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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 10 Page 207

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

(- 4,- 1)

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.

Observing the given equations, it looks like it will be simplest to isolate x in the first equation.

x-5y=1 & (I) - 2x+9y=- 1 & (II)

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

x-5y+5y=1+5y - 2x+9y=- 1

(I): Add terms

x=1+5y - 2x+9y=- 1

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

x=1+5y - 2x+9y=- 1

(II):x= 1+5y

x=1+5y - 2( 1+5y)+9y=- 1

(II): Distribute - 2

x=1+5y - 2(1)-2(5y)+9y=- 1

▼

(II): Simplify left-hand side

(II): Identity Property of Multiplication

x=1+5y - 2-2(5y)+9y=- 1

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

x=1+5y - 2-10y+9y=- 1

(II): Add terms

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

CommutativePropAdd

(II): Commutative Property of Addition

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

(II): LHS+2=RHS+2

x=1+5y - y=1

(II): LHS * (- 1)=RHS* (- 1)

x=1+5y - y(- 1)=1(- 1)

▼

Multiply

MultNegNegOnePar

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

x=1+5y y(1)=1(- 1)

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

x=1+5y y(1)=- 1(1)

(II): Identity Property of Multiplication

x=1+5y y=- 1

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

x=1+5y y=- 1

(I): y= - 1

x=1+5( - 1) y=- 1

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

x=1-5(1) y=- 1

(I): Identity Property of Multiplication

x=1-5 y=- 1

(I): Subtract term

x=- 4 y=- 1

The solution, or point of intersection, to this system of equations is the point (- 4,- 1).

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=1 & (I) - 2x+9y=- 1 & (II)

(I), (II): x= - 4, y= - 1

- 4-5( - 1)? =1 & (I) - 2( - 4)+9( - 1)? =- 1 & (II)

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

- 4+5(1)? =1 & (I) 2(4)+9(- 1)? =- 1 & (II)

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

- 4+5(1)? =1 & (I) 2(4)-9(1)? =- 1 & (II)

(I), (II): Identity Property of Multiplication

- 4+5? =1 & (I) 2(4)-9? =- 1 & (II)

(II): Multiply

- 4+5? =1 & (I) 8-9? =- 1 & (II)

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

1=1 ✓ - 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