The published assessment targets are specific, and knowing them narrows your study list considerably. GED math algebra requires you to solve one-variable linear equations with rational coefficients, including ones needing the distributive property or collection of like terms. It requires solving real-world problems involving linear equations, and writing one-variable and multi-variable equations to represent a context. It requires solving a system of two simultaneous linear equations by graphing, substitution, or linear combination. It also covers linear inequalities in one variable and quadratic equations with real solutions. That is the whole list — it is shorter than most study guides imply, and none of it asks you to recall a formula the test does not already supply.
These targets come from GED Testing Service's own Assessment Guide for Educators, which is the document test items are actually written against. It also confirms that you get an on-screen Mathematics Formula Sheet, so what matters is understanding what an equation represents and knowing which operation moves you toward the answer. Our GED math cheat sheet condenses those operations onto one page.
When an algebra question looks complicated, break it into three stages. Translate: turn the words or given information into an equation. Solve: isolate the unknown by applying the same operation to both sides. Check: substitute your answer back and confirm it answers what the question actually asked. The method scales — it works on a one-line equation like 2x + 1 = 9 and it works on a multi-sentence word problem with two unknowns. Most missed algebra items fail at stage one or stage three rather than stage two, which is worth knowing, because stage two is the part people spend all their practice time on.
Once you have the equation, work toward getting the variable alone. Take 2x + 1 = 9. Subtract 1 from both sides to get 2x = 8, then divide both sides by 2 to get x = 4. The reason you apply every operation to both sides is that an equation is a balance — whatever you do to one side must happen to the other for the statement to stay true. This is worth internalising rather than memorising as a rule about "moving things across the equals sign", because the balance framing is what keeps you correct once equations run to four or five steps.
Try 3(x + 2) − 4 = 17. The parentheses come first, so distribute the 3 across everything inside: 3x + 6 − 4 = 17. Combine like terms: 3x + 2 = 17. Subtract 2 from both sides: 3x = 15. Divide both sides by 3: x = 5. Now check it by substituting back into the original: 3(5 + 2) − 4 becomes 3(7) − 4, which is 21 − 4 = 17. The equation holds, so x = 5 is correct.
Some GED math linear equations need an extra move because the variable sits on both sides. Take 4x + 3 = 2x + 15. Subtract 2x from both sides to gather the variable terms: 2x + 3 = 15. Subtract 3: 2x = 12. Divide by 2: x = 6. Check it — 4(6) + 3 = 27 and 2(6) + 15 = 27, so both sides agree. The rule is simply to move the variable terms to one side first, then isolate as normal.
Word problems feel harder than bare equations because you have to build the equation before you can solve it. The sequence is fixed: identify the unknown, pull out the numbers that matter, work out how they relate, write the equation, solve it, then answer the question that was asked. Not every word problem needs a variable — "Maria has $12 and earns $8 each week; how much after 5 weeks?" is just 12 + (8 × 5) = 52. The goal is representing the situation accurately, not forcing algebra onto arithmetic.
A gym charges a $20 registration fee plus $15 per month. James paid $95 in total. How many months did he pay for? Let m be the number of months. The fixed fee is 20, the monthly cost is 15m, and the total is 95, giving 20 + 15m = 95. Subtract 20: 15m = 75. Divide by 15: m = 5. James paid for five months. Every term in that equation came directly from a phrase in the problem, which is what a clean translation looks like.
A taxi company charges a $4 starting fee plus $2.50 per mile. A second charges $7 plus $2 per mile. At how many miles are the fares equal? Let m be miles. The first is 4 + 2.50m, the second is 7 + 2m, and "equal" means you set them against each other: 4 + 2.50m = 7 + 2m. Subtract 2m: 4 + 0.50m = 7. Subtract 4: 0.50m = 3. Divide by 0.50: m = 6. Verify both — 4 + 2.50(6) = 19 and 7 + 2(6) = 19. At six miles both fares are $19.
Suppose that same taxi problem instead asked what the customer pays at six miles. You would still solve to m = 6, but 6 is a number of miles, not a fare. Substituting gives 4 + 2.50(6) = 19, so the answer is $19. The variable you solved for is frequently an intermediate step rather than the answer, and the wrong-but-plausible value will be sitting in the options. Solve, reread, then answer.
