# solve for the indicated variable: a. temperature formula solve for F: K=5/9(F-32)+273 b....

## Question:

solve for the indicated variable:

a. temperature formula

solve for F:

{eq}K=5/9(F-32)+273 {/eq}

b. annual interest rate

solve for r:

{eq}A=P+Prt {/eq}

## Solving for a Specified Variable:

In mathematics, we are often given equations and formulas with more than one variable, and we want to get one variable in terms of the other variables. When this is the case, we solve for the specified variable by treating the other variables as if they were numbers and isolating the specified variable on one side of the equation.

## Answer and Explanation:

For part *a*, we want to solve {eq}K=\frac{5}{9}(F-32)+273 {/eq} for *F*. Thus, we treat *K* as though it was a number, and isolate *F* on one side of the equation. This goes as follows:

- {eq}K=\frac{5}{9}(F-32)+273 {/eq}

Subtract 273 from both sides of the equation.

- {eq}K-273=\frac{5}{9}(F-32) {/eq}

Multiply both sides of the equation by {eq}\frac{9}{5} {/eq}.

- {eq}\frac{9}{5}(K-273)=F-32 {/eq}

Add 32 to both sides of the equation.

- {eq}\frac{9}{5}(K-273)+32=F {/eq}

Interchange sides of the equation.

- {eq}F=\frac{9}{5}(K-273)+32 {/eq}

Solving for *F* gives that {eq}F=\frac{9}{5}(K-273)+32 {/eq}.

For part *b*, we want to solve {eq}A=P+Prt {/eq} for *r*, so we treat *A*, *P*, and *t* as if they were numbers, and isolate *r* on one side of the equation.

- {eq}A=P+Prt {/eq}

Subtract *P* from both sides of the equation.

- {eq}A-P=Prt {/eq}

Divide both sides of the equation by *Pt*.

- {eq}\frac{A-P}{Pt}=r {/eq}

Interchange sides of the equation.

- {eq}r=\frac{A-P}{Pt} {/eq}

We get that {eq}r=\frac{A-P}{Pt} {/eq}.

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from Algebra I: High School

Chapter 8 / Lesson 11