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Pearson Algebra 1 Common Core, 2011

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Pearson Algebra 1 Common Core, 2011 View details

2. Solving Two-Step Equations

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Exercise 45 Page 92

Gather all of the variable terms on one side of the equation and all of the constant terms on the other side.

0.449

Practice makes perfect

To solve an equation, we should first gather all of the variable terms on one side of the equation and all of the constant terms on the other side using the Properties of Equality.

2.4 + 10m = 6.89

LHS-2.4=RHS-2.4

2.4 - 2.4 + 10m = 6.89 - 2.4

Subtract terms

10m = 4.49

Now we can continue to solve using the Properties of Equality.

10m = 4.49

.LHS /10.=.RHS /10.

10m/10=4.49/10

CancelCommonFac

Cancel out common factors

10m/10=4.49/10

Simplify quotient

m=4.49/10

Calculate quotient

m=0.449

The solution to the equation is m=0.449. We can check our solution by substituting it into the original equation.

2.4 + 10m = 6.89

m= 0.449

2.4 + 10( 0.449) ? = 6.89

▼

Evaluate

Multiply

2.4 + 4.49 ? = 6.89

Add and subtract terms

6.89=6.89

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

Subchapter links

Lesson Check p.91

Practice and Problem-Solving Exercises p.91-93

Standardized Test Prep p.93

Mixed Review p.93