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Systems of two linear equations in two variables — SAT practice questions

Systems of two linear equations in two variables on SAT Math covers finding the ordered pair that satisfies both equations, plus the cases of no solution and infinitely many solutions. A student must solve by substitution or elimination, model a quantity-and-price story as equations, and sometimes find a combination of the unknowns. The SAT often asks which second equation would yield infinitely many solutions, requests a derived value from a solved pair, or hides the system in a sales story. Traps include confusing a unique solution with a dependent system and stopping after one variable. Answer choices are typically other linear equations or single values.

  • Section: Math 44 questions on the paper
  • 82 practice questions in the app

Sample questions

Question 1

4x+3y=534x+3y=53
3x+2y=343x+2y=34

The solution to the given system of equations is (x,y)(x, y). What is the value of x+yx+y?

  1. A-53
  2. B-19
  3. C19
  4. D53
Show answer

Answer: C — 19

Subtracting the second equation from the first gives (4x+3y)−(3x+2y)=53−34(4x+3y)-(3x+2y)=53-34, so x+y=19x+y=19.

Question 2

7x+3y=−117x+3y=-11

The given equation is one equation in a system of two linear equations. If the system has infinitely many solutions, which of the following equations could be the other equation in the system?

  1. A14x+6y=−1114x+6y=-11
  2. B−21x−9y=33-21x-9y=33
  3. C7x−3y=−117x-3y=-11
  4. D21x+9y=3321x+9y=33
Show answer

Answer: B — −21x−9y=33-21x-9y=33

Multiplying 7x+3y=−117x+3y=-11 by -3 produces −21x−9y=33-21x-9y=33. The equations therefore describe the same line, so every point on that line is a solution.

Question 3

A campus shop sold pens for $2 each and notebooks for $3 each. Yesterday the shop sold a combined total of 55 pens and notebooks for $128. How many pens were sold?

  1. A18
  2. B28
  3. C32
  4. D37
Show answer

Answer: D — 37

Let pp be pens and nn notebooks. Then p+n=55p+n=55 and 2p+3n=1282p+3n=128. Substitute n=55−pn=55-p: 2p+3(55−p)=1282p+3(55-p)=128 → 2p+165−3p=1282p+165-3p=128 → −p=−37-p=-37 → p=37p=37.

Practice 82 Systems of two linear equations in two variables questions in the app

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