Exercise 8 Page 527

We will solve the given system of equations using the Substitution Method. y=2x-3 & (I) y=x^2+x-9 & (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-3 for y in Equation (II). Notice that in Equation (II) we have a quadratic equation in terms of only the x-variable. x^2-x-6=0 ⇕ 1x^2+( - 1)x+( - 6)=0We can substitute a= 1, b= - 1, and c= - 6 into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- ( - 1)±sqrt(( - 1)^2-4( 1)( - 6))/2( 1)

▼

Solve for x

NegNeg

- (- a)=a

x=1±sqrt((- 1)^2-4(1)(- 6))/2(1)

CalcPow

Calculate power

x=1±sqrt(1-4(1)(- 6))/2(1)

IdPropMult

Identity Property of Multiplication

x=1±sqrt(1-4(- 6))/2

MultNegNegOnePar

- a(- b)=a* b

x=1±sqrt(1+24)/2

AddTerms

Add terms

x=1±sqrt(25)/2

CalcRoot

Calculate root

x=1± 5/2

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± 5/2
x_1=1+ 5/2	x_2=1- 5/2
x_1=6/2	x_2=- 4/2
x_1=3	x_2=- 2

Now, consider Equation (I). y=2x-3 We can substitute x=3 and x=- 2 into the above equation to find the values for y. Let's start with x=3.

y=2x-3

Substitute

x= 3

y=2( 3)-3

▼

Solve for y

Multiply

Multiply

y=6-3

AddTerms

Add terms

y=3

We found that y=3 when x=3. One solution of the system, which is a point of intersection of the parabola and the line, is (3,3). To find the other solution, we will substitute - 2 for x in Equation (I) again.

y=2x-3

Substitute

x= - 2

y=2( - 2)-3

▼

Solve for y

Multiply

Multiply

y=- 4-3

AddTerms

Add terms

y=- 7

We found that y=- 7 when x=- 2. Therefore, our second solution, which is the other point of intersection of the parabola and the line, is (- 2,- 7).