PERT Algebra Questions: Equations, Inequalities and Word Problems
PERT algebra questions ask you to solve an equation or inequality, or to turn a short situation into one and solve it. They cover linear equations, formulas, inequalities, word problems, slope and lines, systems of two equations and quadratics, all as multiple choice inside the untimed PERT Mathematics subject. The reliable method is the same throughout: write the relationship, solve it one legal step at a time, then check the result in the original condition and confirm it answers the exact question. This guide goes through each family of questions, the traps in the answer choices and two worked examples.
Updated November 5, 2026
PERT Algebra Questions: Equations, Inequalities and Word Problems at a glance
Detail
What to know
Subject
PERT Mathematics, scored from 50 to 150
Questions in the subject
30: 25 scored and 5 field-test; you won’t be told which 5 are unscored, so treat every question as if it counts
Timing
Untimed; testing centers keep their own hours
Navigation
Computer-adaptive; you answer each question to move on and cannot go back
Answer format
Multiple choice; the public study-guide samples show four choices, A to D
Calculator
On-screen four-function calculator on some questions only; no personal calculator unless an approved disability accommodation allows one
Topics in our practice set
Linear equations, formulas, inequalities, word models and percents, slope and lines, systems, quadratics and rational equations
PERT Mathematics has 30 multiple-choice questions. Twenty-five produce your score and five are field-test questions that help build the question bank. You won’t be told which 5 are unscored, so treat every question as if it counts. The test is computer-adaptive, which means each question is chosen from your answers to earlier ones, and you must answer a question before you can move to the next. Once you submit an answer, you cannot go back to change it.
Algebra questions come in a few shapes. Some are bare: solve this equation, or this inequality. Others are one to three sentences about a delivery fee, a phone plan, ticket sales or a rectangle’s perimeter, followed by one question. The answer choices are not always single numbers. In our practice bank they can be intervals such as −2 ≤ x < 5, expressions for y such as 3x − 7, solution sets such as {1/2, −4}, or statements about how many solutions an equation or system has. Reading what kind of answer the choices hold tells you what kind of result to produce.
Wrong choices are usually built from a specific slip: the number you get by ignoring a fixed fee, the value of the other variable in a system, or the right boundary with the inequality pointing the wrong way. Finding your number among the choices does not prove it is right.
You can’t bring your own calculator unless an approved disability accommodation allows one; an on-screen calculator appears only on some math questions. The study guide describes it as a standard four-function calculator that performs operations in the order you enter them: on a calculator that works in the order you enter the keys, 2 + 3 × 4 gives 20, not 14. Practice the algebra by hand and use the tool, when it appears, for arithmetic only.
What they test
The questions on this page share one demand: write or read a relationship, then find the values that keep every original condition true. Most of our 75 practice questions fall into seven families:
Linear equations, including parentheses, fractions and decimals, plus equations that turn out true for every number or for none
Literal equations: solving a formula for a named letter
Linear inequalities and compound inequalities joined by and or or, with endpoints and integer counts
Word models: a fixed fee plus a rate, consecutive integers, perimeters, comparing two plans, and percents as multipliers
Slope, intercepts and equations of lines, including parallel and perpendicular lines
Systems of two linear equations, solved by substitution or elimination, and count-and-value word problems
Quadratic equations solved by factoring, square roots or completing the square, and rational equations with excluded values
How to solve PERT algebra questions
Linear equations: simplify, isolate, then check in the original
Clean up each side before moving anything across the equals sign. Distribute, combine like terms, and only then collect the variable on one side. In 4(3x − 1) = 5x + 17, distributing gives 12x − 4 = 5x + 17. Subtract 5x and add 4 to get 7x = 21, so x = 3. The check goes into the original equation, parentheses and all: 4(9 − 1) = 32 and 15 + 17 = 32.
