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Equations, Inequalities and Quadratics on the GED Math Test

Solving an equation on the GED means finding the number that makes a statement true, and the test asks for it in every shape it can think of: a one-step equation you could solve by looking at it, a multi-step one with brackets and the variable on both sides, an inequality, a pair of equations solved together, a quadratic with two answers, and a formula from the sheet turned around to isolate a different letter. Most arrive as a story about a taxi fare, a phone bill or a rectangular patio, and writing the equation is half the question.

Updated September 21, 2026 · Lessons by Coach Crissy

Questions per set
10
Skills covered
10
Score-report category
Algebraic problem solving with expressions and equations · 30% of the test

Equations and inequalities: 10 questions

Questions
10
Timer
None
No calculator
None
  • Check each answer as you go and read the working before the next question.
  • New numbers every time, so a second run tests the skill rather than your memory.
  • The formula sheet and calculator sit in the toolbar, as they do on the real test.

No account or email needed.

Where do the marks go? Rarely on the arithmetic. They go on a term carried across the equals sign with its sign unchanged, on an inequality divided by a negative number without the sign being flipped, on a quadratic given one of its two solutions, and on a formula where the 2 was divided into one side only. In an April 2026 announcement, GED Testing Service listed solving for variables among the topics learners ask its AI math tutor about most.

This topic is the second half of "algebraic problem solving with expressions and equations", the reporting category GED Testing Service weights at about 30%, the largest of the four; expressions and polynomials are the other half. Thirty percent of about 46 questions is 13 or 14 items shared between the two, so six or seven equation questions is a reasonable guess, and that is our arithmetic on the published weight rather than an official count. The algebraic half of the test as a whole is weighted at about 55%, so nothing here is optional.

Skills in this set: Solving two-step equations · Solving one-step equations · Solving equations with the variable on both sides · Solving equations with fractions or decimals · Solving a linear inequality · Writing and solving an equation from a situation · Solving a quadratic equation by factoring · Modeling a situation with a quadratic and solving it · Solving a system of two linear equations · Rearranging a formula to isolate a variable

What the test asks

The Assessment Guide for Educators covers this topic with indicators A.2.a through A.4.b. A.2.a is solving linear equations in one variable, including ones that need the distributive property or combining like terms, and ones with letters as coefficients, which is where rearranging a formula lives. A.2.b and A.2.c are the word-problem versions: solving a real-life problem that leads to a linear equation, and writing the equation that models a situation. A.2.d is a pair of simultaneous linear equations, solved by graphing, substitution or elimination. A.3.a through A.3.d do the same jobs for one-variable inequalities, including showing the solution set on a number line. A.4.a is solving a quadratic with real roots by factoring, the quadratic formula, completing the square or inspection, and A.4.b is writing the quadratic that models a situation.

The contexts are money and measurement. A repair shop charges a call-out fee plus an hourly rate and the bill is known; a savings account grows by the same amount every week toward a target; two passes cost the same after a certain number of visits; a patio is 5 feet longer than it is wide with a known area, which is where the quadratics come from. Fill-in-the-blank items ask you to type the solution, drag-and-drop items can ask you to put the lines of a solution in order, and a number-line item can ask where the solutions of an inequality begin.

Plenty of this can appear without a calculator. The opening section of about five questions is taken before the on-screen TI-30XS is available, and a pure equation like 3(x − 2) = x + 8 is a natural candidate because the arithmetic is light by design. Our practice set marks the symbol-only equations, inequalities, systems and factored quadratics as no-calculator items and keeps the calculator for the word problems and the quadratic formula. Two things the test does not ask, according to a presentation by a GED Testing Service board member: graphing a quadratic and inequalities in two variables.

The mistakes this set is built to catch

Each wrong answer in the practice set is the result of one specific slip, and the explanation names it. These are the ones that come up most in this topic.

1Dividing an inequality by a negative without flipping the sign

Solving −4x + 3 ≥ 15 goes exactly like an equation until the last step: subtract 3 to get −4x ≥ 12, then divide by −4. Dividing or multiplying by a negative number reverses the inequality, so the answer is x ≤ −3, and x ≥ −3 is the wrong answer waiting in the choices. Adding, subtracting and dividing by a positive number never flip anything, so the rule fires at that one moment only. Test a value from your answer in the original: x = −5 gives 20 + 3 = 23, and 23 ≥ 15 holds.

2Giving a quadratic only one solution

If x² − 3x − 10 = 0 factors as (x − 5)(x + 2) = 0, either bracket can be the zero, so x = 5 or x = −2. Two choices are built to catch a half-finished solution: one lists only x = 5, and the other lists −5 and 2, the numbers inside the brackets with their signs unflipped. A quadratic with two real roots has two solutions until the context throws one away, which a width or a number of hours will do when one solution is negative. Check both by substituting them back.

