Exercise 14 Page 599

We want to solve the given system of equations using the Elimination Method. y=- x +3 & (I) y=x^2+1 & (II) The y-variable is isolated and has the same coefficients in both equations. This allows us to subtract Equation (I) from Equation (II) to eliminate the y-variable.

y=- x+3 y=x^2+1

SysEqnSub

(I): Subtract II

y- y=- x+3-( x^2+1) y=x^2+1

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(I): Simplify

Distr

(I): Distribute - 1

y-y=- x+3-x^2-1 y=x^2+1

SubTerms

(I): Subtract terms

0=- x-x^2+2 y=x^2+1

CommutativePropAdd

(I): Commutative Property of Addition

0=- x^2-x+2 y=x^2+1

Notice that the resultant equation is a quadratic equation in terms of only the x-variable. - x^2-x+2=0 ⇔ - 1x^2+( - 1)x+ 2=0We can substitute a= - 1, b= - 1, and c= 2 into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

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

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Solve for x

NegNeg

- (- a)=a

x=1±sqrt((- 1)^2-4(- 1)(2))/2(- 1)

CalcPow

Calculate power

x=1±sqrt(1-4(- 1)(2))/2(- 1)

MultPropNegOne

Multiplication Property of -1

x=1±sqrt(1+4(2))/- 2

Multiply

x=1±sqrt(1+8)/- 2

AddTerms

Add terms

x=1±sqrt(9)/- 2

CalcRoot

Calculate root

x=1 ± 3/- 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 ± 3/- 2
x_1=1+3/- 2	x_2=1-3/- 2
x_1=4/- 2	x_2=- 2/- 2
x_1=- 2	x_2=1

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

y=x^2+1

Substitute

x= - 2

y=( - 2)^2+1

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Solve for y

CalcPow

Calculate power

y=4+1

AddTerms

Add terms

y=5

We found that y=5 when x=- 2. One solution of the system, which is a point of intersection of the two parabolas, is (- 2,5). To find the other solution we will substitute 1 for x in Equation (II).

y=x^2+1

Substitute

x= 1

y= 1^2+1

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Solve for y

CalcPow

Calculate power

y=1+1

AddSubTerms

Add and subtract terms

y=2

We found that y=2 when x=1. Therefore our second solution, which is the other point of intersection of the two parabolas, is ( 1,2).

Checking Our Answer

Checking the answer

We can check our answers by substituting the points into both equations. If they produce true statements, our solutions are correct. Let's start by checking (- 2, 5). We will substitute - 2 and 5 for x and y, respectively, in Equation (I) and Equation (II).

y=- x+3 y=x^2+1

(I), (II): x= - 2, y= 5

5? =- ( - 2)+3 5? =( - 2)^2+1

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Simplify

NegNeg

(I): - (- a)=a

5? =2+3 5? =(- 2)^2+1

CalcPow

(II): Calculate power

5? =2+3 5? =4+1

(I), (II): Add terms

5=5 ✓ 5=5 ✓

Since both equations produced true statements, the solution (- 2, 5) is correct. Let's now check (1, 2).

y=- x+3 y=x^2+1

(I), (II): x= 1, y= 2

2? =- ( 1)+3 2? = 1^2+1

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Simplify

MultPropNegOne

(I): Multiplication Property of -1

2? =- 1 + 3 2? =1^2+1

BaseOne

(II): 1^a=1

2? =- 1 + 3 2? =1+1

(I), (II): Add terms

2=2 ✓ 2=2 ✓

Since again both equations produce true statements, the solution (1,2) is also correct.

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Checking Our Answer