A number in front of parentheses multiplies every term inside, including its sign. In 9 − 3(x + 2) = 2x − 7, the −3 turns x + 2 into −3x − 6, so the left side is 3 − 3x. Adding 3x and 7 to both sides gives 10 = 5x, and x = 2. Writing −3x + 6 instead leads to a different equation.
Fractions and decimals are easier to clear than to carry. Multiply every term on both sides by a common denominator. For (x + 2)/3 + (x − 1)/2 = 4, multiplying by 6 gives 2(x + 2) + 3(x − 1) = 24. Notice that the 4 on the right becomes 24 as well. That simplifies to 5x + 1 = 24, so x = 23/5; a fraction answer is normal here. For 0.3x + 1.5 = 0.1x + 2.7, multiply by 10 to get 3x + 15 = x + 27, so x = 6.
Sometimes the variable cancels. Then read what is left. In 4(x − 1) = 4x − 4, both sides match, so every real number works. In 3(x + 1) = 3x − 2, you reach 3 = −2, which is false, so nothing works. Formulas follow the same steps: treat the other letters as numbers. To solve y = m + nx for x, subtract m and divide the whole difference by n, giving x = (y − m)/n when n is not 0. If the target letter appears in two terms, gather them and factor it out before dividing.
Inequalities: flip the sign only for a negative, and test a value
Solve an inequality the way you solve an equation, with one extra rule: multiplying or dividing both sides by a negative number reverses the sign. In −3x + 4 ≥ 19, subtracting 4 gives −3x ≥ 15, and dividing by −3 gives x ≤ −5. Then test a value on each side of the boundary: x = −6 gives 22 ≥ 19, which is true, and x = 0 gives 4 ≥ 19, which is false.
Endpoints matter. A strict sign (< or >) leaves the boundary out, and an inclusive sign (≤ or ≥) keeps it in. On a number line that is an open circle versus a filled one. If a question asks for the greatest integer n with 3n − 2 < 17, the inequality gives n < 19/3, about 6.33, and the answer is 6. Going up to 7 breaks the original condition, since 3(7) − 2 = 19.
For a chained inequality, do the same thing to all three parts. In −4 ≤ 3x + 2 < 17, subtract 2 throughout to get −6 ≤ 3x < 15, then divide by 3 to get −2 ≤ x < 5. Two separate conditions joined by and keep only the overlap: x + 1 > 4 and 3x < 18 give 3 < x < 6. Joined by or, they keep either region: x − 3 ≤ −5 or x + 2 ≥ 6 gives x ≤ −2 or x ≥ 4, and the stretch between them satisfies neither.
Harder versions hide the sign of the coefficient in a letter. If b > 5, then 5 − b is negative, so solving (5 − b)x < b means dividing by a negative: x > b/(5 − b). Decide the sign of what you are dividing by before you divide.
Word problems: name the unknown, attach units, and finish the right way
Write what the letter stands for, with its unit, before you write the equation. A taxi ride with a $4 pickup fee plus $1.75 per mile, costing $25 in all, becomes 4 + 1.75m = 25 with m in miles. Subtract the fee once and divide: 1.75m = 21, so m = 12. Dividing 25 by 1.75 skips the fee and gives about 14.3 miles. The same habit sets up other models. Three consecutive even integers are x, x + 2 and x + 4. A rectangle whose length is 3 more than twice its width has width w and length 2w + 3, so a 42-centimeter perimeter gives 2w + 2(2w + 3) = 42 and w = 6. If the question asks for the length, keep going: 2(6) + 3 = 15.
When you weigh two plans, the equal point is a boundary, not the whole answer. A gym’s Plan A costs $24 plus $3 per class and Plan B costs $7 per class. Setting 24 + 3c = 7c gives c = 6, where both cost $42. Below 6 classes Plan B is cheaper (at 4 classes, $36 against $28); above 6, Plan A is. If a question adds conditions, write each as its own inequality. Plan A staying at or under $60 means 24 + 3c ≤ 60, so c ≤ 12. Saving at least $8 over Plan B means 7c − (24 + 3c) ≥ 8, so c ≥ 8. Anywhere from 8 to 12 classes satisfies both.
