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Algebra

Algebra Help: Rearranging Formulas To Highlight Quantities

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

Question 1

The line equation ax+by=cax + by = cax+by=c has constants aaa, bbb, and ccc. Solve for yyy in terms of xxx, aaa, bbb, and ccc.

  1. y=c−axby = \dfrac{c - ax}{b}y=bc−ax​
  2. y=ax−cby = \dfrac{ax - c}{b}y=bax−c​
  3. y=bc−axy = \dfrac{b}{c - ax}y=c−axb​
  4. by=c−xby = c - xby=c−x
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 ax + by = c, we first subtract ax from both sides: ax + by - ax = c - ax, which gives by = c - ax. Then divide both sides by b: by/b = (c - ax)/b, which simplifies to y = (c - ax)/b. Choice A is correct because it properly isolates y using subtraction of ax followed by division by b, giving y = (c - ax)/b. Perfect! Choice B has the wrong sign—it shows ax - c instead of c - ax in the numerator, which would happen if we incorrectly subtracted c from both sides instead of ax. 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!

Question 2

In physics, Newton’s second law is F=maF = maF=ma, where FFF is force, mmm is mass, and aaa is acceleration. Solve for mmm in terms of FFF and aaa.

  1. m=Fam = Fam=Fa
  2. a=mFa = \dfrac{m}{F}a=Fm​
  3. m=Fam = \dfrac{F}{a}m=aF​
  4. m=aFm = \dfrac{a}{F}m=Fa​
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 m, so we divide both sides by a: F/a = ma/a, which simplifies to F/a = m, or m = F/a. Choice C is correct because it properly isolates m using division by a, giving m = F/a. Perfect! Choice A incorrectly multiplies F by a—remember, to undo multiplication by a, we divide by a, not multiply more. When checking your work, substitute back: if you rearranged F = ma to get m = F/a, multiply both sides of your answer by a: a · m = a · (F/a) = F, which gives ma = F—same as the original! This 'does it work backward?' check confirms you rearranged correctly.

Question 3

Temperature can be converted using C=59(F−32)C = \dfrac{5}{9}(F - 32)C=95​(F−32), where CCC is degrees Celsius and FFF is degrees Fahrenheit. Solve the formula for FFF in terms of CCC.

  1. F=95(C−32)F = \dfrac{9}{5}(C - 32)F=59​(C−32)
  2. F=32−95CF = 32 - \dfrac{9}{5}CF=32−59​C
  3. F=59C−32F = \dfrac{5}{9}C - 32F=95​C−32
  4. F=95C+32F = \dfrac{9}{5}C + 32F=59​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 = F - 32, then add 32 to both sides: (9/5)C + 32 = F, or F = (9/5)C + 32. Choice B is correct because it properly isolates F using multiplication by 9/5 and addition of 32, giving F = (9/5)C + 32. Perfect! Choice A incorrectly shows F = (5/9)C - 32, which mixes up the fraction and subtraction—this would give a much colder temperature than intended. 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, remember that multiplying by 9/5 makes numbers bigger (Fahrenheit uses a wider scale), while 5/9 makes them smaller.

Question 4

The perimeter of a rectangle is given by P=2l+2wP = 2l + 2wP=2l+2w, where PPP is perimeter, lll is length, and www is width. Solve for lll in terms of PPP and www.

  1. l=P2−2wl = \dfrac{P}{2} - 2wl=2P​−2w
  2. l=P−2w2l = \dfrac{P - 2w}{2}l=2P−2w​
  3. l=P+2w2l = \dfrac{P + 2w}{2}l=2P+2w​
  4. l=2w−P2l = \dfrac{2w - P}{2}l=22w−P​
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 P = 2l + 2w, we first subtract 2w from both sides: P - 2w = 2l, then divide both sides by 2: (P - 2w)/2 = l, or l = (P - 2w)/2. Choice A is correct because it properly isolates l using subtraction of 2w and division by 2, giving l = (P - 2w)/2. Perfect! Choice B incorrectly shows l = P/2 - 2w, which would mean l = P/2 - 2w, but if we substitute back: P = 2(P/2 - 2w) + 2w = P - 4w + 2w = P - 2w, which doesn't equal P. 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 5

Temperature conversion can be written as the literal equation C=59(F−32)C = \frac{5}{9}(F - 32)C=95​(F−32), where CCC is degrees Celsius and FFF is degrees Fahrenheit. Solve C=59(F−32)C = \frac{5}{9}(F - 32)C=95​(F−32) for FFF.

