Relationships Between Quantities and Reasoning with Equations
Unit 1 of Algebra I. 15 questions below, each with the working. Every answer was checked by a second pass before it was published.
Units and quantities, interpreting expressions, creating equations and inequalities in one and two variables, solving linear equations and inequalities, rearranging formulas.
How this unit is tested
Start by identifying what kind of task the problem is asking for: is it asking you to interpret an existing expression, build a new equation or inequality from a word problem, solve an equation or inequality that's already set up, or isolate a different variable in a formula? Each of these calls for a different first move, and misidentifying the task is the most common way students waste time on this unit.
For word problems, always name your variable with its units before writing anything else ("let m = miles driven"), then translate the sentence piece by piece into symbols, checking that the units on both sides of your equation or inequality actually match. For inequality problems, watch for key phrases — "at least" means ≥, "at most" means ≤, "more than" is strict >, "fewer than" is strict <.
When solving, work systematically: clear parentheses and fractions first, combine like terms on each side, then move variable terms to one side and constants to the other using inverse operations. Every time you multiply or divide both sides by a negative number in an inequality, flip the inequality symbol — build this into your routine rather than trying to remember it only when it looks necessary.
For rearranging formulas, treat the variable you're solving for exactly as you would an unknown in a numerical equation, and treat every other letter as if it were just a number. Finally, once you have a numerical or algebraic answer, always look back at the context: does the sign, size, and unit of the answer make sense as a viable answer to the original question?
What you have to know
Units Guide Problem Solving (N.Q.1)
Use units as a guide in solving multi-step problems; choose and interpret units consistently in formulas, and choose the scale and origin appropriate for graphs and data displays.
Interpreting Expressions (A.SSE.1)
Interpret expressions that represent a quantity in terms of its context: identify individual terms, factors, and coefficients, and interpret complicated expressions by viewing one or more of their parts as a single entity.
Creating Equations and Inequalities (A.CED.1–2)
Create equations and inequalities in one variable to model constraints or relationships, and create equations in two or more variables to represent relationships between quantities, graphing them on coordinate axes with labeled units.
Constraints and Viable Solutions (A.CED.3)
Represent constraints by equations or inequalities, and interpret solutions as viable or nonviable options in the context of the original problem.
Rearranging Formulas / Literal Equations (A.CED.4)
Rearrange formulas to isolate a quantity of interest, using the same reasoning as in solving equations (for example, solve the equation $d = rt$ for $r$).
Properties of Equality and Inequality (A.REI.1, A.REI.3)
Solve linear equations and inequalities in one variable by applying the same operation to both sides; multiplying or dividing an inequality by a negative number reverses the inequality symbol.
15 practice questions
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A car gets 28 miles per gallon and gas costs \$3.50 per gallon. Using units to guide the calculation, find the cost to drive 420 miles.
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Answer. \$52.50
Use units to plan the steps: divide miles by miles-per-gallon to get gallons (420 ÷ 28 = 15 gallons), then multiply gallons by dollars-per-gallon to get dollars (15 × 3.50 = \$52.50). Tracking units shows exactly which operation to perform at each step. -
The expression $200 + 15n$ represents the total cost, in dollars, of a gym membership after n months. What does the term $15n$ represent?
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Answer. The total amount paid in monthly fees over n months (before adding the initial \$200 fee).
The constant 200 is the one-time fee, and 15 is the cost per month, so multiplying by n gives the running total of monthly charges alone. The full expression adds this to the fixed fee to get the overall cost. -
An investment's balance after t years is modeled by $P(1+r)^t$, where P is the initial deposit and r is the annual interest rate. What does the factor $(1+r)$ represent?
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Answer. The growth factor applied each year — the balance is multiplied by $(1+r)$ every year because it keeps 100% of its value and gains r more.
Since the whole balance is multiplied by $(1+r)$ once per year (that's what raising to the power t means), $(1+r)$ must be the single-year growth multiplier, combining the original amount (1) with the rate of increase (r). -
Three consecutive integers have a sum of 72. Write and solve an equation to find the integers.
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Answer. 23, 24, and 25.
Let x be the smallest integer, so the integers are x, x+1, x+2. The equation $x+(x+1)+(x+2)=72$ simplifies to $3x+3=72$, so $x=23$, giving 23, 24, 25. -
A company's fixed costs are $500, and it earns $12 profit per unit sold before subtracting fixed costs. Write an equation for total profit P in terms of units sold u.
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Answer. $P = 12u - 500$
Profit equals the per-unit profit times the number of units, minus the fixed cost that must be covered regardless of sales, giving a linear equation in two variables, P and u. -
A student needs an average of at least 85 across four tests. She scored 82, 90, and 79 on the first three. Write and solve an inequality for the minimum score x needed on the fourth test.
