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Algebra Help: Rearranging Formulas To Highlight Quantities

Review real example questions for Rearranging Formulas To Highlight Quantities in Algebra.

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In physics, distance is modeled by $d = rt$ , where $d$ is distance, $r$ is rate (speed), and $t$ is time. Solve $d = rt$ for $t$ .

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Question 1

In physics, distance is modeled by $d = rt$ , where $d$ is distance, $r$ is rate (speed), and $t$ is time. Solve $d = rt$ for $t$ .

$d = \dfrac{r}{t}$
$t = \dfrac{d}{r}$ (correct answer)
$t = \dfrac{r}{d}$
$t = dr$

Explanation: This question tests your ability to rearrange formulas with multiple variables—an essential skill for working with formulas in science, geometry, and real life. Rearranging formulas (sometimes called solving literal equations) works exactly like solving regular equations, with one difference: instead of finding a number, we're finding a formula that expresses one variable in terms of the others. The same algebraic moves apply—we just keep the variables as letters instead of substituting numbers! To solve d = rt for t, we need to isolate t on one side. Since t is being multiplied by r, we do the opposite operation—divide both sides by r: d/r = rt/r, which simplifies to d/r = t, or t = d/r. Choice A is correct because it properly isolates t using division by r, giving t = d/r. Perfect! Choice B incorrectly shows t = dr (multiplying instead of dividing), while choice C has the fraction flipped as t = r/d—remember, we divide distance by rate to get time, not the other way around. The secret to rearranging formulas: pretend the variable you want to solve for is x (like in regular equations), and treat all the other variables like they're numbers. Use the same steps—add, subtract, multiply, divide, just like normal! For example, solving d = rt for t is just like solving 20 = 5x for x: divide both sides by r (or 5), giving t = d/r. Common formula rearrangements to practice: d = rt becomes t = d/r (divide by rate) and r = d/t (divide by time); A = lw becomes l = A/w (divide by width); P = 2l + 2w becomes l = (P - 2w)/2 (subtract 2w, divide by 2). The same formulas show up repeatedly in math and science, so learning these rearrangements once helps you many times!

Question 2

In the rectangle area formula $A = lw$ (where $A$ is area, $l$ is length, and $w$ is width), solve for $w$ . Treat the other variables like numbers and use inverse operations as you would in a numeric equation.

$w = \dfrac{l}{A}$
$w = \dfrac{A}{l}$ (correct answer)
$A = \dfrac{w}{l}$
$w = Al$

Explanation: This question tests your ability to rearrange formulas with multiple variables—an essential skill for working with formulas in science, geometry, and real life. Rearranging formulas (sometimes called solving literal equations) works exactly like solving regular equations, with one difference: instead of finding a number, we're finding a formula that expresses one variable in terms of the others. The same algebraic moves apply—we just keep the variables as letters instead of substituting numbers! Starting with A = lw, we want to isolate w, so we divide both sides by l: A/l = lw/l, which simplifies to A/l = w, or w = A/l. Choice B is correct because it properly isolates w using division by l, giving w = A/l. Perfect! Choice A incorrectly shows w = l/A, which would mean width equals length divided by area—this reverses the fraction and doesn't match our algebraic steps. The secret to rearranging formulas: pretend the variable you want to solve for is x (like in regular equations), and treat all the other variables like they're numbers. Use the same steps—add, subtract, multiply, divide, just like normal! For example, solving A = lw for w is just like solving 12 = 3x for x: divide both sides by 3 (or l), giving x = 12/3 (or w = A/l).

Question 3

In finance, simple interest is modeled by $I = Prt$ , where $I$ is interest, $P$ is principal, $r$ is annual interest rate, and $t$ is time. Rearrange the formula to solve for $r$ (in terms of $I$ , $P$ , and $t$ ).

$r = \dfrac{IP}{t}$
$r = \dfrac{I}{Pt}$ (correct answer)
$r = \dfrac{Pt}{I}$
$r = I - Pt$

Explanation: This question tests your ability to rearrange formulas with multiple variables—an essential skill for working with formulas in science, geometry, and real life. Rearranging formulas (sometimes called solving literal equations) works exactly like solving regular equations, with one difference: instead of finding a number, we're finding a formula that expresses one variable in terms of the others. The same algebraic moves apply—we just keep the variables as letters instead of substituting numbers! For I = P r t, isolate r by dividing both sides by (P t), since r is multiplied by both P and t, giving r = I / (P t). Choice B is correct because it properly isolates r using division by the product P t, giving r = I / (P t). Perfect! Choice C flips the fraction, but remember, to undo multiplication by P t, we divide I by P t—it's easy to mix up, but verifying with numbers helps! The secret to rearranging formulas: pretend the variable you want to solve for is x (like in regular equations), and treat all the other variables like they're numbers. Use the same steps—add, subtract, multiply, divide, just like normal! For example, solving V = I R for R is just like solving 12 = 3x for x: divide both sides by I, giving R = V / I. When checking your work, substitute back: plug in values like I=10, P=100, t=1, r=0.05 into original (I=5) and your formula to confirm it matches.

