Educerie

Educerie · SAT · Math

Algebra · ALG.4 Systems of two linear equations in two variables

Where it is examined
both modules, and often as the harder end of the Algebra block. Part of the ≈35% of the section that is Algebra.
The question this unit answers
two equations, two unknowns. Which method, and what is the test really after when it asks how many solutions there are?
Before you start
linear equations in one variable, and writing a line from a context. A system is two of those at once.

What you must be able to do

You must be able toWhat it looks like on the test
Solve by substitutionWhen one variable is already alone, or easy to isolate
Solve by eliminationWhen neither variable is alone
Say how many solutions a system hasOne, none, or infinitely many
Build a system from a word problemTwo sentences, two equations
Solve for an expression like x + yOften faster than solving for each

1The idea in one paragraph

Two lines in a plane either cross once, never cross, or are the same line. That is the whole theory: one solution, no solution, infinitely many. Solving means finding where they cross, and the test's favourite question is not the crossing point at all but the count — which you answer by comparing slopes, not by solving. Different slopes cross once; the same slope with different intercepts never crosses; the same slope and the same intercept is one line written twice.

When the answer choices are the number of solutions, put both equations in y = mx + b or compare coefficients. Do not solve.

2Choosing a method

Substitution when one equation already has a variable alone, or a coefficient of 1. y = 2x − 5 goes straight into the other equation.

Elimination when neither does. Line the equations up, scale one or both so a variable's coefficients match, then add or subtract.

3x + 2y = 16
5x − 2y = 8
8x = 24add: the y terms cancel
x = 3
3(3) + 2y = 16 → y = 3.5

Elimination is faster than students expect and is the method to default to when both equations are in standard form.

3How many solutions, without solving

Write both as ax + by = c.

ConditionSolutions
a₁/a₂ ≠ b₁/b₂One — different slopes
a₁/a₂ = b₁/b₂ but ≠ c₁/c₂None — parallel
a₁/a₂ = b₁/b₂ = c₁/c₂Infinitely many — same line

For what value of k does 6x + ky = 10 and 9x + 12y = 15 have infinitely many solutions?

6/9 = 2/3the x coefficients
10/15 = 2/3the constants agree
k/12 = 2/3so the y coefficients must too
k = 8

Half the "no solution" questions on the test are that middle row in disguise.

4Word problems: two sentences, two equations

Twelve items were bought, some at 40 lira and some at 65 lira, for 605 lira in total.

a + b = 12the count
40a + 65b = 605the value
40(12 − b) + 65b = 605substitute a = 12 − b
480 + 25b = 605
b = 5, a = 7

SAT numbers come out clean. If yours do not, check the translation before you check the arithmetic — a decimal answer to a question about items almost always means one of the two equations is wrong.

Always write what each letter means. Two of the four answer choices are usually the other variable's value.

5Solving for an expression

If x + 3y = 14 and 2x − y = 7, what is the value of 3x + 2y?

x + 3y = 14
2x − y = 7
3x + 2y = 21add them as they stand

No solving needed.

Whenever a question asks for a combination rather than for x or y, try adding or subtracting the equations first. When it works it saves a minute.

6The calculator solves any of them

Put both equations in y = form and graph. The intersection is the solution, parallel lines show no crossing, and a single visible line means infinitely many. For a messy system this is faster and safer than elimination.


Where points are lost

  • Answering with the wrong variable. Write down which is which, and check what was asked.
  • Solving when the question wanted the number of solutions.
  • Sign errors in elimination — subtracting a negative term.
  • Scaling only part of an equation when multiplying to match coefficients.
  • Forgetting to find the second variable when the question needs both.
  • Ignoring the ratio test and guessing at parallel.

Work it right

  1. Read what is wanted: x, y, a combination, or a count.
  2. If it is a count, compare coefficients and stop.
  3. Otherwise pick a method: substitution if a variable is alone, elimination if not.
  4. Solve for one, substitute back for the other.
  5. Check both equations with your pair before answering.

Try it

Q1. If 4x + 3y = 30 and 2x − 3y = 6, what is the value of x?

A) 2 B) 3 C) 5 D) 6

Q2. The system 5x − 2y = 11 and 15x − 6y = 30 has how many solutions?

A) Zero B) Exactly one C) Exactly two D) Infinitely many

Q3. For what value of c does the system 3x + 4y = 9 and 6x + 8y = c have infinitely many solutions?

A) 9 B) 12 C) 18 D) No value of c

Q4. A cinema sold 240 tickets for 3,720 lira. Adult tickets cost 20 lira and child tickets 12 lira. How many adult tickets were sold? (Type your answer.)

Q5. If a + 2b = 13 and 3a − 2b = 7, what is the value of 4a?

A) 5 B) 10 C) 20 D) 26

In one breath

Two lines cross once, never, or everywhere, and the test asks which far more often than it asks where. Compare the coefficients: different ratios of x and y coefficients mean one solution, equal ratios with a different constant mean none, and equal all the way through means the same line twice. When you do have to solve, use substitution if a variable is already alone and elimination if it is not — and check which letter the question actually wanted.

Answers

Q1. D — 6. add the equations to eliminate y

4x + 3y = 30
2x − 3y = 6
6x = 36
x = 6, y = 2

A is y. B and C come from subtracting the equations instead of adding, which eliminates nothing.

Q2. A — Zero. same slope, different constant Multiplying the first by 3 gives 15x − 6y = 33, which contradicts 15x − 6y = 30. The lines are parallel. D is the trap for a student who sees 15 = 3 × 5 and 6 = 3 × 2 and stops before checking the constant.

Q3. C — 18. every coefficient must scale by the same factor

3x + 4y = 9
6x + 8y = c
6/3 = 8/4 = 2the second is twice the first
c = 2 × 9 = 18

A keeps the constant unchanged; B is unrelated; D ignores that a consistent scaling exists.

Q4. 105. two equations, then substitute

a + c = 240
20a + 12c = 3720
20a + 12(240 − a) = 3720
8a + 2880 = 3720
a = 105and 135 child tickets

Typing 135 is the error the question is built for: a correct value of the wrong variable.

Q5. C — 20. add the equations

a + 2b = 13
3a − 2b = 7
4a = 20which is what was asked for
a = 5which was not

A is a. B and D come from solving for b or adding the constants.


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

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