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Answer. $x \ge 89$
The average condition gives $\frac{82+90+79+x}{4}\ge 85$. Multiplying both sides by 4 gives $251+x\ge 340$, so $x\ge 89$; she needs at least an 89. -
A theater sells adult tickets for \$10 and student tickets for \$6, and wants at least \$1200 in revenue from a showing. Write an inequality relating the number of adult tickets a and student tickets s that models this goal.
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Answer. $10a + 6s \ge 1200$
Total revenue is the sum of the money from each ticket type, 10 dollars times a adult tickets plus 6 dollars times s student tickets, and this total must meet or exceed the \$1200 goal. -
Using the constraint $10a + 6s \ge 1200$ from the ticket-sales problem, explain why $a = -5$ is not a viable solution even if it makes the inequality true for some value of s.
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Answer. Because a represents a count of tickets sold, it cannot be negative; only nonnegative whole-number values of a (and s) are viable in this context.
Algebra allows negative numbers as solutions to an inequality, but the real-world meaning of the variable restricts which of those solutions actually make sense — this is the difference between an algebraic solution and a viable one. -
Solve for x: $5(2x-3) = 3x+6$
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Answer. $x = 3$
Distribute to get $10x-15=3x+6$. Subtract 3x from both sides: $7x-15=6$. Add 15: $7x=21$. Divide by 7: $x=3$. Checking: $5(2\cdot3-3)=15$ and $3\cdot3+6=15$. -
Solve for x: $\dfrac{x-2}{3}+5=\dfrac{x+4}{2}$
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Answer. $x = 14$
Multiply every term by the least common denominator, 6: $2(x-2)+30=3(x+4)$, giving $2x+26=3x+12$. Subtract 2x: $26=x+12$, so $x=14$. Check: $(14-2)/3+5=9$ and $(14+4)/2=9$. -
Solve for x: $-3(x+2) \ge 4x-27$
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Answer. $x \le 3$
Distribute: $-3x-6\ge 4x-27$. Add 3x to both sides: $-6\ge 7x-27$, then add 27: $21\ge 7x$. Divide by 7 (positive, no flip needed here): $3\ge x$, i.e. $x\le 3$. -
Solve the compound inequality: $-5 < 2x+3 \le 9$
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Answer. $-4 < x \le 3$
Subtract 3 from all three parts: $-8 < 2x \le 6$. Divide all three parts by 2 (positive, so inequality directions stay the same): $-4 < x \le 3$. -
The formula for the perimeter of a rectangle is $P = 2l+2w$. Solve this formula for w.
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Answer. $w = \dfrac{P-2l}{2}$
Subtract $2l$ from both sides to isolate the term with w: $P-2l=2w$. Then divide the entire side $(P-2l)$ by 2, not just one part of it, giving $w=(P-2l)/2$. -
Solve the equation $F=\dfrac{9}{5}C+32$ for C.
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Answer. $C = \dfrac{5}{9}(F-32)$
Subtract 32 from both sides: $F-32=\dfrac{9}{5}C$. Multiply both sides by the reciprocal $\dfrac{5}{9}$ to isolate C: $C=\dfrac{5}{9}(F-32)$. -
Solve $3x-4y=12$ for y in terms of x, and state the slope of the resulting line.
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Answer. $y = \dfrac{3}{4}x-3$; slope $= \dfrac{3}{4}$
Subtract $3x$ from both sides: $-4y=-3x+12$. Divide every term by $-4$: $y=\dfrac{3}{4}x-3$. In slope-intercept form $y=mx+b$, the coefficient of x is the slope, so the slope is $\dfrac{3}{4}$.
What people get wrong
- Forgetting to reverse the inequality symbol when multiplying or dividing both sides by a negative number. Instead, flip the sign every time you multiply or divide by a negative, and check with a test value if unsure.
- Distributing a negative sign incorrectly, for example treating $-3(x+2)$ as $-3x+2$ instead of $-3x-6$. Instead, multiply every term inside the parentheses by the full coefficient, sign included.
- When rearranging a formula, dividing only one term on a side instead of the entire side by the divisor, writing $w = P - 2l/2$ instead of $w = (P-2l)/2$. Instead, treat the whole side being divided as one quantity in parentheses.
- Accepting any algebraic solution to a constraint without checking it makes sense in context, such as a negative number of tickets or a fractional number of people. Instead, always test the solution against the real-world limits of the variable.
- Mixing units within an equation, such as adding minutes and hours directly. Instead, convert all quantities to the same unit before combining them, and let the units of the final answer confirm the setup was correct.
Drill this unit until it sticks
These questions come back on a schedule built from what you get wrong, alongside the rest of Algebra I. Free, and no account needed to start.