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Pearson Algebra 1 Common Core, 2011

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Pearson Algebra 1 Common Core, 2011 View details

6. Trigonometric Ratios

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Exercise 48 Page 651

Solve for the variable by using the Cross Products Property.

3

Practice makes perfect

To solve the proportion, we will start by using the Cross Products Property. Remember that we will need to treat 4b+2 as a single quantity in the cross multiplication process.

3/7=2b/4b+2

Cross multiply

3(4b+2)=7(2b)

Multiply

3(4b+2)=14b

From here, we will continue solving for x by using the Distributive Property and the Properties of Equality.

3(4b+2)=14b

Distribute 3

12b+6=14b

LHS-12b=RHS-12b

6=14b-12b

Subtract term

6=2b

.LHS /2.=.RHS /2.

3=b

Rearrange equation

b=3

Subchapter links

Lesson Check p.649

Practice and Problem-Solving Exercises p.649-651

Standardized Test Prep p.651

Mixed Review p.651