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Core Connections: Course 3
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Core Connections: Course 3 View details
2. Section 9.2
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Exercise 117 Page 433

Practice makes perfect
In this system of equations, at least one of the variables has a coefficient of 1. This means that we can approach its solution with the Equal Values Method. When solving a system of equations using this system, there are four steps.

  1. Rewrite both equations into slope-intercept form, if necessary.
  2. Set both expressions that are equal to y equal to each other to form a new equation.
  3. Solve this equation for x.
  4. Substitute this solution into one of the equations and solve for y. For this system, y is already isolated in both equations, so we can skip straight to solving!

    y=2x+1 & (I) y=- 3x-4 & (II)
    Substitute

    (II):y= 2x+1

    y=2x+1 2x+1=- 3x-4
    AddEqn

    (II): LHS+3x=RHS+3x

    y=2x+1 5x+1=- 4
    SubEqn

    (II): LHS-1=RHS-1

    y=2x+1 5x=- 5
    DivEqn

    (II): .LHS /5.=.RHS /5.

    y=2x+1 5x5= - 55
    SimpQuot

    (II): Simplify quotient

    y=2x+1 x= - 55
    CalcQuot

    (II): Calculate quotient

    y=2x+1 x=- 1

    Great! Now, to find the value of y, we need to substitute x=- 1 into either of the equations in the given system. Let's use the first equation.

    y=2x+1 x=- 1
    Substitute

    (I):x= - 1

    y=2( - 1)+1 x=- 1
    MultPosNeg

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

    y=- 2+1 x=- 1
    AddTerms

    (I): Add terms

    y=- 1 x=- 1

    The solution, or point of intersection, to this system of equations is the point (- 1,- 1).

    Checking Our Answer

     Let's Check!
    To check our answer, we will substitute our solution into both equations. If doing so results in true statements, then our solution is correct.

    y=2x+1 & (I) y=- 3x-4 & (II)

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

    - 1? =2( - 1)+1 - 1? =- 3( - 1)-4
    MultPosNeg

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

    - 1? =- 2+1 - 1? =- 3(- 1)-4
    MultNegNegOnePar

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

    - 1? =- 2+1 - 1? =3-4

    (I), (II): Add and subtract terms

    - 1=- 1 ✓ - 1=- 1 ✓

    Because both equations are true statements, we know that our solution is correct.

Like in Part A, since both equations are already written in slope-intercept form, let's set them equal to each other and solve for x.

y= 13x+4 & (I) y= 12x-2 & (II)
Substitute

(II):y= 1/3x+4

y= 13x+4 13x+4= 12x-2
SubEqn

(II): LHS-1/3x=RHS-1/3x

y= 13x+4 4= 12x-2- 13x
ExpandFrac

(II): a/b=a * 3/b * 3

y= 13x+4 4= 1* 32* 3x-2- 13x
ExpandFrac

(II): a/b=a * 2/b * 2

y= 13x+4 4= 1* 32* 3x-2- 1* 23* 2x
Multiply

(II): Multiply

y= 13x+4 4= 36x-2- 26x
FactorOut

(II): Factor out x

y= 13x+4 4=( 36- 26)x-2
SubFrac

(II): Subtract fractions

y= 13x+4 4= 3-26x-2
SubTerm

(II): Subtract term

y= 13x+4 4= 16x-2
AddEqn

(II): LHS+2=RHS+2

y= 13x+4 6= 16x
MultEqn

(II): LHS * 6=RHS* 6

y= 13x+4 6(6)=6( 16x)
DenomMultFracToNumber

(II): 6 * a/6= a

y= 13x+4 6(6)=x
Multiply

(II): Multiply

y= 13x+4 36=x
RearrangeEqn

(II): Rearrange equation

y= 13x+4 x=36

Next, let's find the value of y by substituting x=36 into the first equation.

y= 13x+4 x=36
Substitute

(I):x= 36

y= 13( 36)+4 x=36
MoveRightFacToNumOne

(I): 1/b* a = a/b

y= 363+4 x=36
CalcQuot

(I): Calculate quotient

y=12+4 x=36
AddTerms

(I): Add terms

y=16 x=36

The solution to the system of equations is the point (36,16).

Checking Our Answer

 Let's Check!
We can check our answer by substituting our solution into both equations. If doing so results in true statements, then our solution is correct.

y= 13x+4 & (I) y= 12x-2 & (II)

(I), (II): x= 36, y= 16

16? = 13( 36)+4 16? = 12( 36)-2

(I), (II): 1/b* a = a/b

16? = 363+4 16? = 362-2

(I), (II): Calculate quotient

16? =12+4 16? =18-2

(I), (II): Add and subtract terms

16=16 ✓ 16=16 ✓

Since the left-hand side is equal to the right-hand side, our solution is correct.

Subchapter links expand_more
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Side Lengths and Triangles 63 (Page 415)
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Pythagorean Theorem in Three Dimensions 142 (Page 441)
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Pythagorean Theorem Proofs 154 (Page 445)