Question 4

In physics, distance traveled is modeled by $d = rt$ , where $d$ is distance, $r$ is speed (rate), and $t$ is time. Solve for $t$ in terms of $d$ and $r$ using the same steps you would use to isolate a variable in a numeric equation.

$t = d - r$
$t = \dfrac{r}{d}$
$t = dr$
$t = \dfrac{d}{r}$ (correct answer)

Explanation: This question tests your ability to rearrange formulas with multiple variables—an essential skill for working with formulas in science, geometry, and real life. Rearranging formulas (sometimes called solving literal equations) works exactly like solving regular equations, with one difference: instead of finding a number, we're finding a formula that expresses one variable in terms of the others. The same algebraic moves apply—we just keep the variables as letters instead of substituting numbers! Starting with d = r t, to isolate t, divide both sides by r (treating d and r like constants), resulting in t = d / r. Choice C is correct because it properly isolates t using division, giving t = d / r. Perfect! Something like choice A multiplies instead, but that's the opposite of what we need—since r and t are multiplied, division is the inverse to undo it, so double-check those operations! Common formula rearrangements to practice: d = r t becomes t = d / r (divide by rate) and r = d / t (divide by time); A = l w becomes l = A / w (divide by width). The same formulas show up repeatedly in math and science, so learning these rearrangements once helps you many times! When checking your work, substitute back: if you rearranged d = r t to get t = d / r, multiply both sides by r: r · t = r · (d / r) = d, matching the original.

Question 5

Temperature conversion is given by $C = \dfrac{5}{9}(F - 32)$ , where $C$ is degrees Celsius and $F$ is degrees Fahrenheit. Solve for $F$ in terms of $C$ .

$F = \dfrac{5}{9}C + 32$
$F = \dfrac{9}{5}C - 32$
$F = \dfrac{9}{5}C + 32$ (correct answer)
$F = \dfrac{5}{9}(C - 32)$

Explanation: This question tests your ability to rearrange formulas with multiple variables—an essential skill for working with formulas in science, geometry, and real life. Rearranging formulas (sometimes called solving literal equations) works exactly like solving regular equations, with one difference: instead of finding a number, we're finding a formula that expresses one variable in terms of the others. The same algebraic moves apply—we just keep the variables as letters instead of substituting numbers! Starting with C = (5/9)(F - 32), we first multiply both sides by 9/5: (9/5)C = (9/5) · (5/9)(F - 32), which simplifies to (9/5)C = F - 32. Then we add 32 to both sides: (9/5)C + 32 = F - 32 + 32, giving us F = (9/5)C + 32. Choice C is correct because it properly isolates F using multiplication by 9/5 followed by addition of 32, giving F = (9/5)C + 32. Perfect! Choice A incorrectly shows F = (5/9)C + 32, keeping the original fraction 5/9 instead of using its reciprocal 9/5—when we multiply both sides by 9/5, we're undoing the original multiplication by 5/9. The secret to rearranging formulas: pretend the variable you want to solve for is x (like in regular equations), and treat all the other variables like they're numbers. Use the same steps—add, subtract, multiply, divide, just like normal! For temperature conversion, think of it as solving 20 = (5/9)(x - 32) for x: multiply by 9/5, then add 32.

Question 6

The compound interest formula is $A = P(1 + r)^t$ , where $A$ is the final amount, $P$ is the principal, $r$ is the interest rate, and $t$ is time. A financial advisor needs to determine what interest rate is required to reach a target amount in a given time. Which equation correctly solves for $r$ ?

$r = \left(\frac{A}{P}\right)^{\frac{1}{t}} - 1$ (correct answer)
$r = \frac{A}{P(1)^t} - 1$
$r = \left(\frac{A}{P}\right)^t - 1$
$r = \frac{1}{t}\sqrt{\frac{A}{P}} - 1$

Explanation: Starting with $A = P(1 + r)^t$ , divide by $P$ : $\frac{A}{P} = (1 + r)^t$ . Take the $t$ th root: $\left(\frac{A}{P}\right)^{\frac{1}{t}} = 1 + r$ . Subtract 1: $r = \left(\frac{A}{P}\right)^{\frac{1}{t}} - 1$ . Choice B incorrectly simplifies $(1+r)^t$ to just 1. Choice C uses $t$ as the exponent instead of $\frac{1}{t}$ . Choice D uses a square root instead of the $t$ th root.

Question 7

The kinetic energy formula is $KE = \frac{1}{2}mv^2$ , where $m$ is mass and $v$ is velocity. If a physics student wants to find the velocity required to achieve a specific kinetic energy for an object of known mass, which rearranged formula should they use?

$v = \sqrt{\frac{2KE}{m}}$ (correct answer)
$v = \frac{2KE}{m}$
$v = \sqrt{\frac{KE}{2m}}$
$v = \frac{\sqrt{2KE}}{m}$

Explanation: Starting with $KE = \frac{1}{2}mv^2$ , multiply both sides by 2: $2KE = mv^2$ . Divide by $m$ : $\frac{2KE}{m} = v^2$ . Take the square root: $v = \sqrt{\frac{2KE}{m}}$ . Choice B omits the square root operation. Choice C incorrectly places the 2 in the denominator instead of the numerator. Choice D incorrectly places $m$ outside the square root in the denominator.

