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Big Ideas Math: Modeling Real Life, Grade 8 View details

2. Solving Systems of Linear Equations by Substitution

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Exercise 15 Page 209

Does either of the equations have an isolated variable in it?

(6,- 3)

Practice makes perfect

When solving a system of equations using the Substitution Method, there are three steps.

Isolate a variable in one of the equations.
Substitute the expression for that variable into the other equation and solve.
Substitute this solution into one of the equations and solve for the value of the other variable.

It looks like it will be simplest to isolate x in the first equation.

- 2x-5y=3 & (I) 3x+8y=- 6 & (II)

AddEqn

(I): LHS+5y=RHS+5y

- 2x=3+5y 3x+8y=- 6

DivEqn

(I): .LHS /(-2).=.RHS /(-2).

- 2x- 2= 3+5y- 2 3x+8y=- 6

▼

(I): Simplify left-hand side

MovePartNumRight

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

- 2- 2x= 3+5y- 2 3x+8y=- 6

QuotOne

(I): a/a=1

1x= 3+5y- 2 3x+8y=- 6

IdPropMult

(I): Identity Property of Multiplication

x= 3+5y- 2 3x+8y=- 6

▼

(II): Simplify right-hand side

WriteSumFrac

(II): Write as a sum of fractions

x= 3- 2+ 5y- 2 3x+8y=- 6

MoveNegDenomToFrac

(II): Put minus sign in front of fraction

x=- 32- 5y2 3x+8y=- 6

MovePartNumRight

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

x=- 32- 52y 3x+8y=- 6

WriteDec

(II): Write as a decimal

x=- 1.5-2.5y 3x+8y=- 6

Now that we have isolated x, we can solve the system by substitution.

x=- 1.5-2.5y & (I) 3x+8y=- 6 & (II)

Substitute

(II):x= (- 1.5-2.5y)

x=- 1.5-2.5y 3 (- 1.5-2.5y)+8y=- 6

Distr

(II): Distribute 3

x=- 1.5-2.5y - 4.5-7.5y+8y=- 6

AddTerms

(II): Add terms

x=- 1.5-2.5y - 4.5+0.5y=- 6

AddEqn

(II): LHS+4.5=RHS+4.5

x=- 1.5-2.5y 0.5y=- 1.5

MultEqn

(II): LHS * 2=RHS* 2

x=- 1.5-2.5y 0.5y(2)=- 1.5(2)

▼

(II): Multiply

CommutativePropMult

(II): Commutative Property of Multiplication

x=- 1.5-2.5y 0.5(2)y=- 1.5(2)

Multiply

(II): Multiply

x=- 1.5-2.5y 1y=- 3

IdPropMult

(II): Identity Property of Multiplication

x=- 1.5-2.5y y=- 3

Great! Now, to find the value of x, we need to substitute y=- 3 into the first equation.

x=- 1.5-2.5y y=- 3

Substitute

(I): y= - 3

x=- 1.5-2.5( - 3) y=- 3

MultNegNegOnePar

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

x=- 1.5+2.5(3) y=- 3

Multiply

(I): Multiply

x=- 1.5+7.5 y=- 3

AddTerms

(I): Add terms

x=6 y=- 3

The solution, or point of intersection, to this system of equations is the point (6,- 3).

Checking Our Answer

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

- 2x-5y=3 & (I) 3x+8y=- 6 & (II)

(I), (II): x= 6, y= - 3

- 2( 6)-5( - 3)? =3 3( 6)+8( - 3)? =- 6

MultNegNegOnePar

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

- 2(6)+5(3)? =3 3(6)+8(- 3)? =- 6

MultPosNeg

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

- 2(6)+5(3)? =3 3(6)-8(3)? =- 6

(I), (II): Multiply

- 12+15? =3 18-24? =- 6

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

3=3 ✓ - 6=- 6 ✓

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

Subchapter links

Try It p.206-207

Self-Assessment for Concepts & Skills p.207

Self-Assessment for Problem Solving p.208

Review & Refresh p.209

Concepts, Skills, & Problem Solving p.209-210

Exercise 15 Page 209

Hints

Check the answer

Practice exercises

Assess your skill

Checking Our Answer