Linear equations in one variable

Official Digital SAT skill — Algebra domain.

What this question tests

Linear equations in one variable test your ability to find the single value that makes an equation true by isolating the variable using equivalent transformations. On the Digital SAT, this can appear as a direct “solve the equation” item, a multiple-choice “same solution as” item, or a short word problem where you translate a sentence into an equation and then solve it. You will often see equations that require combining like terms, distributing over parentheses, and handling fractions or negatives carefully. This skill is tested because it reflects a core algebra idea: equations represent a balance, and valid operations preserve truth while moving you closer to the unknown. When you can solve these reliably, you can also check your work quickly and avoid common algebra traps that create tempting wrong answers.

The forms you should expect include simple two-step equations, equations with parentheses such as $a(bx+c)=d$, and equations with fractions that require clearing denominators. You may also see comparison language like “how many more” or “difference,” which must be translated precisely into subtraction in the correct order. The goal is not memorizing a special formula, but applying a small set of reliable properties consistently: distribute correctly, combine like terms, and use inverse operations to isolate the variable. A strong habit is to verify by substitution, because it catches sign mistakes and arithmetic slips that otherwise look “algebraically fine.”

What to know

A linear equation in one variable has the form $ax+b=c$ or can be rearranged into that form, and its solution is the value of $x$ that makes both sides equal.
Equivalent equations have exactly the same solution set, and you create them by performing the same valid operation on both sides, such as adding/subtracting the same expression or multiplying/dividing both sides by the same nonzero number.
The Distributive Property is $a(b+c)=ab+ac$, and it also works with subtraction as $a(b-c)=ab-ac$; correct distribution is essential before combining like terms.
Like terms are terms with the same variable part and exponent, so you can combine $3x$ and $-7x$ into $-4x$, but you cannot combine $3x$ and $5$ into a single term.
To isolate a variable, use inverse operations in a logical order: undo addition/subtraction first, then undo multiplication/division, keeping track of signs at each step.
If an equation contains fractions, you can multiply both sides by a common denominator to clear fractions, but you must apply that multiplication to every term on both sides to preserve equivalence.

How to approach it

First, simplify each side of the equation by distributing and combining like terms so you can see the true structure and avoid trying to solve while the expression is still messy.
Next, move all variable terms to one side by adding or subtracting the appropriate term on both sides, because collecting the $x$ terms together makes it possible to factor or isolate $x$ cleanly.
Then move constant terms to the other side by adding or subtracting, because separating constants from variable terms is what turns the equation into a simple one-step isolation.
After that, isolate $x$ by dividing or multiplying by the coefficient of $x$, since the coefficient tells you how the variable is being scaled.
If fractions are present, consider clearing denominators early by multiplying both sides by a common denominator, because it reduces fraction arithmetic errors and makes the remaining steps more straightforward.
Finally, check your solution by substituting it back into the original equation, because a quick verification catches sign errors and distribution mistakes that can produce a plausible but wrong value.
If the variable cancels out and you get a statement like $0=5$, conclude there is no solution, and if you get a true statement like $0=0$, conclude there are infinitely many solutions, because the algebraic outcome tells you whether any value can work.

Common traps

Sign-flip mistakes when moving terms across the equals sign are common because students mentally “move” a term without writing the inverse operation; avoid this by explicitly adding or subtracting the same term on both sides line by line.
Incorrect distribution, such as applying a multiplier to the first term inside parentheses but not the second, happens because parentheses look like a single object; prevent this by marking both terms and distributing to each before combining.
Combining unlike terms, like merging $x$ terms with constants, occurs when students focus on numbers and ignore variable parts; stop this by labeling terms as variable or constant and only combining matching types.
Clearing fractions incorrectly, such as multiplying only one side or only one term by the common denominator, happens under time pressure; avoid it by rewriting the equation with parentheses around each side and applying the multiplier to every term.
Misreading comparison language in word-based equations, especially “how many more” or “difference,” leads to reversed subtraction; avoid it by defining variables and writing the relationship in words first, then translating in the same order.

Tips & shortcuts

If an equation looks fraction-heavy, clear denominators early to turn it into a cleaner linear equation and reduce arithmetic strain.
Write each operation on its own line and keep the equals signs aligned, because neat structure makes mistakes much easier to spot quickly.
Use a fast mental check by plugging your solution into the simplified form you created; if it works there, do one spot-check in the original to confirm no distribution error was introduced.
Watch the coefficient of $x$ carefully at the end; dividing by a negative flips the sign of the solution, and dividing by a fraction is the same as multiplying by its reciprocal.

Worked example

What value of $x$ satisfies $3(x - 4) = \frac{1}{2}(2x + 6) + x$?

A. $x = 18$
B. $x = 7$
C. $x = 9$
D. $x = 15$ ✓ (correct answer)

Why: Expand each side. On the left, $3(x - 4) = 3x - 12$. On the right, $\frac{1}{2}(2x + 6) = x + 3$, so $\frac{1}{2}(2x + 6) + x = (x + 3) + x = 2x + 3$. Set them equal: $3x - 12 = 2x + 3$. Subtract $2x$ from both sides to get $x - 12 = 3$. Add $12$ to both sides to get $x = 15$. Therefore, the correct choice is C.

Use the Practice Questions for this skill to drill it, then attempt a Timed Practice Test.