Isolate a variable • Rearrange formulas • Express one quantity in terms of others
Isolate a variable or rewrite a formula (the “in terms of” questions inside Advanced Math: Nonlinear equations in one variable and systems of equations in two variables). Search each ID at satsuitequestionbank.collegeboard.org for the full stem, choices, and rationale.
Difficulty labels below are College Board’s. Use them as the target for the original practice set on this page:
Easy = one inverse operation;
Medium = two-step isolate, distribute, or a short formula;
Hard = nested fractions, square roots, or multi-reciprocal formulas.
| Question ID | Difficulty | Topic Focus |
|---|---|---|
| ad03127d | Easy | Isolate one variable |
| 70f98ab4 | Easy | Add a term: isolate |
| c1964c11 | Easy | Subtract a constant: isolate |
| b8c4a1cd | Easy | Divide both sides: isolate |
| 7a8cb72a | Easy | Divide to isolate |
| b76a2815 | Easy | Context formula: isolate |
| 63c00e2b | Easy | Clear a denominator: isolate |
| 98f735f2 | Easy | Context formula: isolate |
| d964bc26 | Easy | Add a constant: isolate |
| fcb78856 | Easy | Divide a product: isolate |
| bf704c34 | Easy | Add a constant: isolate \ |
| 4e18fc5d | Medium | Isolate from a quotient: |
| ff2c1431 | Medium | Distribute, then isolate |
| 652054da | Medium | Context formula: isolate |
| 95ed0b69 | Medium | Isolate a binomial expression |
| 11ccf3e1 | Medium | Two-step isolate: |
| 29ed5d39 | Medium | Isolate the denominator variable |
| 8f65cddc | Medium | Proportion: isolate \(x\) |
| c77ef2fb | Medium | Context formula: isolate |
| a1262cdb | Medium | Express a combination |
| 2683b5db | Medium | Context formula: isolate |
| 2f958af9 | Medium | Isolate |
| 6acdcece | Medium | Clear a denominator: isolate \(x\) |
| be1b8c74 | Medium | Divide by a binomial factor: isolate |
| 17d0e87d | Hard | Isolate inside a square root: |
| b03adde3 | Hard | Isolate from a reciprocal: |
| 133f3e41 | Hard | Multi-reciprocal formula: isolate |
| 5910bfff | Hard | Context formula with a fraction coefficient: isolate \(H\) |
| c8e9a011 | Hard | Combine reciprocals, then isolate \(n\) |
Goal
Get the requested letter alone on one side. Everything else becomes “in terms of” the remaining variables.
Undo operations in reverse
Add/subtract first to peel off extra terms, then multiply/divide to clear a coefficient or a denominator.
$$ax + b = c \;\Rightarrow\; x = \dfrac{c – b}{a}$$
Do the same thing to both sides
Every add, subtract, multiply, divide, square, or take-reciprocal step must happen on both sides.
Clear a denominator
Multiply both sides by the denominator that contains (or sits next to) the target variable.
$$y = \dfrac{k}{x} \;\Rightarrow\; x = \dfrac{k}{y}$$
Distribute before isolating
If the target is inside parentheses, either divide by the outside factor first or distribute, then finish isolating.
Watch restricted values
You cannot divide by zero. If a variable is given as positive, that usually means it is safe to multiply or divide by it.
One inverse operation — add, subtract, multiply, or divide — matching the official Easy SAT items.
T \( y + 9 = 4x \)
Solve for y
Subtract 9 from both sides:
$$y = 4x – 9$$
Answer: \( y = 4x – 9 \)
\( 6a = b \)
Solve for a?
Divide both sides by 6:
$$a = \dfrac{b}{6}$$
Answer: \( a = \dfrac{b}{6} \)
\( p – 15 = r + s \)
Solve for p
Add 15 to both sides:
$$p = r + s + 15$$
Answer: \( p = r + s + 15 \)
\( \dfrac{k}{5} = m \)
Solve for k
Multiply both sides by 5:
$$k = 5m$$
Answer: \( k = 5m \)
The work \( W \) done by a constant force can be written \( W = Pt \), where \( P \) is power and \( t \) is time. Which equation correctly expresses \( t \) in terms of \( W \) and \( P \)?
