Algebra Basics and Equations: Practice Questions & Explanations

Linear equation solving: isolate the variable. Step 1: subtract 7 from both sides: 3x + 7 − 7 = 22 − 7, giving 3x = 15. Step 2: divide both sides by 3: 3x ÷ 3 = 15 ÷ 3, giving x = 5. Verify: 3(5) + 7 = 15 + 7 = 22 ✓. The two-step linear equation is the foundation of algebra. Strategy: undo operations in reverse order of operations — first reverse addition/subtraction, then reverse multiplication/division. Many ASVAB MK problems use this structure. Common errors: dividing before subtracting; sign errors; arithmetic errors. Always verify by substituting the answer back into the original equation. Linear equations in one variable always have exactly one solution unless the variable terms cancel.

Two approaches both work. Approach 1 — distribute first: 2(x − 4) = 14 becomes 2x − 8 = 14; add 8: 2x = 22; divide by 2: x = 11. Approach 2 — divide first: divide both sides by 2: x − 4 = 7; add 4: x = 11. Verify: 2(11 − 4) = 2(7) = 14 ✓. When the variable appears once inside parentheses with a coefficient, dividing first is often cleaner. When the equation involves multiple variable terms, distributing first is necessary. Order of operations matters during solving just as in evaluation. Parentheses with subtraction are a common source of sign errors — distribute carefully.

Distribute each coefficient through its parentheses, watching signs. 4(2x + 3) = 8x + 12. −2(x − 5) = −2x + 10 (note: −2 times −5 = +10). Combine: 8x + 12 − 2x + 10. Combine like terms: (8x − 2x) + (12 + 10) = 6x + 22. The most common error is sign mistakes when distributing a negative coefficient — −2(x − 5) is −2·x + (−2)·(−5) = −2x + 10, NOT −2x − 10. Always distribute the sign with the coefficient. Like terms (terms with the same variable raised to the same power) can be combined; constants are like terms with each other; x terms are like terms with each other.

System of two equations in two unknowns. Elimination method (often easiest for ASVAB): add the equations to eliminate y. (x + y) + (x − y) = 10 + 4, giving 2x = 14, so x = 7. Substitute back: 7 + y = 10, so y = 3. Verify both: 7 + 3 = 10 ✓ and 7 − 3 = 4 ✓. Substitution method: solve first equation for y: y = 10 − x; substitute into second: x − (10 − x) = 4; simplify: 2x − 10 = 4; solve: x = 7; then y = 3. Both methods give the same answer. Elimination works well when adding or subtracting eliminates a variable (coefficients are opposites or equal); substitution works when one equation is easily solved for a variable. On multiple-choice tests, you can also test answers by substituting — efficient for systems.

Two-step linear equation. Add 9 to both sides: 4y − 9 + 9 = 11 + 9, giving 4y = 20. Divide both sides by 4: 4y ÷ 4 = 20 ÷ 4, giving y = 5. Verify: 4(5) − 9 = 20 − 9 = 11 ✓. The algorithm: undo operations in reverse order of operations. The original expression performs (1) multiply variable by 4, then (2) subtract 9. To solve, reverse: first add 9 (undo subtraction), then divide by 4 (undo multiplication). For more complex equations with variables on both sides: first move all variables to one side (by adding/subtracting), then constants to the other, then solve. Example: 3x + 5 = 2x − 8 → x + 5 = −8 → x = −13. Variables in equations represent unknown quantities; solving identifies the value(s) that satisfy the equation.

Solve inequalities like equations, but reverse the inequality sign when multiplying or dividing by a negative. Step 1: add 5: 2x > 12. Step 2: divide by 2 (positive, so don't flip): x > 6. Verify: try x = 7: 2(7) − 5 = 9 > 7 ✓; try x = 5: 2(5) − 5 = 5, not > 7 ✗. Inequality symbols: (1) > greater than (strict); (2) < less than (strict); (3) ≥ greater than or equal; (4) ≤ less than or equal. Number line representation: open circle for strict (>, <); closed circle for ≥, ≤; arrow indicates direction. Critical rule: multiplying or dividing both sides by a negative REVERSES the inequality. Example: −2x > 10 becomes x < −5 (divide both sides by −2, flip the >). Adding/subtracting and multiplying/dividing by positives does NOT flip. Compound inequalities: 'and' (intersection, both must hold) vs 'or' (union, either holds). Example: x > 2 AND x < 5 means 2 < x < 5.

3x = 24 - 9 = 15. x = 15 ÷ 3 = 5. Two-step linear equation: undo the addition first (subtract 9 from both sides), then undo the multiplication (divide both sides by 3). Check by substituting back: 3(5) + 9 = 15 + 9 = 24 ✓. The distractors come from slipping a step — 4 and 6 come from dividing 24 by the wrong number (24 ÷ 6, 24 ÷ 4) and 3 is 9 ÷ 3.