Question 8

The ideal gas law is $PV = nRT$ , where $P$ is pressure, $V$ is volume, $n$ is the number of moles, $R$ is the gas constant, and $T$ is temperature. A chemist needs to find the number of moles when pressure decreases by half and volume doubles, while temperature remains constant. If the original number of moles was $n_0$ , what is the new number of moles?

$\frac{n_0}{2}$
$2n_0$
$n_0$ (correct answer)
$4n_0$

Explanation: When you encounter problems involving the ideal gas law with changing conditions, the key is to set up equations for both the initial and final states, then compare them to see what happens to each variable. Let's start with the original conditions: $P_0 V_0 = n_0 RT_0$ . Now we need to identify what changes. The pressure decreases by half (so $P = \frac{P_0}{2}$ ), the volume doubles (so $V = 2V_0$ ), and temperature stays constant (so $T = T_0$ ). For the new conditions: $P \cdot V = n \cdot R \cdot T$ . Substituting our changed values: $\frac{P_0}{2} \cdot 2V_0 = n \cdot R \cdot T_0$ . This simplifies to $P_0 V_0 = n R T_0$ . Since we know from the original equation that $P_0 V_0 = n_0 R T_0$ , we can conclude that $n = n_0$ . Looking at the wrong answers: Choice A ( $\frac{n_0}{2}$ ) incorrectly assumes the pressure decrease alone determines the mole change, ignoring the volume doubling. Choice B ( $2n_0$ ) makes the opposite error, focusing only on the volume doubling while ignoring the pressure halving. Choice D ( $4n_0$ ) represents a complete misunderstanding, perhaps multiplying the pressure and volume changes instead of recognizing they cancel out. The correct answer is C: $n_0$ . Study tip: In ideal gas problems with multiple changing variables, always write out both the initial and final state equations completely. When two changes have opposite effects (like pressure decreasing while volume increases proportionally), they often cancel each other out.

Question 9

In physics, $F = ma$ (force equals mass times acceleration). Solve for $a$ in terms of $F$ and $m$ .

$a = \dfrac{m}{F}$
$a = \dfrac{F}{m}$ (correct answer)
$a = Fm$
$F = \dfrac{a}{m}$

Explanation: This question tests your ability to rearrange formulas with multiple variables—an essential skill for working with formulas in science, geometry, and real life. Rearranging formulas (sometimes called solving literal equations) works exactly like solving regular equations, with one difference: instead of finding a number, we're finding a formula that expresses one variable in terms of the others. The same algebraic moves apply—we just keep the variables as letters instead of substituting numbers! Starting with F = ma, we want to isolate a, so we divide both sides by m: F/m = ma/m, which simplifies to F/m = a, or a = F/m. Choice B is correct because it properly isolates a using division by m, giving a = F/m. Perfect! Choice C incorrectly shows a = Fm, which would mean acceleration equals force times mass—this multiplies instead of divides and contradicts physics (more mass with same force means less acceleration, not more). Common formula rearrangements to practice: d = rt becomes t = d/r (divide by rate) and r = d/t (divide by time); A = lw becomes l = A/w (divide by width); F = ma becomes a = F/m (divide by mass). The same formulas show up repeatedly in math and science, so learning these rearrangements once helps you many times!

Question 10

The formula for electrical power is $P = I^2R$ , where $P$ is power, $I$ is current, and $R$ is resistance. An electrical engineer needs to determine the current when the power is quadrupled and the resistance is doubled. If the original current was $I_0$ , what is the new current?

$I_0\sqrt{8}$
$2I_0$
$I_0\sqrt{2}$ (correct answer)
$4I_0$

Explanation: When you encounter problems involving formulas where one variable appears squared or raised to a power, pay close attention to how changes in other variables affect the relationship. This question tests your ability to manipulate algebraic equations when multiple variables change simultaneously. Start with the original power formula: $P_0 = I_0^2R_0$ . Now you need to find the new current when power becomes $4P_0$ and resistance becomes $2R_0$ . Set up the new equation: $4P_0 = I_{new}^2(2R_0)$ . Since $P_0 = I_0^2R_0$ , substitute this into your new equation: $4(I_0^2R_0) = I_{new}^2(2R_0)$ . Simplifying: $4I_0^2R_0 = 2I_{new}^2R_0$ . Divide both sides by $R_0$ : $4I_0^2 = 2I_{new}^2$ . Divide by 2: $2I_0^2 = I_{new}^2$ . Taking the square root: $I_{new} = I_0\sqrt{2}$ . Choice A ( $I_0\sqrt{8}$ ) likely comes from incorrectly multiplying the factors: $4 \times 2 = 8$ , then taking the square root. Choice B ( $2I_0$ ) results from thinking the current doubles when power quadruples, ignoring the resistance change and the squared relationship. Choice D ( $4I_0$ ) assumes current increases by the same factor as power, completely missing the inverse relationship with resistance and the squared term. Remember: when variables in a formula change simultaneously, substitute all changes into the equation before solving. Don't try to handle each change separately—the relationships between variables matter.