Divide both sides by \( P \):
$$t = \dfrac{W}{P}$$
Answer: \( t = \dfrac{W}{P} \)
\( 8j = k + 12m \)
Solve for j
Divide both sides by 8:
$$j = \dfrac{k + 12m}{8}$$
Answer: \( j = \dfrac{k + 12m}{8} \)
\( 4m = 3(n + p) \)
Solve for m
Divide both sides by 4:
$$m = \dfrac{3(n + p)}{4}$$
Answer: \( m = \dfrac{3(n + p)}{4} \)
\( b = 18cf \)
Solve for c
Divide both sides by \( 18f \):
$$c = \dfrac{b}{18f}$$
Answer: \( c = \dfrac{b}{18f} \)
\( \dfrac{r}{12} = s + t \)
Solve for r
Multiply both sides by 12:
$$r = 12(s + t)$$
Answer: \( r = 12(s + t) \)
\( c – 11 = 7p + k \)
Solve for c
Add 11 to both sides:
$$c = 7p + k + 11$$
Answer: \( c = 7p + k + 11 \)
Two-step rearrangements, distributing, simple formulas, and isolating a variable in a denominator — matching official Medium SAT items.
Solve for w
$$v = \dfrac{w}{40x}$$
Multiply both sides by \( 40x \):
$$40xv = w$$
Answer: \( w = 40xv \)
\( 8m = 3(n + q) \)
Solve for n
Divide both sides by 3:
$$\dfrac{8m}{3} = n + q$$
Subtract \( q \):
$$n = \dfrac{8m}{3} – q$$
Answer: \( n = \dfrac{8m}{3} – q \)
A kayak’s distance \( d \), in miles, after \( t \) hours is modeled by \( d = \dfrac{5}{4}t \). Which equation represents the time in terms of the distance?
Multiply both sides by \( \dfrac{4}{5} \):
$$t = \dfrac{4}{5}d$$
Answer: \( t = \dfrac{4}{5}d \)
\( p = \dfrac{k}{3j + 2} \)
Solve for 3j+2
Multiply both sides by \( 3j + 2 \):
$$p(3j + 2) = k$$
Divide both sides by \( p \):
$$3j + 2 = \dfrac{k}{p}$$
Answer: \( 3j + 2 = \dfrac{k}{p} \)
\( 9x + 4y = z \)
Solve for y?
Subtract \( 9x \) from both sides, then divide by 4:
$$4y = z – 9x$$
$$y = \dfrac{z – 9x}{4}$$
Answer: \( y = \dfrac{z – 9x}{4} \)
\( p = 8 + \dfrac{12}{n} \) Solve for n
Subtract 8 from both sides:
$$p – 8 = \dfrac{12}{n}$$
Multiply both sides by \( n \), then divide by \( p – 8 \):
$$n = \dfrac{12}{p – 8}$$
Answer: \( n = \dfrac{12}{p – 8} \)
\( \dfrac{1}{5b} = \dfrac{3x}{y} \)
Solve for x
Multiply both sides by \( y \):
$$\dfrac{y}{5b} = 3x$$
Divide both sides by 3:
$$x = \dfrac{y}{15b}$$
Answer: \( x = \dfrac{y}{15b} \)
A discounted price is given by \( C = \dfrac{P}{1 – r} \), where \( P \) is the sale price and \( r \) is the discount rate. Which equation correctly expresses \( r \) in terms of \( C \) and \( P \)?
Multiply both sides by \( 1 – r \):
$$C(1 – r) = P$$
$$1 – r = \dfrac{P}{C}$$
$$r = 1 – \dfrac{P}{C}$$
Answer: \( r = 1 – \dfrac{P}{C} \)
A shipping fee is calculated with \( T = 0.02(P – 2{,}500) \), where \( P \) is the package value. Which equation expresses \( P \) in terms of \( T \)?
Divide both sides by 0.02 (equivalently, multiply by 50):
$$50T = P – 2{,}500$$
$$P = 50T + 2{,}500$$
Answer: \( P = 50T + 2{,}500 \)
The formula \( v^{2} = \dfrac{RT}{m} \) relates the positive numbers \( v \), \( R \), \( T \), and \( m \). What is \( R \) in terms of \( m \), \( v \), and \( T \)?
Multiply both sides by \( m \):
$$mv^{2} = RT$$
Divide both sides by \( T \):
$$R = \dfrac{mv^{2}}{T}$$
Answer: \( R = \dfrac{mv^{2}}{T} \)
Nested fractions, square roots, and multi-reciprocal formulas
\( \dfrac{5x}{y} = 3\sqrt{w + 8} \)
Solve for w
Divide both sides by 3:
$$\dfrac{5x}{3y} = \sqrt{w + 8}$$
Square both sides (all values are positive):
$$\left(\dfrac{5x}{3y}\right)^{2} = w + 8$$
$$w = \left(\dfrac{5x}{3y}\right)^{2} – 8$$
Answer: \( w = \left(\dfrac{5x}{3y}\right)^{2} – 8 \)
\( u – 4 = \dfrac{10}{t + 1} \)
Solve for t?
