of Multiplication

x=- 4±sqrt(16-4(- 5))/2

MultNegNegOnePar

- a(- b)=a* b

x=- 4±sqrt(16+20)/2

AddTerms

Add terms

x=- 4±sqrt(36)/2

CalcRoot

Calculate root

x=- 4± 6/2

CalcQuot

Calculate quotient

x=- 2 ± 3

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=- 2 ± 3
x_1=- 2 + 3	x_2=- 2 - 3
x_1=1	x_2=- 5

Now, consider Equation (I). y=x^2-x+12 We can substitute x=1 and x=- 5 into the above equation to find the values for y. Let's start with x=1.

y=x^2-x+12

Substitute

x= 1

y= 1^2- 1+12

▼

Solve for y

BaseOne

1^a=1

y=1-1+12

AddSubTerms

Add and subtract terms

y=12

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.

y=x^2-x+12

Substitute

x= - 5

y=( - 5)^2-( - 5)+12

▼

Solve for y

CalcPow

Calculate power

y=25-(- 5)+12

SubNeg

a-(- b)=a+b

y=25+5+12

AddTerms

Add terms

y=42

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).