How to solve SAT linear equations in one variable

Linear equations in one variable are equations like $4(x - 3) = 2x + 10$, where the only unknown is $x$ and it is never squared or raised to any other power. On the Digital SAT they show up as plain equations to solve, as word problems you have to translate, and as questions about how many solutions an equation has. Algebra is about 35% of the Math section, or 13 to 15 of the 44 questions, and you can expect roughly 2 to 4 of those to test this skill. They are usually among the faster points on the test, which makes careless errors here expensive.

What the questions look like

Solve the equation. You get an equation with parentheses, fractions, or decimals and must find $x$. The harder versions bury the variable on both sides or put it inside several fractions.

Translate a situation. A short scenario describes a fixed amount plus a rate, such as a phone plan that costs $12 per month plus $0.15 per minute. You either pick the equation that models it or solve for the unknown quantity.

Count the solutions. The equation contains a constant like $k$, and the question asks for the value of $k$ that gives no solution or infinitely many solutions. Some versions ask how many solutions a given equation has.

A variation on the first format asks for an expression instead of $x$. If $3x - 5 = 10$, the question might ask for $6x - 10$. You can double both sides and get $20$ without ever finding $x$.

Step-by-step method

1. Clear fractions first. Multiply every term on both sides by the least common denominator, including whole-number terms that have no denominator.
2. Distribute and combine. Expand parentheses, then combine like terms on each side separately.
3. Collect the variable. Move all $x$ terms to one side and all constants to the other.
4. Divide by the coefficient of $x$.
5. Reread the question. Check whether it asks for $x$, for $2x + 1$, or for some other quantity.
6. For solution-count questions, simplify both sides to the form $ax + b = cx + d$ and compare:
   • If $a \neq c$, there is exactly one solution.
   • If $a = c$ and $b \neq d$, there is no solution.
   • If $a = c$ and $b = d$, there are infinitely many solutions.

Worked example

Example 1. What value of $x$ satisfies $\dfrac{2x + 1}{3} - \dfrac{x - 4}{2} = 5$?

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

$$2(2x + 1) - 3(x - 4) = 30$$

Distribute carefully. The $-3$ multiplies both the $x$ and the $-4$:

$$4x + 2 - 3x + 12 = 30$$

Combine like terms: $x + 14 = 30$, so $x = 16$.

Check it in the original equation. The first fraction is $\frac{33}{3} = 11$ and the second is $\frac{12}{2} = 6$. Since $11 - 6 = 5$, the answer $x = 16$ is correct.

Example 2. In the equation $4(3x - 2) - 2x = kx + 7$, $k$ is a constant. For what value of $k$ does the equation have no solution?

Simplify the left side: $12x - 8 - 2x = 10x - 8$. The equation is now

$$10x - 8 = kx + 7$$

For no solution, the $x$ coefficients must match while the constants differ. If $k = 10$, the equation becomes $10x - 8 = 10x + 7$. Subtracting $10x$ from both sides leaves $-8 = 7$, which is never true. So $k = 10$.

Notice that no value of $k$ gives infinitely many solutions here, because the constants $-8$ and $7$ can never match. If the question had asked for infinitely many solutions, you would know a choice was wrong just by checking constants.

Common traps

• Dropping a negative when distributing. $-3(x - 4)$ equals $-3x + 12$, not $-3x - 12$. When a minus sign sits in front of parentheses, write out the distributed terms on their own line before combining.
• Clearing fractions on only some terms. If you multiply the fractions by 6 but leave the 5 on the right side alone, you get the wrong equation. Multiply the whole-number terms too.
• Answering for $x$ when the question wants something else. The SAT often lists the value of $x$ as a wrong choice when it asks for $x + 3$ or $2x$. Underline the final quantity in the question before you start.
• Mixing up no solution and infinitely many. Same coefficient with different constants means no solution. Same coefficient and same constant means every $x$ works. If you get a true statement like $5 = 5$ after the variables cancel, the answer is infinitely many.
• Putting the rate and the fixed fee in the wrong places. In a word problem, the amount that is charged per item gets multiplied by the variable. The one-time amount stands alone. For the phone plan above, the cost for $m$ minutes is $12 + 0.15m$, not $12m + 0.15$.

Desmos tip

Type the left side as $y = \dfrac{2x+1}{3} - \dfrac{x-4}{2}$ and the right side as $y = 5$. Click the intersection point, and its $x$-coordinate is the solution. This also works for solution-count questions. Two parallel lines mean no solution, and a single line drawn on top of itself means infinitely many. For an equation with $k$, type the right side with $k$ in it and Desmos will offer a slider. Drag it until the lines become parallel.