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Big Ideas Math Integrated I, 2016

BI

Big Ideas Math Integrated I, 2016 View details

5. Rewriting Equations and Formulas

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Exercise 31 Page 41

Isolate $C$ on one side through inverse arithmetic operations.

$C = A (\frac{R}{5} + 0.3)$

Practice makes perfect

Let's solve for

C

by applying inverse operations until we have isolated

C .

R = 5 (\frac{C}{A} - 0.3)

$LHS / 5 = RHS / 5$

\frac{R}{5} = \frac{C}{A} - 0.3

$LHS + 0.3 = RHS + 0.3$

\frac{R}{5} + 0.3 = \frac{C}{A}

$LHS \cdot A = RHS \cdot A$

A (\frac{R}{5} + 0.3) = C

Rearrange equation

C = A (\frac{R}{5} + 0.3)

Since

C

is isolated on one side, we have solved the given equation for

C

.

Subchapter links

Communicate Your Answer p.35

Monitoring Progress p.36-39

Exercises p.40-42