Exercise 23 Page 599

We want to solve the given system of equations using the Substitution Method. y=3x^2+21x-5 & (I) - 10x+y=- 1 & (II) The y-variable is isolated in Equation (I). This allows us to substitute its value 3x^2+21x-5 for y in Equation (II).

y=3x^2+21x-5 - 10x+y=- 1

Substitute

(II): y= 3x^2+21x-5

y=3x^2+21x-5 -10x+ 3x^2+21x-5=- 1

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(II): Simplify

AddEqn

(II): LHS+1=RHS+1

y=3x^2+21x-5 -10x+3x^2+21x-4=0

SubTerm

(II): Subtract term

y=3x^2+21x-5 3x^2+11x-4=0

Notice that in Equation (II) we have a quadratic equation in terms of only the x-variable. 3x^2+11x-4=0 ⇕ 3x^2+ 11x+( - 4)=0We can substitute a= 3, b= 11, and c= - 4 into the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- 11±sqrt(11^2-4( 3)( - 4))/2( 3)

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Solve for x

CalcPow

Calculate power

x=- 11±sqrt(121-4(3)(- 4))/2(3)

Multiply

x=- 11±sqrt(121-12(- 4))/6

MultPosNeg

a(- b)=- a * b

x=- 11±sqrt(121-(- 48))/6

SubNeg

a-(- b)=a+b

x=- 11±sqrt(169)/6

CalcRoot

Calculate root

x=- 11 ± 13/6

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=- 11 ± 13/6
x_1=- 11+13/6	x_2=- 11-13/6
x_1=2/6	x_2=- 24/6
x_1=1/3	x_2=-4

Now, consider Equation (I). y=3x^2+21x-5 We can substitute x= 13 and x=- 4 into the above equation to find the values for y. Let's start with x= 13.

y=3x^2+21x-5

Substitute

x= 1/3

y=3( 1/3)^2+21( 1/3)-5

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Evaluate right-hand side

PowQuot

(a/b)^m=a^m/b^m

y=3(1/9)+21(1/3)-5

MoveLeftFacToNumOne

a* 1/b= a/b

y=3/9+21/3-5

ReduceFrac

a/b=.a /3./.b /3.

y=1/3+21/3-5

NumberToFrac

a = 3* a/3

y=1/3+21/3-15/3

AddSubFrac

Add and subtract fractions

y=7/3

We found that y= 73 when x= 13. One solution of the system, which is a point of intersection of the two parabolas, is ( 13, 73). To find the other solution, we will substitute - 4 for x in Equation (I).

y=3x^2+21x-5

Substitute

x= - 4

y=3( - 4)^2+21( - 4)-5

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Evaluate right-hand side

CalcPow

Calculate power

y=3(16)+21(- 4)-5

Multiply

y=48+21(- 4)-5

MultPosNeg

a(- b)=- a * b

y=48-84-5

SubTerms

Subtract terms

y=- 41

We found that y=- 41 when x=- 4. Therefore, our second solution, which is the other point of intersection of the two parabolas, is ( - 4, - 41).

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 (- 4,- 41). We will substitute - 4 and - 41 for x and y, respectively, in Equation (I) and Equation (II).

y=3x^2+21x-5 - 10x+y=- 1

(I), (II): x= - 4, y= - 41

- 41? =3( - 4)^2+21( - 4)-5 - 10( - 4)+( - 41)? =- 1

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Simplify

CalcPow

(I): Calculate power

- 41? =3(16)+21(- 4)-5 - 10(- 4)+(- 41)? =-1

MultPosNeg

(I): a(- b)=- a * b

- 41? =3(16)-84-5 - 10(- 4)+(- 41)? =-1

Multiply