An equation states that two expressions have the same value. Solving keeps that equality true while isolating the unknown. A clear first equation often saves more time than faster arithmetic on a poorly chosen setup.
Suggested pace: 10 minutes on examples · 10–15 minutes on questions · 5 minutes on review. Take a second session if a method needs more work.
Keep in mind: Perform the same operation on both sides, then verify in the original statement.
Method 1 of 4
Give the unknown a name and a unit. Separate a one-time fixed amount from an amount charged or produced repeatedly. “Three more than twice x” is 2x + 3; “three times the sum of x and 2” is 3(x + 2). Parentheses preserve that difference.
A rental costs $9 plus $4 per hour. The total is $37. How many hours were rented?
Answer: 7 hours
Check it: The hourly charge is $4/hour × 7 hours = $28. Adding the $9 fixed fee gives the stated $37 total.
A $5 fee plus $3 per hour totals $20. Write an equation and find the hours.
5 + 3h = 20, so 3h = 15 and h = 5 hours. Check: 5 + 3 × 5 = 20.
Common mistake: Dividing 37 by 4 before removing the fixed fee treats the fee as another hourly charge.
Method 2 of 4
Distribution multiplies every term inside a pair of parentheses. Once parentheses are gone, combine terms with the same variable and power. Then move variable terms to one side and constants to the other using equal operations.
Solve 3(x − 2) + 4 = 2x + 9.
Answer: x = 11
Check it: Left side: 3(11 − 2) + 4 = 31. Right side: 2(11) + 9 = 31.
Expand 2(x − 3). Which terms get multiplied by 2?
Both terms: 2x − 6. The factor outside the parentheses applies to the whole expression inside.
Common mistake: Writing 3(x − 2) as 3x − 2 distributes to only one term. The entire parenthesized expression is multiplied by 3.
Method 3 of 4
Multiplying every term on both sides by a common denominator turns a fractional equation into an equivalent equation with simpler arithmetic. Multiplication must reach each term, including any whole-number term.
Solve x/3 + x/2 = 10.
Answer: x = 12
Check it: 12/3 + 12/2 = 4 + 6 = 10.
Multiply every term of x/2 + 1 = 4 by 2. What equation results?
x + 2 = 8, so x = 6. Multiplying only the fraction would change the equality.
Common mistake: Multiplying only the left side by 6 changes the equation. The 10 must become 60 as well.
Method 4 of 4
An inequality compares order, not equality. Adding the same amount preserves order; multiplying or dividing by a negative reverses it. Test a value from your proposed solution set and keep track of whether the endpoint is included.
Solve −2x + 5 < 13.
Answer: x > −4
Check it: x = 0 gives 5 < 13, which is true. At x = −4 the sides are both 13, so the boundary is excluded.
Solve −3x > 12. Does x = −5 satisfy the result?
x < −4. Dividing by −3 reverses the sign, and x = −5 works because 15 > 12.
Common mistake: Reversing the sign when merely adding or subtracting a negative is unnecessary. The reversal is tied to multiplication or division by a negative.
Keep one operation per line until the process is reliable. Checking the original equation can catch distribution, sign, and transcription errors together.
Work the first six questions with the explanations closed. Review any missed or guessed answers before continuing with the other six. If you cannot set up a problem, return to its method above and try again from a blank page.
Work through one question at a time. Explain your approach, check the worked answer, then move on. Use scratch paper for calculations.
0 of 12 checked · 0 correct.
Progress is saved in this browser. These original study questions do not produce an official AFQT score.
Answer: B. −2
The filled endpoint includes −2, whereas the open endpoint excludes 3. The shaded set is −2≤x<3. Of the four choices, only −2 is in that set; −3 and 4 lie outside it.
Answer: C. −2
Distribute to get −6x + 8 ≤ 20, then subtract 8: −6x ≤ 12. Dividing by −6 reverses the inequality, giving x ≥ −2. Therefore −2 is the least permitted integer. Two also satisfies the inequality but is not the least, while −3 and −4 do not satisfy it.
Answer: B. −3
Distribute to obtain 6x − 3 = 6x + k. Subtracting 6x leaves −3 = k, so k must be −3. At that value the two sides match for every real x. The 3 choice loses the minus sign, while k = 0 would omit the constant term entirely.
Answer: A. 5
The positive factor pair 7 and 12 has both the required product, 84, and the required sum, 19. Their positive difference is 12 − 7 = 5. Another method uses (a − b)² = (a + b)² − 4ab = 19² − 4(84) = 25, so |a − b| = 5. The choices 7 and 12 give the individual numbers rather than their difference.
Answer: A. 0
A simultaneous solution would be a point on both lines. Distinct parallel lines do not intersect, so there is no common ordered pair. Each individual line has many points, but none belongs to both.
Answer: A. −1.6
Substitute −2 for x: −0.6 + b = −1.4 − 0.8 = −2.2. Adding 0.6 to both sides gives b = −1.6. The constant −0.8 is already present in the equation and does not by itself balance the different x terms.
Answer: D. 26
Write the integers as n and n + 2. Then n² + 2n − 168 = 0, which factors as (n − 12)(n + 14) = 0. Positivity selects n = 12, so the integers are 12 and 14 and their sum is 26. The choices 12 and 14 are individual integers rather than the requested sum.
Answer: A. 6
The inequality gives −7/2 < x − 1/2 < 7/2, so −3 < x < 4. The integers are −2, −1, 0, 1, 2, and 3: six values. The strict inequality excludes both −3 and 4. Counting one or both endpoints incorrectly gives seven or eight.
Answer: A. vt − q
Multiply by t to get vt = p + q, then subtract q: p = vt − q. The plus-q choice undoes the addition in the wrong direction. Multiplying (v − q) by t also multiplies q unnecessarily, and dividing v by t repeats the original division instead of undoing it.
Answer: B. 6
The first inequality gives x < 8. The second gives x ≥ 2. Their intersection contains the integers 2, 3, 4, 5, 6 and 7, for a total of 6. Including 8 ignores the strict first inequality; excluding 2 ignores the inclusive second inequality.
Answer: C. 4
Doubling the first equation gives 6x + 4y = 14. When k = 4, the second equation has exactly the same left side but requires it to equal 9, which is impossible. For each other listed k, subtracting the doubled first equation leaves a nonzero coefficient of y, so the equations have one solution. A zero determinant alone would not distinguish no solution from identical equations; the constants differ here.
Answer: B. 2
Let the smaller integer be n. Then 1/n + 1/(n + 1) = 5/6. Multiplication by 6n(n + 1) gives 5n² − 7n − 6 = 0, or (5n + 3)(n − 2) = 0. The positive integer solution is n = 2, and 1/2 + 1/3 = 5/6 checks it. The other root, −3/5, is neither positive nor an integer; 3 is the larger integer, not the requested one.
A remembered choice letter is different from a method you can reconstruct. Keep one correction in your error log and use it on your next attempt.