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Pearson Algebra 1 Common Core, 2011 View details

3. Solving Multi-Step Equations

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Exercise 15 Page 98

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

5 47

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. In this case, the variable terms are already on the left-hand side of the equation, so we will start by combining like terms and then isolating b-term.

b-9+6b=30

AddTerms

Add terms

7b-9=30

AddEqn

LHS+9=RHS+9

7b-9+9=30+9

AddTerms

Add terms

7b=39

DivEqn

.LHS /7.=.RHS /7.

b=39/7

Since 39 and 7 share no common factors, this is as simplified as our solution can be. The solution to the equation is b= 397. We can check our solution by substituting it into the original equation.

b-9+6b=30

Substitute

b= 39/7

39/7-9+6( 39/7)? =30

▼

Evaluate left-hand side

MoveLeftFacToNum

a*b/c= a* b/c

39/7-9+234/7? =30

NumberToFrac

a = 7* a/7

39/7-63/7+234/7? =30

SubFrac

Subtract fractions

-24/7+234/7? =30

AddFrac

Add fractions

210/7? =30

CalcQuot

Calculate quotient

30=30

Since the left-hand side is equal to the right-hand side, our solution is correct. Although b= 397 is a perfectly valid solution, we can also rewrite it as a mixed number.

b=39/7

WriteSum

Write as a sum

b=35+4/7

WriteSumFrac

Write as a sum of fractions

b=35/7+4/7

CalcQuot

Calculate quotient

b=5+4/7

Rewrite

Rewrite 5+4/7 as 5 47