Exercise 121 Page 593

We will solve the given system of equations using the Substitution Method. y=- 5x-7 & (I) y=x^2-3x-10 & (II) Notice that the y-variable is isolated in both equations. Since the expression equal to y in Equation (I) is simpler, we will substitute its value - 5x-7 for y in Equation (II). Notice that in Equation (II) we have a quadratic equation in terms of only the x-variable. x^2+2x-3=0 ⇕ 1x^2+ 2x+( - 3)=0We can substitute a= 1, b= 2, and c= - 3 into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- 2±sqrt(2^2-4( 1)( - 3))/2( 1)

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Solve for x

CalcPow

Calculate power

x=- 2±sqrt(4-4(1)(- 3))/2(1)

IdPropMult

Identity Property of Multiplication

x=- 2±sqrt(4-4(- 3))/2

MultNegNegOnePar

- a(- b)=a* b

x=- 2±sqrt(4+12)/2

AddTerms

Add terms

x=- 2±sqrt(16)/2

CalcRoot

Calculate root

x=- 2± 4/2

CalcQuot

Calculate quotient

x=- 1± 2

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

Now, consider Equation (I). y=- 5x-7 We can substitute x=- 3 and x=1 into the above equation to find the values for y. Let's start with x=- 3.

y=- 5x-7

Substitute

x= - 3

y=- 5( - 3)-7

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Solve for y

MultNegNegOnePar

- a(- b)=a* b

y=15-7

SubTerm

Subtract term

y=8

We found that y=8 when x=- 3. One solution of the system, which is a point of intersection of the parabola and the line, is (- 3,8). To find the other solution we will substitute 1 for x in Equation (I) again.

y=- 5x-7

Substitute

x= 1

y=- 5( 1)-7

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Solve for y

IdPropMult

Identity Property of Multiplication

y=- 5-7

SubTerm

Subtract term

y=- 12

We found that y=- 12 when x=1. Therefore, our second solution, which is the other point of intersection of the parabola and the line, is (1,- 12).