Exercise 91 Page 679

We will solve the given system of equations using the Substitution Method. y=2x+9 & (I) y=x^2-2x-3 & (II) Notice that y-variable is isolated in both equations. Since the expression equal to y in Equation (I) is simpler, we will substitute its value 2x+9 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-12=0 ⇕ 1x^2+( - 4)x+( - 12)=0We can substitute a= 1, b= - 4, and c= - 12 into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- ( - 4)±sqrt(( - 4)^2-4( 1)( - 12))/2( 1)

▼

Solve for x

NegNeg

- (- a)=a

x=4±sqrt((- 4)^2-4(1)(- 12))/2(1)

CalcPow

Calculate power

x=4±sqrt(16-4(1)(- 12))/2(1)

IdPropMult

Identity Property of Multiplication

x=4±sqrt(16-4(- 12))/2

MultNegNegOnePar

- a(- b)=a* b

x=4±sqrt(16+48)/2

AddTerms

Add terms

x=4±sqrt(64)/2

CalcRoot

Calculate root

x=4± 8/2

CalcQuot

Calculate quotient

x=2± 4

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± 4
x_1=2+ 4	x_2=2- 4
x_1=6	x_2=- 2

Now, consider Equation (I). y=2x+9 We can substitute x=6 and x=- 2 into the above equation to find the values for y. Let's start with x=6.

y=2x+9

Substitute

x= 6

y=2( 6)+9

▼

Solve for y

Multiply

Multiply

y=12+9

AddTerms

Add terms

y=21

We found that y=21 when x=6. One solution of the system, which is a point of intersection of the parabola and the line, is (6,21). To find the other solution we will substitute - 2 for x in Equation (I) again.

y=2x+9

Substitute

x= - 2

y=2( - 2)+9

▼

Solve for y

Multiply

Multiply

y=- 4+9

AddTerms

Add terms

y=5

We found that y=5 when x=- 2. Therefore our second solution, which is the other point of intersection of the parabola and the line, is (- 2,5).