3Dividing only part of a side

Rearranging P = 2l + 2w for w means subtracting 2l from both sides and then dividing everything that is left by 2, giving w = (P − 2l) ÷ 2. The slip is w = P ÷ 2 − 2l, where the 2 only reached the P. The quadratic formula sets the same trap at its last step, because the whole of −b ± √(b² − 4ac) is divided by 2a, and forgetting that gives an answer exactly twice the right size. Whenever you divide, draw a line under the entire side first.

4Moving a term across without changing its sign

In 5x + 7 = 2x − 8 the 2x comes across as −2x and the 7 as −7, giving 3x = −15 and x = −5. Carry the 2x over as +2x and you get 7x = −15 and a fraction that should tell you something is wrong. Systems lean on the same move: solving x + 3y = 11 for x gives x = 11 − 3y, and 11 + 3y is the version that produces the wrong ordered pair. Say the operation, subtract 2x from both sides, rather than picturing terms as jumping the equals sign.

A worked example

A climbing gym offers two ways to pay. The Basic plan costs $50 to set up and then $14 for each visit. The Flex plan costs $20 to set up and then $20 for each visit. After how many visits will the two plans have cost the same amount?

  1. Write an expression for each plan

    Let v be the number of visits. Each setup fee is paid once and each per-visit price is paid v times, so the cost of a plan is its setup fee plus its rate times v.

    Basic: 50 + 14v; Flex: 20 + 20v

  2. Set the two costs equal

    The question asks when the plans cost the same, so the two expressions are equal at the number of visits you are looking for.

    50 + 14v = 20 + 20v

  3. Collect the v-terms on one side and the numbers on the other

    Subtract 14v from both sides so the variable appears only on the right, then subtract 20 from both sides so the plain number appears only on the left. Both moves keep the equation balanced.

    50 = 20 + 6v, so 30 = 6v

  4. Divide, then check in the original

    Dividing both sides by 6 gives v = 5. Substitute 5 back into both original expressions: the Basic plan costs 50 + 70 = 120 dollars and the Flex plan costs 20 + 100 = 120 dollars, so the answer holds.

    v = 30 ÷ 6 = 5; check: 50 + 14(5) = 120 and 20 + 20(5) = 120

Answer: 5 visits (both plans cost $120)

How to practice it

  1. On a multiple-choice equation, try the choices. There are four, and only one makes the original equation balance, so substituting each one is a legitimate way to answer and a fast way to check work you have already done. For a quadratic, the correct choice lists two numbers and both must work; for a word problem, put the choice back into the story and see whether the total comes out right.
  2. Write down what is being done to the variable, then undo it in reverse order. In (x + 6)/3 = 4 the x had 6 added and was then divided by 3, so multiply by 3 first and subtract 6 second, giving x = 6. The same reading turns a formula around: in P = 2l + 2w the w is doubled and then has 2l added, so subtract 2l before dividing by 2.
  3. Solve every inequality twice: once with the algebra and once with a test number. After you get x ≤ −3, pick something inside that range, such as −5, and put it into the original inequality. If it comes out false, you either flipped a sign you should not have or failed to flip one you should. The number test takes seconds and does not depend on remembering the rule under pressure.
  4. For a quadratic, get everything on one side equal to zero and write a, b and c with their signs before anything else. If the left side factors, set each bracket to zero and expect two answers. If it does not, the formula on the sheet needs those three numbers, and the sign of c is where it usually goes wrong. Finish by throwing out the answer the story cannot use and substituting the one you keep.

Frequently asked questions

Is the quadratic formula on the GED formula sheet?

Yes. The formula sheet gives x = (−b ± √(b² − 4ac)) ÷ 2a and, just above it, the standard form y = ax² + bx + c. What the sheet cannot do is tell you which number is a, b or c, keep the sign of c straight, or remind you that the ± means two answers. Practice the formula on equations you can also solve by factoring, so you know what a correct pair of answers looks like.

Does the GED care which method I use to solve an equation?

No. The math test has no short-answer or extended-response items, so nobody sees your working, and the indicator for quadratics explicitly allows factoring, the quadratic formula, completing the square or inspection. The same goes for systems, where graphing, substitution and elimination are all named. Use the method you trust, and on multiple-choice items feel free to test the choices instead of solving from scratch.

Are systems of equations on the GED math test?

Yes. Indicator A.2.d covers a pair of simultaneous linear equations solved by graphing, substitution or elimination, and says the pair can come dressed as a real situation, such as two plans that cost the same after a certain number of months. When the answer is an ordered pair, a wrong choice with the two numbers swapped is a standard distractor in our set, so keep x first.

Do I have to graph inequalities on the GED?

Only on a number line. Indicator A.3.b asks you to identify or draw the solution set of a one-variable inequality on a number line, which the test can deliver as a hot-spot item where you click the right place. Inequalities in two variables, the kind shaded on a coordinate grid, are on the list of topics a GED Testing Service board member has said the test does not assess.

Sources

Figures on this page were checked against these sources on the dates shown.

  1. Assessment Guide for Educators: Mathematical Reasoning · GED Testing Service · checked September 13, 2026
  2. Mathematics Formula Sheet (2026-02 revised) · GED Testing Service · checked September 13, 2026

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