2.4 Expressions, equations and inequalities

Introduction

This is the largest single category on the test, about 30% of it, and for many adult candidates the least familiar. If school algebra ended for you around "solve for x" and never explained what x was doing there, slow down here.

Start with what algebra is for. Arithmetic answers a question about particular numbers: this many hours at this rate. Algebra answers a question about a relationship holding for any numbers: what happens to the total when the hours change, or what the hours must be to reach a target. A variable is a placeholder for a quantity that varies, and an equation is a claim that two expressions have the same value.

The skill the test rewards is translation: turn a paragraph about a rental agreement into symbols, operate on them, and translate the result back into a sentence about the rental agreement. The middle step can be learned mechanically; the two ends are where the thinking is, and where DOK level 2 items live.

Expressions: building them and simplifying them

An expression combines numbers, variables and operations with no equals sign: 3x + 7, 2(l + w). An equation sets two expressions equal. The distinction is not pedantic: you evaluate an expression and you solve an equation, and a question asking you to "write an expression for the total cost" is not asking you to solve anything.

Translating English into symbols

Most of the work is a small vocabulary, and the entries worth care are the ones people reverse.

"Seven less than x" reverses the order you read it in, and the last two rows differ only in where the parentheses sit — which changes the answer for every value of x. Name the quantity with its units — "let h = hours worked in week one" — so you do not solve for a number and then wonder what it counts.

Combining like terms and distributing

Like terms have the same variable to the same power. 4x and 7x combine to 11x; 4x and 4x² do not, and neither do 4x and 4y — four boxes plus seven crates is just four boxes and seven crates.

The distributive property says a(b + c) = ab + ac — the multiplier reaches every term inside. Two failures are common and both are expensive: distributing to only the first term, so that 3(2x − 5) becomes 6x − 5, and mishandling a negative multiplier, since −2(x − 4) is −2x + 8, not −2x − 8. Subtracting a parenthesized expression is distribution by −1, so (5x + 3) − (2x − 7) is 3x + 10. Every sign inside flips.

Exponent rules

Four rules cover this level, and each is just counting factors: xᵃ · xᵇ = xᵃ⁺ᵇ; xᵃ / xᵇ = xᵃ⁻ᵇ, because factors cancel in pairs; (xᵃ)ᵇ = xᵃᵇ; and x⁰ = 1 for nonzero x. A negative exponent means a reciprocal: x⁻² = 1/x². It does not mean a negative number, and that misreading is a standard distractor.

Solving linear equations

An equation is a balance: whatever you do to one side you must do to the other. Solving means peeling operations off the variable in reverse order of operations — undo addition and subtraction first, then multiplication and division, like unwrapping a parcel.

The procedure, in order: clear parentheses by distributing; clear fractions by multiplying every term on both sides by the common denominator; combine like terms on each side; move variable terms to one side and constants to the other; divide by the coefficient; and substitute your answer back into the original equation.

That last step is why algebra is safer than arithmetic — your answer is checkable, exactly. On a multiple-choice item you can also work backward, substituting each choice until one balances; on a fill-in-the-blank item you must solve forward and then check. When variables appear on both sides, move them to whichever side keeps the coefficient positive, which eliminates a class of sign errors for free.

Sometimes the variables cancel entirely. A true statement left over, like 7 = 7, means every number is a solution — the two sides were the same expression in disguise. A false one, like 7 = 3, means no number works. Neither means you made a mistake, and items are written to see whether you know that.

Rearranging formulas: the skill the formula sheet demands

Here the sheet stops helping. It gives you A = ½h(b₁ + b₂); suppose you have the area, the height and one base and need the other base. Nobody hands you a rearranged version.

Rearranging uses the same moves as solving — the other letters just come along for the ride. Treat the letter you want as "the variable" and every other letter as "a number whose value you happen not to know." If the target is multiplied by something, divide both sides by it: I = Prt solved for r is r = I/(Pt). If the target is added to something, subtract it: y = mx + b solved for b is b = y − mx.

There is a second route, easier for many candidates: substitute the numbers first, then solve the resulting simple equation. Use the general rearrangement when the item asks for a formula, and substitute-then-solve when it asks for a number.

Inequalities

An inequality says one side is larger, or at least as large, rather than equal. They are the language of constraints — budgets, capacities, weight limits — which is why they appear in workforce contexts constantly. "At most" and "no more than" are ≤; "at least" and "no less than" are ≥; "more than" and "exceeds" are strict >.

You solve one exactly like an equation, with one rule that has no equivalent in equation solving: multiplying or dividing both sides by a negative reverses the direction of the sign. The reason is visible on a number line — 3 is less than 5, but −3 is greater than −5. Note the precise condition: the flip comes from multiplying or dividing by a negative, not from a negative merely appearing. Solving x + 7 < 2 gives x < −5 with no flip.