Treat percents as multipliers on a base you can name. A 15% discount leaves 85%, so a $60 jacket sells for 0.85 × 60 = $51. A price after 7% tax is 1.07 times the original, so an $85.60 total came from 85.60 ÷ 1.07 = $80; taking 7% off $85.60 uses the wrong base. A 10% increase followed by a 10% decrease multiplies by 1.10 and then 0.90, leaving 99% of the start. A percent increase compares the change with the original: growth from 50 to 60 is 10/50 = 20%. For a map scale of 2 centimeters to 15 kilometers, find the unit rate, 7.5 kilometers per centimeter, so 6 centimeters stand for 45.
Slope and lines: keep the order, then pin down the intercept
Slope is the change in y divided by the change in x, with both subtractions in the same order. Through (−4, 7) and (2, −5), the slope is (−5 − 7)/(2 − (−4)) = −12/6 = −2. Here y drops as x grows, so a positive answer would signal a sign slip. Two points with the same y, such as (1, −3) and (6, −3), give slope 0 and a horizontal line. Two points with the same x, such as (−5, 1) and (−5, 8), give a vertical line, x = −5, whose slope is undefined.
To write a line from a slope and a point, put the point into y = mx + b and solve for b. A slope of 3 through (2, −1) gives −1 = 6 + b, so b = −7 and y = 3x − 7. The point’s y-coordinate is the intercept only when its x-coordinate is 0. Lines in standard form need rearranging: 4x − 3y = 12 becomes −3y = 12 − 4x, and dividing every term by −3 gives y = (4/3)x − 4. For an intercept alone, set the other variable to 0. In 3x + 4y = 24, x = 0 gives the y-intercept 6, and y = 0 gives the x-intercept 8.
Parallel lines share a slope. Perpendicular lines have slopes that multiply to −1, so you flip the fraction and change the sign: 3/4 becomes −4/3. A line perpendicular to y = 2x − 5 has slope −1/2, and through (4, 3) that gives 3 = −2 + b, so y = −x/2 + 5. Vertical and horizontal lines are their own pair: the line through (−1, 4) perpendicular to x = 6 is y = 4.
Systems: pick the faster method and check both equations
If one equation already has a variable alone, substitute. With y = 3x − 2 and x + y = 18, replacing y gives 4x − 2 = 18, so x = 5 and y = 13. If the coefficients line up, eliminate. For 3x + 2y = 19 and 2x − 3y = −9, multiply the first equation by 3 and the second by 2, add, and get 13x = 39, so x = 3 and y = 5. Check the pair in both original equations: 3(3) + 2(5) = 19 and 2(3) − 3(5) = −9. A pair that works in only one equation is not a solution.
Word systems usually pair a count with a value. A school concert takes in $240 from 25 tickets, adults at $12 and children at $7. Then a + k = 25 and 12a + 7k = 240. Subtracting 7 times the count equation from the money equation gives 5a = 65, so 13 adult tickets and 12 children’s tickets. With coins, work in cents to avoid decimals.
Some questions ask how many solutions a system has. Compare whole equations, constants included. In 2x − y = 3 and 6x − 3y = 9, the second is exactly three times the first, so they are the same line and every point on it works. In 2x − y = 3 and 4x − 2y = 7, doubling the first gives 4x − 2y = 6, which conflicts with 7, so the lines are parallel and never meet. Lines with different slopes meet exactly once, and two lines never meet in exactly two points.
Quadratic and rational equations: set up zero, and screen every candidate
The zero-product property works only when a product equals zero. In 2x² + 7x = 4, move the 4 first: 2x² + 7x − 4 = 0 factors as (2x − 1)(x + 4) = 0, so x = 1/2 or x = −4. When a question asks for the larger solution of x² − 10x + 21 = 0, factor to (x − 3)(x − 7) and choose 7. The 10 in the equation is the sum of the roots, not a root.
