Exercise 34 Page 536

We want to solve the given system of equations using the Substitution Method. y=x^2-9 & (I) y=2x+5 & (II) The y-variable is isolated in Equation (II). This allows us to substitute its value, 2x+5, for y in Equation (I). Notice that in Equation (I), we have a quadratic equation in terms of only the x-variable. x^2-2x-14=0 ⇔ 1x^2+( - 2)x+( - 14)=0Now, recall the Quadratic Formula. x=- b±sqrt(b^2-4ac)/2a We can substitute a= 1, b= - 2, and c= - 14 into this formula to solve the quadratic equation.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- ( - 2)±sqrt(( - 2)^2-4( 1)( - 14))/2( 1)

▼

Solve for x

NegNeg

- (- a)=a

x=2±sqrt((- 2)^2-4(1)(- 14))/2(1)

CalcPow

Calculate power

x=2±sqrt(4-4(1)(- 14))/2(1)

MultByOne

a * 1=a

x=2±sqrt(4-4(- 14))/2

MultNegNegOnePar

- a(- b)=a* b

x=2±sqrt(4+56)/2

AddTerms

Add terms

x=2±sqrt(60)/2

SplitIntoFactors

Split into factors

x=2± sqrt(4* 15)/2

RootProd

sqrt(a* b)=sqrt(a)*sqrt(b)

x=2± sqrt(4)sqrt(15)/2

CalcRoot

Calculate root

x=2± 2sqrt(15)/2

FactorOut

Factor out 2

x=2(1± sqrt(15))/2

CancelCommonFac

Cancel out common factors

x=1± sqrt(15)

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. Therefore, we have that x_1=1+sqrt(15) and x_2=1-sqrt(15). Now, consider Equation (II). y=2x+5 We can substitute x_1=1+sqrt(15) and x_2=1-sqrt(15) in the above equation to find the values for y. Let's start with x_1=1+sqrt(15).

y=2x+5

Substitute

x= 1+sqrt(15)

y= 2( 1+sqrt(15))+5

Distr

Distribute 2

y= 2+2sqrt(15)+5

AddTerms

Add terms

y= 7+2sqrt(15)

We found that y= 7+2sqrt(15) when x=1+sqrt(15). One solution to the system is (1+sqrt(15),7+2sqrt(15)). To find the other solution, we will substitute 1-sqrt(15) for x in Equation (II) again.

y=2x+5

Substitute

x= 1-sqrt(15)

y= 2( 1-sqrt(15))+5

Distr

Distribute 2

y= 2-2sqrt(15)+5

AddTerms

Add terms

y= 7-2sqrt(15)

We found that y= 7-2sqrt(15) when x=1-sqrt(15). Therefore, our second solution is (1-sqrt(15),7-2sqrt(15)). Finally, if desired, we can approximate our solutions to two decimal places. (1+sqrt(15),7+2sqrt(15)) &≈ (4.87,14.75) (1-sqrt(15),7-2sqrt(15)) &≈ (- 2.87, - 0.75)