Linear equations in one variable
These are the most common questions on the SAT Math section. Most are a few steps of careful algebra — the points are usually lost to small mistakes, misreading what's asked, or missing a "no solution" setup.
What College Board tests
Solving an equation with one variable, often after expanding parentheses or clearing fractions. Many questions set the equation inside a real-world situation, ask for the value of an expression rather than the variable, or test whether an equation has one solution, no solution, or infinitely many.
The method
Every linear equation is solved by the same sequence. Do the steps in order and most of these become routine.
One, none, or infinitely many
When the variable terms are identical on both sides, the variable cancels. What's left tells you the answer: a true statement like means infinitely many solutions; a false statement like means no solution.
Four worked examples in SAT format. Read the approach, try it yourself, then tap Show the full solution.
1 · Solve within a situation
A technician charges a flat fee of $35 for a house call, plus $8 for each quarter-hour of work. The total charge for one visit was $131. The equation represents this situation, where is the number of quarter-hours worked. How many quarter-hours did the technician work?
Approach The flat fee is added on once, and the is attached to the variable. Peel them off in reverse order: first undo the thing being added (the 35), then undo the thing being multiplied (the 8).
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Answer: B
The rate "$8 per quarter-hour" is really a fraction. The word "per" means "for each one," so the denominator is 1 quarter-hour — the cost of a single quarter-hour:
Multiplying by the 12 quarter-hours worked, the "quarter-hour" unit cancels — just like :
That leaves 96 dollars of labor; adding the $35 flat fee gives the $131 total.
Why the other choices are wrong
A Divides 131 by 8 and ignores the flat fee entirely.
C Adds the 35 instead of subtracting it before dividing.
D Results from a setup error while isolating .
2 · Find the value of an expression
If , what is the value of ?
Approach Notice what's being asked: the value of , not . That whole expression is sitting inside the parentheses, multiplied by 3. So you don't need to find — just undo the multiplication by 3.
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Answer: B
The expression equals 11. Recognizing that the question wanted the whole expression saved the work of solving for and substituting back.
Why the other choices are wrong
A Keeps going to solve — the value of , not .
C Adds 3 to the result instead of recognizing the expression is already found.
D Results from distributing the 3 incorrectly before dividing.
3 · One, none, or infinitely many
In the equation , is a constant. For what value of does the equation have no solution?
Approach "No solution" happens when both sides have the same variable term but different constants — the variable cancels and leaves something impossible. So make the -terms match by setting equal to the coefficient on the other side, then check what's left.
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Answer: C
Why the other choices are wrong
A Gives , which has exactly one solution.
B Gives , which has one solution.
D Matches a constant, not the coefficient; the equation still has one solution.
4 · Translate a percentage into an equation
After a 20% discount, the price of a jacket is $52. What was the original price, in dollars, before the discount?
Approach Taking 20% off means you keep 80%. So the sale price is 80% of the original — write that as an equation with the original price as the unknown, then solve.
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Answer: C
Why the other choices are wrong
A Takes 20% of 52 and subtracts — discounting the sale price a second time.
B Adds 20% of 52 to 52, applying the percent to the wrong base.
D Uses the wrong divisor; , not 72.