Taking a square root gives two signs. From (x + 1)² = 25, x + 1 = 5 or −5, so x = 4 or x = −6. When the quadratic will not factor neatly, complete the square: x² + 6x + 4 = 0 becomes (x + 3)² = 5, so x = −3 ± √5. The same form counts solutions: x² − 10x + c = 0 becomes (x − 5)² = 25 − c, which has exactly one solution when c = 25.
With fractions that have x in the denominator, list the forbidden values before you clear anything. In (2x − 1)/(x − 3) = 5/(x − 3), x cannot be 3. Multiplying by x − 3 gives 2x − 1 = 5, so the only candidate is 3, which is forbidden, and the equation has no solution. In 2/x + 1/(x − 2) = 1, exclude 0 and 2, then clear denominators to get x² − 5x + 4 = 0, so x = 1 or x = 4. Neither is excluded, and both check: 2 − 1 = 1 and 1/2 + 1/2 = 1.
Common mistakes
Six slips that the answer choices in this question type are built to catch, and the habit that prevents each one.
Mistake: Solving −3x + 4 ≥ 19 and writing x ≥ −5.
Better: Reverse the sign when you divide by −3: x ≤ −5. Then test −6 and 0 in the original.
Why: Dividing by a negative flips the order of the number line. The boundary −5 is right in both versions, which is why this slip is so easy to miss.
Mistake: Expanding 9 − 3(x + 2) as 9 − 3x + 6.
Better: Multiply both terms by −3: 9 − 3x − 6, which is 3 − 3x.
Why: The −3 applies to everything in the parentheses, so the 2 becomes −6. Losing that sign changes the whole equation.
Mistake: Dividing $25 by $1.75 to find the miles on a ride that also has a $4 pickup fee.
Better: Subtract the fee once, then divide: (25 − 4) ÷ 1.75 = 12 miles.
Why: The fee is charged once, not per mile. Writing 4 + 1.75m = 25 before calculating keeps the two parts separate.
Mistake: Using −1, the y-coordinate of the point (2, −1), as the intercept of a line with slope 3.
Better: Substitute the point into y = 3x + b: −1 = 6 + b, so b = −7.
Why: The intercept is the value of y where x = 0. A point with x = 2 is two steps away from the y-axis.
Mistake: Solving a system for x and choosing that value when the question asks for y.
Better: Finish both coordinates, check the pair in both equations, then reread which one is asked for.
Why: The other coordinate is almost always one of the choices, so a correct calculation can still land on the wrong letter.
Mistake: Accepting x = 3 as the solution of (2x − 1)/(x − 3) = 5/(x − 3).
Better: List x ≠ 3 first. The only candidate is forbidden, so there is no solution.
Why: Clearing denominators can produce a value that makes the original fractions undefined. Only the original equation decides what counts.
Working without a clock
No PERT subject has an official time limit. In 2015, FDOE reported that the Mathematics subtest took about 30 minutes on average, and colleges’ own estimates range from about two hours to more than three for all three subjects, and our PERT test dates and registration guide lists them by college. Check your testing center’s hours before you go.
Without a clock, the risk shifts from rushing to over-checking. Because you cannot return to a question after you submit it, do one deliberate check before you click: put your answer into the original equation, test one value for an inequality, or confirm both equations in a system. Miami Dade College notes that you can change an answer before you move on, so this is the moment for second thoughts. Then commit. Reworking the same steps three times rarely finds anything new and drains focus you need later.
Two questions from our practice bank: a plan comparison that rewards checking both costs, and a harder inequality that flips, has a strict end and an extra condition. Each has a tempting wrong choice.
Example 1 · Medium · Building and comparing linear models
Plan A costs $18 plus $4 per visit. Plan B costs $6 per visit. At how many visits do the plans have equal cost?