The answer is a range, and in context it usually needs one more translation. If a mileage constraint solves to m ≤ 428.7, the practical answer is 428. Read the context to decide whether the boundary itself is allowed and whether fractional values make sense.

Quadratics

A quadratic equation has a squared variable and no higher power: y = ax² + bx + c in standard form. Squared relationships describe real things — areas, projectile heights, some revenue models. Before anything else, get the equation into the form ax² + bx + c = 0. Every method assumes it, and the most common error is applying a method to an equation that still has terms on both sides.

Factoring works when the numbers are friendly. For a leading coefficient of 1, find two numbers that multiply to c and add to b, then use the fact that makes factoring worth anything: if a product is zero, one of the factors must be zero.

The quadratic formula works always, and it is on your sheet: x = (−b ± √(b² − 4ac))/(2a). Identify a, b and c with their signs and compute the part under the root first. Errors cluster around signs: b² is positive even when b is negative, −4ac becomes a plus when a and c have opposite signs, and −b becomes positive when b is negative — in 2x² − 7x + 3 = 0, b is −7, so −b is +7.

Applied quadratics almost always produce two solutions of which only one makes sense. A width of −8 feet is arithmetically valid and physically meaningless — discard it explicitly and say why, since that discarding is part of what a reasoning-focused item is grading.

Worked examples

Example 1 — variables on both sides. Solve 3(2x − 5) + 4 = 5x + 1.

• Distribute: 6x − 15 + 4 = 5x + 1.
• Combine on the left: 6x − 11 = 5x + 1.
• Subtract 5x: x − 11 = 1, so x = 12.

Check: Substitute into the original. Left: 3(24 − 5) + 4 = 61. Right: 5(12) + 1 = 61. Balanced.

Example 2 — rearranging a formula from the sheet. A trapezoidal bed has area 162 sq ft, height 9 ft, and one parallel side of 22 ft. Find the other parallel side.

• Setup: A = ½h(b₁ + b₂) with A = 162, h = 9, b₂ = 22.
• Rearrange: multiply by 2 → 2A = h(b₁ + b₂); divide by h → 2A/h = b₁ + b₂; subtract b₂ → b₁ = 2A/h − b₂.
• Substitute: b₁ = 324/9 − 22 = 36 − 22 = 14 ft.

Check: Run the formula forward with b₁ = 14: ½ × 9 × (14 + 22) = 162. The substitute-first route lands in the same place: 162 = 4.5(b₁ + 22) → 36 = b₁ + 22 → b₁ = 14.

Example 3 — an inequality with a negative coefficient. Solve 5 − 3x ≥ 17.

• Subtract 5: −3x ≥ 12.
• Divide by −3 and reverse the sign: x ≤ −4.

Check: Test three values. At x = −4: 5 + 12 = 17, and 17 ≥ 17 is true, so the boundary belongs. At x = −5: 20 ≥ 17, inside. At x = −3: 14 is not ≥ 17, outside. The direction is right.

Example 4 — a budget constraint. A van rents for $49.95 per day plus $0.35 per mile. With $200 for a one-day rental, what is the greatest whole number of miles?

• Setup: let m = miles. 49.95 + 0.35m ≤ 200.
• Subtract 49.95: 0.35m ≤ 150.05, so m ≤ 428.71… and the answer is 428 miles.

Check: Price both ends. At 428 miles: 149.80 + 49.95 = $199.75, within budget. At 429: 150.15 + 49.95 = $200.10, over. Rounding 428.71 up is the trap — the round-up rule from chapter 2.3 is for things you must buy enough of, not for a ceiling you must stay under.

Example 5 — factoring. Solve x² + 3x − 40 = 0.

• Already in standard form, so find two numbers multiplying to −40 and adding to +3.
• The factor pairs of 40 are 1 and 40, 2 and 20, 4 and 10, 5 and 8. The pair 5 and 8 differs by 3, so take +8 and −5.
• (x + 8)(x − 5) = 0, so x = −8 or x = 5.

Check: For x = 5: 25 + 15 − 40 = 0. For x = −8: 64 − 24 − 40 = 0. Both satisfy it.

Example 6 — a quadratic in context. A rectangular garden is 3 feet longer than twice its width and has an area of 65 square feet. Find its dimensions.

• Setup: let w = width. Length = 2w + 3, so w(2w + 3) = 65.
• Set to zero: 2w² + 3w − 65 = 0.
• Formula with a = 2, b = 3, c = −65: discriminant 9 + 520 = 529, and √529 = 23.
• w = (−3 ± 23)/4, giving w = 5 or w = −6.5. A width cannot be negative, so width 5 ft, length 13 ft.

Check: 5 × 13 = 65 square feet, and 13 is 3 more than twice 5. Both conditions hold.