Task Sets: Rearrange formulas - Common Core Math - HS Algebra

Concepts

A student solves the equation $3x + 6 = 15$ by first subtracting 6
from both sides, then dividing both sides by 3, to get $x = 3$ .

The same student now needs to rearrange the formula $V = IR$ to find
$R$ . Which statement correctly describes the relationship between these
two tasks?

The formula requires a completely different approach because $V$ ,
$I$ , and $R$ are not numbers.

Rearranging $V = IR$ is the same type of move as solving $3x = 15$
— divide both sides by the coefficient of the target variable.

Formulas with three or more letters cannot be rearranged without
additional information about the values.

Only formulas with exactly one operation can be rearranged; $V = IR$
has multiplication, so it needs a different method.

When rearranging the formula $A = \frac{1}{2}bh$ to solve for $h$ ,
what should be done first?

Subtract $b$ from both sides.

Multiply both sides by 2 (to undo the $\frac{1}{2}$ ).

Divide both sides by $A$ .

Subtract $h$ from both sides.

The formula for simple interest is $I = Prt$ , where $I$ is interest,
$P$ is principal, $r$ is annual rate, and $t$ is time in years.

Rearrange for $P$ : divide both sides of $I = Prt$ by $rt$ to get
$P = \frac{I}{rt}$ .

If $I = 360$ , $r = 0.06$ , and $t = 5$ , what is the value of $rt$ ?

Procedures

Ohm's Law states $V = IR$ , where $V$ = voltage (volts), $I$ = current
(amps), and $R$ = resistance (ohms).

Rearranging for $R$ : divide both sides by $I$ to get $R = \frac{V}{I}$ .

If a circuit has voltage $V = 24$ volts and current $I = 6$ amps,
what is the resistance $R$ in ohms?

The area of a circle is $A = \pi r^2$ . To rearrange this formula to
solve for $r$ , which sequence of steps is correct?

Subtract $\pi$ from both sides: $A - \pi = r^2$ ; then take the
square root: $r = \sqrt{A - \pi}$ .

Divide both sides by $\pi$ : $\frac{A}{\pi} = r^2$ ; then take the
square root: $r = \sqrt{\frac{A}{\pi}}$ .

Divide both sides by $\pi$ : $\frac{A}{\pi} = r^2$ ; then the answer
is $r = \frac{A}{\pi}$ (no square root needed).

Take the square root of both sides first: $\sqrt{A} = \sqrt{\pi} r$ ;
then divide by $\sqrt{\pi}$ : $r = \frac{\sqrt{A}}{\sqrt{\pi}}$ .

The total amount with simple interest is $A = P + Prt$ , where $A$ is
total amount, $P$ is principal, $r$ is rate, and $t$ is time.

To solve for $P$ , a student writes:

"Divide both sides by $rt$ : $\frac{A}{rt} = 1 + P$ , so $P = \frac{A}{rt} - 1$ ."

What error did the student make?

The student correctly isolated $P$ . The answer $P = \frac{A}{rt} - 1$
is right.

The student divided only the $Prt$ term by $rt$ , not the $P$ term.
The correct first step is to factor: $A = P(1 + rt)$ , then divide
by $(1 + rt)$ .

The student should have subtracted $Prt$ from both sides first.

The student should have multiplied both sides by $rt$ instead of
dividing.

HSA.CED.A.4 Tutor Intake — Rearranging Formulas

Concepts

Procedures