A system is two equations sharing the same unknowns, and the solution is the pair of values that makes both true at once. Assessment target A.2.d requires solving these by graphing, substitution, or linear combination, plus real-world problems that lead to a system. On the GED they nearly always arrive as a two-quantity scenario — two ticket types, two plans, two sizes — where you can write one equation for the count and one for the total. Recognizing that shape is most of the work; once both equations are written correctly, the algebra is ordinary. Substitution is usually the fastest of the three permitted methods on a timed test, because the count equation rearranges to a single variable in one step. Graphing is the slowest and is best reserved for items that hand you the graph already. The tell that you are looking at a system is simple: the problem gives you two different totals about the same group of things.
Adult tickets cost $8 and student tickets cost $5. Twenty tickets sold for $130 total. How many of each? Let a be adult tickets and s be student tickets. The count gives a + s = 20. The money gives 8a + 5s = 130. Rearrange the first to s = 20 − a and substitute into the second: 8a + 5(20 − a) = 130. Distribute: 8a + 100 − 5a = 130. Combine: 3a + 100 = 130. Subtract 100: 3a = 30, so a = 10. Then s = 20 − 10 = 10. Ten adult and ten student tickets — and checking, 8(10) + 5(10) = 130.
Work these before reading the solutions. Each one targets a different failure point — pure procedure, translation, variables on both sides, and a relationship you have to build before you can solve. If you miss one, note which stage of the three-step method broke down rather than just noting the wrong answer, because that diagnosis is what tells you which drill to do next. The distractors matter as much as the answers here. Every wrong option on a real GED algebra item is the result of a specific plausible mistake — stopping a step early, dropping a fixed fee, mishandling a variable term — rather than a random number. Learning to recognize your own error inside the option list is a genuine test-taking skill, and it is why the solutions below name what each trap option represents instead of only confirming the correct letter.
Options: A) 3 B) 5 C) 7 D) 25. Answer: B. Add 7 to both sides to get 5x = 25, then divide both sides by 5 to get x = 5. Option D is what you get by stopping one step early, which is the trap.
A plan costs $30 per month plus a one-time $15 activation fee. What does the customer pay after 4 months? Options: A) $120 B) $135 C) $150 D) $165. Answer: B. The monthly cost is 30 × 4 = 120, then add the one-time fee: 120 + 15 = 135. Option A is the answer to a question that forgot the activation fee.
Solve 3x + 8 = x + 20. Options: A) 4 B) 6 C) 8 D) 14. Answer: B. Subtract x from both sides: 2x + 8 = 20. Subtract 8: 2x = 12. Divide by 2: x = 6. Option D is 20 − 8 + 2, the result of mishandling the variable term.
A rectangle's length is 4 inches greater than its width, and its perimeter is 28 inches. What is the width? Let w be the width, so the length is w + 4. Perimeter gives 2w + 2(w + 4) = 28. Distribute: 2w + 2w + 8 = 28. Combine: 4w + 8 = 28. Subtract 8: 4w = 20. Divide by 4: w = 5 inches. This one is harder than it looks because the relationship between length and width has to be translated before any solving starts.
Strong algebra performance comes from recognizing the structure behind a problem, not from memorising isolated tricks. The most useful habit when you miss a question is diagnosing which stage broke rather than simply reading the correct solution. Each failure mode has a different fix, and applying the wrong fix is why people practice for weeks without moving.
Could not understand the wording? Drill translating phrases into expressions, not solving.
Wrote the wrong equation? Label every quantity before you calculate anything.
Right equation, could not solve it? Review inverse operations and the balance principle.
Solved for the variable but picked the wrong option? Reread the final sentence — every time.
Arithmetic slip? Slow down before reaching for the calculator, not after.
You get an on-screen TI-30XS MultiView for the calculator-permitted items and may bring your own handheld TI-30XS to a test center — see can you use a calculator on the GED for the exact rules. For pacing across the whole 115-minute test rather than algebra alone, how to pass the GED math test builds the wider plan.
