Educerie

Educerie · SAT · Math

Algebra · ALG.1 Linear equations in one variable

Where it is examined
anywhere in either module. Algebra is ≈35% of the Math section — 13 to 15 questions — and this is the skill the rest of the domain is built on.
The question this unit answers
one unknown, one equals sign. How do you solve it quickly, and what is the test really asking when it wants the number of solutions instead of the solution?
Before you start
you need to be able to distribute a bracket and add the same thing to both sides. Everything else here is built from those two moves.

What you must be able to do

You must be able toWhat it looks like on the test
Solve a linear equation, including with fractionsIf 3(x − 4) = 5x + 2, what is the value of x?
Solve for an expression, not just the variable…what is the value of 2x + 1?
Say when an equation has no solution…for what value of a does the equation have no solution?
Say when it has infinitely manyBoth sides identical after simplifying
Turn a sentence into an equationMost of the hard questions in this unit are word problems

1The idea in one paragraph

A linear equation in one variable has exactly one solution unless the variable disappears. Solve it by clearing brackets and fractions, gathering the variable on one side and the numbers on the other, and dividing. The only genuinely SAT-specific ideas are the two ways the variable can disappear: if the sides end up identical, every number works; if they end up contradicting each other, none does. When a question asks about the number of solutions, it is asking you to compare the coefficients of x, not to solve.

Clear fractions first by multiplying every term by the lowest common denominator. A question with fractions is usually easy underneath and is set that way to cost you time.

2The order that keeps you out of trouble

  1. Multiply out brackets.
  2. Multiply through by the LCD to clear fractions.
  3. Collect the x terms on the side that keeps x positive.
  4. Collect the numbers on the other side.
  5. Divide.
3(x − 4) = 5x + 2
3x − 12 = 5x + 2multiply out the bracket
−12 − 2 = 5x − 3xx terms right, numbers left
−14 = 2x
x = −7

3No solution and infinitely many

Write both sides in the form (number)x + (number).

IfThen
The x coefficients differOne solution
Same x coefficient, different constantsNo solution — the lines are parallel
Same x coefficient, same constantInfinitely many — it is the same equation twice
4(x + a) = 4x + 9
4x + 4a = 4x + 9the x terms are identical
4a = 9so the constants decide it

No solution when 4a ≠ 9; infinitely many when a = 9/4. The test asks this constantly, and it takes five seconds once you stop solving and start comparing.

4It asks for an expression more often than for x

If 2x − 7 = 11, what is the value of 4x − 14?

4x − 14 = 2(2x − 7)the expression asked for is twice the one given
2x − 7 = 11
4x − 14 = 2 × 11 = 22

The slow route solves for x = 9 and substitutes. The fast route never finds x at all.

Look for that relationship before you solve. When it is there, it saves half a minute; when it is not, you have lost three seconds.

5Word problems: name the variable in writing

Write x = number of hours after 9 am before you write anything else. Most wrong answers on context questions are correct values of a variable the student never defined, and the test offers those values as options.

Then translate in the order the sentence gives you:

a fee of 40 plus 12 per hour → 40 + 12h
three fewer than twice the number → 2n − 3never 3 − 2n

6Use the calculator when the algebra turns ugly

Graph the left side and the right side as two functions and read the x of their intersection. It is exact enough for every answer choice on this test, it catches sign errors, and on a "no solution" question the two graphs are visibly parallel.


Where points are lost

  • Answering for x when the question asked for 3x, x + 2, or 2x − 1.
  • Dropping a negative when distributing: −2(x − 5) is −2x + 10.
  • Solving a number-of-solutions question instead of comparing coefficients.
  • Multiplying only some terms by the LCD when clearing fractions.
  • Translating three fewer than twice n as 3 − 2n.
  • Not defining the variable on a word problem.
  • Grinding by hand when graphing both sides would take fifteen seconds.

Work it right

  1. Read what is being asked for — x, or an expression in x.
  2. Brackets out, fractions cleared.
  3. Variables one side, numbers the other.
  4. Solve, then look back at the question before you answer.
  5. If it asks how many solutions, compare coefficients instead.

Try it

Q1. If 5(x − 3) = 2x + 9, what is the value of x?

A) 2 B) 6 C) 8 D) 12

Q2. If 3x + 7 = 22, what is the value of 6x + 14?

A) 15 B) 29 C) 44 D) 88

Q3. In the equation 6(x + 2) = ax + 12, a is a constant. For what value of a does the equation have infinitely many solutions?

A) 2 B) 6 C) 12 D) There is no such value.

Q4. A technician charges a 45-lira call-out fee plus 30 lira for each half hour of work. If a job costs 165 lira in total, how many hours did the technician work? (Type your answer.)

Q5. If (x/3) + (x/4) = 14, what is the value of x?

A) 12 B) 21 C) 24 D) 28

In one breath

Clear brackets, clear fractions, gather the variable on one side and the numbers on the other, then check what the question actually asked for — it wants an expression as often as it wants x. When it asks how many solutions an equation has, stop solving and compare the coefficients of x: different coefficients means one solution, the same coefficient with different constants means none, and identical sides mean every number works.

Answers

Q1. C — 8. distribute, then gather

5(x − 3) = 2x + 9
5x − 15 = 2x + 9
3x = 24subtract 2x, add 15
x = 8

A and B come from sign slips when moving terms; D from adding 15 to 9 and dividing by 2.

Q2. C — 44. 6x + 14 is exactly twice 3x + 7

3x + 7 = 22
6x + 14 = 2(3x + 7)
6x + 14 = 2 × 22 = 44no need to find x at all

A is 3x, the left side minus its constant. B adds 7 to 22 instead of doubling. D doubles twice.

Q3. B — 6. infinitely many means the two sides are identical

6(x + 2) = ax + 12
6x + 12 = ax + 12expand the left
a = 6the constants already match

A and C are the other numbers visible in the equation. D is the trap for a student who tests only whether there is no solution.

Q4. 2. define the variable before translating

h = number of half hours
45 + 30h = 165
30h = 120
h = 4four half hours
answer = 2 hoursthe question asked in hours

Typing 4 is the error the question is built for: a correct value of the variable you defined, in the wrong unit.

Q5. C — 24. multiply every term by the LCD

x/3 + x/4 = 14
4x + 3x = 168multiply every term by 12
7x = 168
x = 24

A and B come from multiplying only one fraction; D from adding the denominators.


Educerie · written from the published College Board* Assessment Framework for the Digital SAT Suite *(Math, Algebra, skill/knowledge testing point "Linear equations in one variable"). All questions and explanations are original Educerie text. Last reviewed 12 September 2026.

Check your understanding

The main ideas of this note. Tick each one you could do now, in an exam, without looking back up. Anything you cannot tick yet is the part to read again.

Practise this skillMocks: in the future, hold tight!