Multiply both sides by \( t + 1 \):
$$(t + 1)(u – 4) = 10$$
$$t + 1 = \dfrac{10}{u – 4}$$
$$t = \dfrac{10}{u – 4} – 1 = \dfrac{10 – (u – 4)}{u – 4} = \dfrac{14 – u}{u – 4}$$
Answer: \( t = \dfrac{14 – u}{u – 4} \)
Solve for q
$$\dfrac{12}{p} = \dfrac{12}{q} – \dfrac{12}{r}$$
Add \( \dfrac{12}{r} \) to both sides:
$$\dfrac{12}{q} = \dfrac{12}{p} + \dfrac{12}{r} = 12\left(\dfrac{r + p}{pr}\right)$$
Divide both sides by 12, then take reciprocals:
$$\dfrac{1}{q} = \dfrac{p + r}{pr} \quad\Rightarrow\quad q = \dfrac{pr}{p + r}$$
Answer: \( q = \dfrac{pr}{p + r} \)
The formula \( D = T – \dfrac{3}{8}(80 – H) \) approximates a dew-point-style index. Which equation expresses \( H \) in terms of \( D \) and \( T \)?
Subtract \( T \):
$$D – T = -\dfrac{3}{8}(80 – H)$$
Multiply both sides by \( -\dfrac{8}{3} \):
$$-\dfrac{8}{3}(D – T) = 80 – H$$
$$H = 80 + \dfrac{8}{3}(D – T)$$
Answer: \( H = 80 + \dfrac{8}{3}(D – T) \)
The equation \( \dfrac{8}{n} – \dfrac{3}{t} = -\dfrac{3}{w} \) relates the variables \( n \), \( t \), and \( w \), where \( n > 0 \), \( t > 0 \), and \( w > t \). Which expression is equivalent to \( n \)?
Add \( \dfrac{3}{t} \) to both sides:
$$\dfrac{8}{n} = \dfrac{3}{t} – \dfrac{3}{w} = 3\left(\dfrac{w – t}{tw}\right)$$
Take reciprocals and multiply by 8:
$$n = \dfrac{8tw}{3(w – t)}$$
Answer: \( n = \dfrac{8tw}{3(w – t)} \)
If \( \dfrac{x + 2}{y – 3} = \dfrac{5}{z} \) and \( y \neq 3 \), which equation correctly expresses \( y \) in terms of \( x \) and \( z \)?
Cross-multiply:
$$z(x + 2) = 5(y – 3)$$
$$y – 3 = \dfrac{z(x + 2)}{5}$$
$$y = \dfrac{z(x + 2)}{5} + 3$$
Answer: \( y = \dfrac{z(x + 2)}{5} + 3 \)
Simple interest is given by \( A = P(1 + rt) \), where \( A \) is the accumulated amount, \( P \) is the principal, \( r \) is the interest rate, and \( t \) is time. What is \( t \) in terms of \( A \), \( P \), and \( r \)?
Divide both sides by \( P \):
$$\dfrac{A}{P} = 1 + rt$$
$$\dfrac{A}{P} – 1 = rt$$
$$t = \dfrac{A – P}{Pr}$$
Answer: \( t = \dfrac{A – P}{Pr} \)
The thin-lens equation is \( \dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{q} \). Which equation correctly expresses \( p \) in terms of \( f \) and \( q \)?
Subtract \( \dfrac{1}{q} \) from both sides:
$$\dfrac{1}{p} = \dfrac{1}{f} – \dfrac{1}{q} = \dfrac{q – f}{fq}$$
Take the reciprocal:
$$p = \dfrac{fq}{q – f}$$
Answer: \( p = \dfrac{fq}{q – f} \)
If \( \dfrac{4}{x – 1} + 3 = \dfrac{2}{y} \), which equation correctly expresses \( x \) in terms of \( y \)?
Subtract 3 from both sides:
$$\dfrac{4}{x – 1} = \dfrac{2}{y} – 3 = \dfrac{2 – 3y}{y}$$
Take reciprocals and multiply by 4:
$$x – 1 = \dfrac{4y}{2 – 3y}$$
$$x = \dfrac{4y}{2 – 3y} + 1 = \dfrac{4y + (2 – 3y)}{2 – 3y} = \dfrac{y + 2}{2 – 3y}$$
Answer: \( x = \dfrac{y + 2}{2 – 3y} \)
The equation \( ax^{2} + by = c \) relates the numbers \( a \), \( b \), \( c \), and \( x \), where \( a > 0 \) and \( c – by \ge 0 \). Which equation correctly expresses the nonnegative value of \( x \) in terms of \( a \), \( b \), and \( c \)?
Subtract \( by \) and divide by \( a \):
$$ax^{2} = c – by$$
$$x^{2} = \dfrac{c – by}{a}$$
Take the nonnegative square root:
$$x = \sqrt{\dfrac{c – by}{a}}$$
Answer: \( x = \sqrt{\dfrac{c – by}{a}} \)
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