Solve linear equations in one variable, including those requiring the distributive property and those with variables on both sides, and identify equations with one solution, no solution, or infinitely many (TEKS A.5A).

What this topic is asking

Solving one-variable linear equations is the most heavily sampled skill in the most heavily weighted reporting category. TEKS A.5A asks you to solve equations that may need the distributive property and may have variables on both sides, and to recognize the three solution types: one solution, no solution, or infinitely many. Fluency here secures a large block of points and underpins everything that follows.

The inverse-operations routine

Every linear equation is solved by the same sequence, working operations in reverse of the order of operations.

1. Distribute to clear any parentheses.
2. Combine like terms on each side separately.
3. Move the variable to one side (add or subtract a variable term).
4. Move the constants to the other side.
5. Divide by the coefficient of the variable.

For $3(x + 4) = 21$: distribute to $3x + 12 = 21$, subtract 12 to get $3x = 9$, divide to get $x = 3$.

Distributing correctly

The distributive property multiplies the outside factor by every term inside the parentheses. The classic error is hitting only the first term.

A negative outside factor flips every inside sign: $-2(x - 4) = -2x + 8$. Watch this sign change carefully, because it is a frequent STAAR distractor.

Variables on both sides

When the variable appears on both sides, collect it on one side first, usually the side that keeps the coefficient positive.

One solution, no solution, or infinitely many

After simplifying, the variable terms may cancel, and the leftover statement tells you the solution type.

• One solution. A single value of $x$ remains, as in $5x = 15 \Rightarrow x = 3$.
• No solution. The variables cancel and a false statement remains, as in $2x + 1 = 2x + 5 \Rightarrow 1 = 5$ (never true).
• Infinitely many. The variables cancel and a true statement remains, as in $3(x + 2) = 3x + 6 \Rightarrow 6 = 6$ (always true).

Recognizing these is itself an assessed skill, often as an inline-choice item. The test is whether the two sides become identical (infinite) or contradictory (none) once simplified.

How STAAR examines this topic

• Multiple choice. Solve an equation with distribution and variables on both sides; "distribute to first term only" and sign-error distractors are standard.
• Number entry and equation editor. Enter the solution value, or the equation after a step.
• Inline choice. Classify the number of solutions, and state why (sides equal or contradictory).

A clarifying idea is that solving keeps the equation balanced: whatever you do to one side you do to the other, so the truth of the equation is preserved at every step, which is why substituting your answer back must reproduce a true statement.

Clearing fractions to simplify

When an equation contains fractions, multiplying every term by the least common denominator first turns it into a whole-number equation that is far easier to solve cleanly. For $\frac{x}{3} + \frac{1}{2} = \frac{5}{6}$, the least common denominator is 6: multiplying every term by 6 gives $2x + 3 = 5$, so $2x = 2$ and $x = 1$. The key discipline is multiplying every term, including those without a fraction, by the same denominator; multiplying only the fractional terms unbalances the equation. This clearing step is optional but it removes a common source of arithmetic slips, especially on number-entry items where the final value must be exact rather than a rounded decimal.

Try this

Q1. Solve $2(3x - 1) = 4x + 8$. [1 point]

• Cue. $6x - 2 = 4x + 8 \Rightarrow 2x = 10 \Rightarrow x = 5$.

Q2. How many solutions does $3x + 5 = 3x - 2$ have? [1 point]

• Cue. Variables cancel, $5 = -2$ false, so no solution.