A.3
B.6
C.9
D.18
Answer: C (9)
How to solve it
Name the unknown: let v be the number of visits. Plan A costs 18 + 4v dollars, with the $18 charged once. Plan B costs 6v dollars.
Equal cost means 18 + 4v = 6v. Subtract 4v from both sides to get 18 = 2v, so v = 9.
Check in both plans: 18 + 4(9) = 54 and 6(9) = 54. The costs match, so the answer is C, 9.
A, 3, comes from dividing the $18 fee by Plan B’s $6 rate, which ignores that Plan A also charges $4 per visit. D, 18, is the fee itself, not a number of visits.
Why it works: Set 18 + 4x = 6x. Subtract 4x to get 18 = 2x, so x = 9. Both plans then cost 54 dollars. Dividing 18 by 6 neglects the per-visit charge in Plan A.
Example 2 · Harder · Compound inequalities
How many integers x satisfy −6 < −2x ≤ 4 and x ≠ 0?
A.3
B.4
C.5
D.6
Answer: B (4)
How to solve it
Solve the chain −6 < −2x ≤ 4 by dividing all three parts by −2. Dividing by a negative reverses both signs: 3 > x ≥ −2.
Rewrite it smallest first so it reads naturally: −2 ≤ x < 3. The −2 is included and the 3 is not.
List the integers in that range: −2, −1, 0, 1 and 2. The extra condition x ≠ 0 removes 0, leaving 4 integers, choice B.
Check the edges in the original: x = −2 gives −2x = 4, which meets ≤ 4; x = 3 gives −6, which fails −6 < −2x. C, 5, is the count you get if you forget to remove 0.
Why it works: Dividing the chain by −2 reverses both comparisons: 3 > x ≥ −2. Thus the integer candidates are −2, −1, 0, 1, 2. The additional restriction removes 0, leaving 4 values. Including 3 would violate the strict lower comparison on −2x.
Practice algebra and mathematical models questions
Here are 20 more algebra and mathematical models questions from our practice bank. The set is untimed. Submit it when you are done to see the answer and a full explanation for every question. Unlike the real PERT, this practice set lets you move back and forth between questions. For mixed sets from every part of the test, use the question bank.
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Lessons on this question type
Each lesson teaches one idea with a worked example, then gives you a question to try.
Expressions, polynomials and factoring: Expression questions test whether two forms of an expression mean the same thing. Work with signed numbers and fractions, exponents, substitution, distribution, polynomial arithmetic, factoring and rational expressions. A correct simplification keeps the original meaning, including values that make a denominator zero.
The study guide’s Mathematics outline lists equations (linear, quadratic and literal), evaluating expressions, polynomials, dividing by monomials and binomials, coordinate planes and pairs of simultaneous linear equations. Our algebra practice questions cover equations and formulas, inequalities, word problems with fees, rates and percents, slope and lines, systems, and quadratic and rational equations.
You can’t bring your own calculator unless an approved disability accommodation allows one; an on-screen calculator appears only on some math questions. The study guide describes it as a four-function calculator that works in the order you enter operations.
It is untimed, so there is no official limit. FDOE reported in 2015 that Mathematics took about 30 minutes on average, and colleges’ own estimates for all three subjects run from about two hours to more than three.
Florida’s college-ready (standard) score for PERT Mathematics is 114 on the 50 to 150 scale, set by State Board of Education Rule 6A-10.0315. Colleges set their own course charts above that. At Santa Fe College, for example, 114 to 122 may qualify you for a college-level math elective, and 123 or higher requires additional placement testing. Ask your college which score your planned math course uses.
Math has 30 questions, 25 scored and 5 field-test. The test is adaptive, so the mix depends on your answers, and the sources we rely on give no count by question type.
No. No official conversion from a number of correct answers to a PERT score is published, and our questions are practice items, not real PERT questions. Use your results to find which families of questions need more work.