2.2 Solving Linear Equations

2.2 Solving Linear Equations

Once the equation exists, AAF asks you to solve it. Solving a linear equation means applying the same operation to both sides until the unknown stands alone. The operations you need on this exam are distribution, combining like terms, moving variable terms to one side, and — when denominators appear — multiplying through by a common denominator. The algebra is first-year, but the CAT items are built to catch sign slips and incomplete distribution, not to reward calculator punching.

/practice/accuplacer-advanced-algebraPractice questions with detailed explanations

A reliable solving order

1. Clear grouping symbols with the distributive property: a(b + c) = ab + ac and a(b − c) = ab − ac.
2. Clear fractions or decimals if they slow you down, by multiplying every term by the LCD (or by a power of 10).
3. Combine like terms on each side separately.
4. Collect variable terms on one side and constants on the other, adding or subtracting the same quantity on both sides.
5. Divide (or multiply) by the coefficient of the unknown.
6. Check by substituting back into the original equation — not into a later, possibly corrupted, line.

If the coefficient of x becomes 0, stop and classify the equation as an identity or a contradiction (below). Do not divide by zero.

Worked example: 3(2x − 4) + 5 = 4x + 7

Distribute first. 3 hits both terms inside the parentheses:

3 · 2x − 3 · 4 + 5 = 4x + 7

6x − 12 + 5 = 4x + 7

Combine the constants on the left: −12 + 5 = −7, so

6x − 7 = 4x + 7.

Subtract 4x from both sides: 2x − 7 = 7.

Add 7 to both sides: 2x = 14.

Divide by 2: x = 7.

Check in the original, not in a simplified middle step: left side 3(2 · 7 − 4) + 5 = 3(14 − 4) + 5 = 3(10) + 5 = 35. Right side 4 · 7 + 7 = 28 + 7 = 35. Both sides match, so x = 7 is the solution.

The trap is the sign on −12. Students who write 3(2x − 4) as 6x − 4 or 6x + 12 never recover. Write the two products separately (6x and −12) before you combine anything else.

Variables on both sides, with a second distribution

Solve 2(3x − 1) − 4 = 5x + 3(x − 6).

Left: 6x − 2 − 4 = 6x − 6.

Right: 5x + 3x − 18 = 8x − 18.

So 6x − 6 = 8x − 18.

Subtract 6x: −6 = 2x − 18.

Add 18: 12 = 2x.

Divide by 2: x = 6.

Check: left 2(18 − 1) − 4 = 2(17) − 4 = 30; right 5(6) + 3(6 − 6) = 30 + 0 = 30.

Notice that you can also collect x on the right. Either direction is legal; pick the side that keeps the coefficient of x positive if that helps you avoid a later sign error.

Clearing fractions

Solve (2/3)x + 1/4 = (1/2)x − 1/6.

The denominators are 3, 4, 2, and 6. The least common denominator is 12. Multiply every term by 12:

12 · (2/3)x + 12 · (1/4) = 12 · (1/2)x − 12 · (1/6)

8x + 3 = 6x − 2.

Subtract 6x: 2x + 3 = −2.

Subtract 3: 2x = −5.

Divide by 2: x = −5/2.

A partial multiply — clearing only the left side, or forgetting the −1/6 — produces a wrong integer that looks plausible. If an on-screen calculator appears, you can still clear fractions by hand faster than you can retype every mixed term. Decimals work the same way: 0.2x + 1.4 = 0.5x − 0.7 becomes 2x + 14 = 5x − 7 after multiplying by 10, then 21 = 3x, so x = 7.

Identities versus contradictions

Sometimes the variable disappears.

Identity (infinitely many solutions). Compare 3(2x − 4) + 5 with 6x − 7:

3(2x − 4) + 5 = 6x − 12 + 5 = 6x − 7.

So 6x − 7 = 6x − 7 is true for every real x. After subtracting 6x you have −7 = −7, a true statement with no variable left. Report infinitely many solutions (all real numbers), not “x = 0” and not “no solution.”

Contradiction (no solution). Compare the same left side with 6x + 1:

6x − 7 = 6x + 1 → subtract 6x → −7 = 1, which is false. Report no solution. The empty set is the solution set.

How AAF may word the options

AAF will not always use the words “identity” or “contradiction.” The options may say “infinitely many solutions,” “no solution,” or a specific value such as x = 7. Read the options before you start so you know whether classification is on the table.

Arithmetic slips the CAT is built to catch

• Dropped signs: −(2x − 5) is −2x + 5, not −2x − 5.
• Combining unlike terms: 6x − 7 is not −x and not −1.
• Dividing only one side: from 2x = 14, both sides divide by 2.
• Dividing by a coefficient but forgetting it applies to a sum: (2x + 6)/2 is x + 3, not x + 6.
• Checking a corrupted line: substitute into the original equation.

If you stall, rebuild from the previous correct line rather than “adjusting” a suspicious constant. Linear solving is reversible; if the check fails, an earlier sign is almost always the cause.

One-variable solving is the mechanical core of Table 11’s linear-equations slice. Inequalities use the same moves with one extra rule about reversing the inequality symbol. Systems (Section 2.4) reuse substitution and elimination on two of these equations at once. Graphing the resulting line is covered in slope-intercepts-graphs.