  1. F=95(C−32)F = \dfrac{9}{5}(C - 32)F=59​(C−32)
  2. F=32−95CF = 32 - \dfrac{9}{5}CF=32−59​C
  3. F=59C+32F = \dfrac{5}{9}C + 32F=95​C+32
  4. F=95C+32F = \dfrac{9}{5}C + 32F=59​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! To solve C = (5/9)(F - 32) for F, we first multiply both sides by 9/5 to clear the fraction: (9/5)C = F - 32. Then we add 32 to both sides: (9/5)C + 32 = F, or F = (9/5)C + 32. Choice C is correct because it properly isolates F using multiplication by 9/5 and addition of 32, giving F = (9/5)C + 32. Perfect! Choice A incorrectly keeps the 5/9 fraction, while choice B has the wrong order of operations—we need to multiply C by 9/5 first, then add 32, not subtract 32 from C before multiplying. 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 C = (5/9)(F - 32) for F is like solving 20 = (5/9)(x - 32) for x: multiply by 9/5, then add 32. 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 6

In physics, distance is modeled by d=rtd = rtd=rt, where ddd is distance, rrr is rate (speed), and ttt is time. Solve d=rtd = rtd=rt for ttt.

  1. d=rtd = \dfrac{r}{t}d=tr​
  2. t=drt = \dfrac{d}{r}t=rd​
  3. t=rdt = \dfrac{r}{d}t=dr​
  4. t=drt = drt=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 7

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

  1. w=lAw = \dfrac{l}{A}w=Al​
  2. w=Alw = \dfrac{A}{l}w=lA​
  3. A=wlA = \dfrac{w}{l}A=lw​
  4. w=Alw = Alw=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 8

Simple interest is modeled by I=PrtI = PrtI=Prt (where III is interest, PPP is principal, rrr is annual rate, and ttt is time). Rearrange to solve for rrr.

  1. r=IPtr = \dfrac{I}{Pt}r=PtI​
  2. r=PtIr = \dfrac{Pt}{I}r=IPt​
  3. r=IPtr = IPtr=IPt
  4. r=IP−tr = \dfrac{I}{P} - tr=PI​−t
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 I = Prt, we want to isolate r, so we divide both sides by Pt (the product of P and t): I/(Pt) = Prt/(Pt), which simplifies to I/(Pt) = r, or r = I/(Pt). Choice A is correct because it properly isolates r using division by the product Pt, giving r = I/(Pt). Perfect! Choice B incorrectly shows r = Pt/I, which inverts the fraction—this would mean rate equals principal times time divided by interest, which doesn't match our algebraic steps. When checking your work, substitute back: if you rearranged I = Prt to get r = I/(Pt), multiply both sides by Pt: Pt · r = Pt · [I/(Pt)] = I, which gives Prt = I—same as the original! This 'does it work backward?' check confirms you rearranged correctly.

Question 9

Temperature conversion can be written as C=59(F−32)C = \dfrac{5}{9}(F - 32)C=95​(F−32), where CCC is degrees Celsius and FFF is degrees Fahrenheit. Solve for FFF in terms of CCC.

  1. F=59C+32F = \dfrac{5}{9}C + 32F=95​C+32
  2. F=95(C−32)F = \dfrac{9}{5}(C - 32)F=59​(C−32)
  3. F=95C+32F = \dfrac{9}{5}C + 32F=59​C+32
  4. F=32−95CF = 32 - \dfrac{9}{5}CF=32−59​C
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 add 32 to both sides: (9/5)C + 32 = F - 32 + 32, which gives (9/5)C + 32 = F, or 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 B incorrectly applies the 9/5 to (C - 32) instead of just C—the parentheses placement matters! 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 10

Ohm’s law is V=IRV = IRV=IR (where VVV is voltage, III is current, and RRR is resistance). Rearrange the formula to solve for RRR.

  1. R=IVR = \dfrac{I}{V}R=VI​
  2. R=VIR = VIR=VI
  3. R=VIR = \dfrac{V}{I}R=IV​
  4. V=RIV = \dfrac{R}{I}V=IR​
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 V = IR, we want to isolate R, so we divide both sides by I: V/I = IR/I, which simplifies to V/I = R, or R = V/I. Choice C is correct because it properly isolates R using division by I, giving R = V/I. Perfect! Choice A incorrectly shows R = I/V, which reverses the fraction—this would mean resistance equals current divided by voltage, which doesn't match our algebraic steps or the physics (higher voltage with same current means higher resistance, not lower). 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 = IR for R is just like solving 12 = 3x for x: divide both sides by 3 (or I), giving x = 12/3 (or R = V/I).