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Big Ideas Math: Modeling Real Life, Grade 8

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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 3 Page 206

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

(0.5,- 0.5)

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.

For this exercise, x is already isolated in one equation, so we can skip straight to solving!

x=5y+3 & (I) 2x+4y=- 1 & (II)

(II):x= 5y+3

x=5y+3 2( 5y+3)+4y=- 1

(II): Distribute 2

x=5y+3 2(5y)+2(3)+4y=- 1

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(II): Simplify left-hand side

(II): Multiply

x=5y+3 10y+6+4y=- 1

CommutativePropAdd

(II): Commutative Property of Addition

x=5y+3 10y+4y+6=- 1

(II): Add terms

x=5y+3 14y+6=- 1

(II): LHS-6=RHS-6

x=5y+3 14y=- 7

(II): .LHS /14.=.RHS /14.

x=5y+3 14y/14=- 7/14

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(II): Calculate quotient

MovePartNumRight

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

x=5y+3 1414y= - 714

(II): a/a=1

x=5y+3 1y= - 714

(II): Identity Property of Multiplication

x=5y+3 y= - 714

MoveNegNumToFrac

(II): Put minus sign in front of fraction

x=5y+3 y=- 714

(II): a/b=.a /7./.b /7.

x=5y+3 y=- 12

(II): Write as a decimal

x=5y+3 y=- 0.5

Great! Now, to find the value of y, we need to substitute y=- 0.5 into either one of the equations in the given system. Let's use the first equation.

x=5y+3 y=- 0.5

(I): y= - 0.5

x=5( - 0.5)+3 y=- 0.5

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

x=- 5(0.5)+3 y=- 0.5

(I): Multiply

x=- 2.5+3 y=- 0.5

(I): Add terms

x=0.5 y=- 0.5

The solution, or point of intersection, to this system of equations is the point (0.5,- 0.5).

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.

x=5y+3 & (I) 2x+4y=- 1 & (II)

(I), (II): x= 0.5, y= - 0.5

0.5? =5( - 0.5)+3 2( 0.5)+4( - 0.5)? =- 1

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

0.5? =-5(0.5)+3 2(0.5)-4(0.5)? =- 1

(I), (II): Multiply

0.5? =-2.5+3 1-2? =- 1

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

0.5=0.5 ✓ - 1=- 1 ✓

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