Difference of squares: a² - b² = (a-b)(a+b). Here: x² - 9 = x² - 3² = (x-3)(x+3). Recognise the pattern: a square term minus a perfect-square constant with no middle term is always a difference of squares. The distractors fail when expanded: (x − 3)(x − 3) = x² − 6x + 9 and (x + 3)(x + 3) = x² + 6x + 9 both have a middle term; (x − 9)(x + 1) = x² − 8x − 9. Expanding your chosen factors (FOIL) is the fastest check on the test.

f(7) = 2(7) - 3 = 14 - 3 = 11. Function notation f(7) just means 'replace every x in the rule with 7, then evaluate'. Follow order of operations: multiply first (2 × 7 = 14), then subtract 3. The distractor 17 adds 3 instead of subtracting; 8 is 7 + 1 or 2 × 7 − 6; 13 would be f(8). Writing the substitution out, 2(7) − 3, prevents sign slips.

Distribute: 2x + 8 = 22. Subtract 8: 2x = 14. Divide by 2: x = 7. Either distribute first, as shown, or divide both sides by 2 straight away: x + 4 = 11, so x = 7 — the second route is quicker when the number outside the bracket divides the other side evenly. Check: 2(7 + 4) = 2 × 11 = 22 ✓. The distractor 9 comes from forgetting to multiply the 4 by 2 (2x + 4 = 22), 5 from a subtraction slip, and 6 from dividing 22 by 2 and then subtracting 5 instead of 4.

Slope = (y₂-y₁)/(x₂-x₁) = (11-3)/(6-2) = 8/4 = 2. Slope is rise over run: the change in y divided by the change in x between any two points on the line. Keep the subtraction order consistent in both numerator and denominator ((11 − 3) over (6 − 2)); reversing only one of them gives −2, the wrong sign. The distractor 4 comes from using the change in x alone, 3 from a subtraction slip, and 1 from assuming equal changes.

Add 5 to both sides: 2x > 14. Divide by 2: x > 7. Solve an inequality exactly like an equation, with one exception: if you multiply or divide both sides by a negative number, flip the inequality sign. Here you divide by positive 2, so the sign stays as '>'. Check with a value: x = 8 gives 2(8) − 5 = 11 > 9 ✓, while x = 6 gives 7, which is not greater than 9, ruling out the distractors x > 6, x > 4 and x > 2.

Use the DISTRIBUTIVE property: 4(x + 3) = 4x + 12. Then subtract 2x: 4x + 12 − 2x = 2x + 12. ASVAB Mathematics Knowledge tests simplifying expressions. Strategy: (1) distribute the 4 across the parentheses: 4·x + 4·3 = 4x + 12; (2) combine like terms: 4x − 2x = 2x, and keep the +12. Result: 2x + 12.

Common error: forgetting to distribute the 4 to BOTH terms in the parentheses, or trying to combine 12 with the x-terms (you can't combine unlike terms).

If x² = 49, then x = ±7 (both 7 AND −7), because 7² = 49 AND (−7)² = 49. ASVAB Mathematics Knowledge tests square roots and equations.

Key concept: when solving x² = a number, there are usually TWO solutions — a positive and a negative root — because squaring a negative number gives a positive result.

Common error: giving only the positive root (7). The square root symbol √ alone denotes the positive root, but solving the EQUATION x² = 49 yields both +7 and −7. Check: (−7)² = (−7)(−7) = 49. ✓

To solve a linear equation, isolate x. Start with 3x − 7 = 14. Add 7 to both sides: 3x = 21. Then divide both sides by 3: x = 7. The strategy is to undo operations in reverse order — first undo the subtraction (add 7), then undo the multiplication (divide by 3). You can check by substituting back: 3(7) − 7 = 21 − 7 = 14. ✓ Linear equations are a staple of Mathematics Knowledge, and the reliable method is to perform the same operation on both sides until x stands alone.

First distribute or divide. Dividing both sides by 2: x + 3 = 8. Then subtract 3: x = 5. Alternatively, distribute first: 2x + 6 = 16, then 2x = 10, so x = 5. Either order works. Check: 2(5 + 3) = 2(8) = 16. ✓ Equations with parentheses can be solved by distributing the multiplier or by dividing both sides by it first; dividing first is often quicker when the right side is divisible. The value of x is 5.

Solve an inequality much like an equation: subtract 5 from both sides of x + 5 > 12 to get x > 7. The solution is all values greater than 7. Note that the inequality sign stays the same when adding or subtracting; it only flips when you multiply or divide both sides by a negative number (which doesn't happen here). So the answer is x > 7. Inequalities are common on Mathematics Knowledge, and the steps mirror solving equations, with that one rule about flipping the sign for negative multiplication or division.