Systems of Linear Equations · Sub-skill drill
Solving Systems by Substitution
Substitution is the right method when one of the equations is already solved for a single variable, or when one variable has a coefficient of 1 or −1 that lets you isolate it in one step. The procedure is mechanical: solve one equation for one variable, substitute the resulting expression into the other equation, solve the resulting single-variable equation, then back-substitute to find the second variable. The most common error is forgetting the back-substitution step and reporting only the first variable found.
How this sub-skill is tested on the SAT
Substitution is the right method when one of the equations is already solved for a single variable, or when one variable has a coefficient of 1 or −1 that lets you isolate it in one step. The procedure is mechanical: solve one equation for one variable, substitute the resulting expression into the other equation, solve the resulting single-variable equation, then back-substitute to find the second variable. The most common error is forgetting the back-substitution step and reporting only the first variable found.
This sub-skill sits inside the broader Systems of Linear Equations topic, which is part of the College Board's Heart of Algebra content domain. Heart of Algebra accounts for roughly a third of every SAT Math section. The College Board frames it as the ability to analyze, fluently solve, and create linear equations and inequalities — and to interpret what their solutions mean in context. If you walk into test day weak here, no amount of advanced math fluency will compensate, because Heart of Algebra questions appear in both the calculator and no-calculator modules and are usually front-loaded so they set the tempo for the entire section. Mastery looks like solving a two-step linear equation in under twenty seconds, recognizing parall
Practice questions in this drill set
Below are 7 practice questions targeting this exact sub-skill, ordered from easier to harder. Each question is tagged with its target score band so you can focus on questions that match the band you are working out of. Worked solutions are open by default — read each one even if you got the question right, because the way the solution is structured often reveals a faster path than the one you used.
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If 2x + 4y = -10 and 3x - 2y = -15, what is the value of x + y?
- A -4
- B -5
- C 0
- D 5
Worked solution
Answer: B — -5
Solve by elimination. Multiply the first equation by 3 and the second by 2 so that the x-coefficients match, then subtract to isolate y. The system has unique solution x = -5, y = 0, so x + y = -5.
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If 3x + 2y = 12 and 4x - 3y = 33, what is the value of x + y?
- A 9
- B -18
- C -3
- D 3
Worked solution
Answer: D — 3
Solve by elimination. Multiply the first equation by 4 and the second by 3 so that the x-coefficients match, then subtract to isolate y. The system has unique solution x = 6, y = -3, so x + y = 3.
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If 5x + 5y = 30 and 4x - 3y = 24, what is the value of x + y?
- A 6
- B -6
- C 7
- D 0
Worked solution
Answer: A — 6
Solve by elimination. Multiply the first equation by 4 and the second by 5 so that the x-coefficients match, then subtract to isolate y. The system has unique solution x = 6, y = 0, so x + y = 6.
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If 5x + 2y = -40 and 1x + 5y = -31, what is the value of x + y?
- A 30
- B 11
- C -1
- D -11
Worked solution
Answer: D — -11
Solve by elimination. Multiply the first equation by 1 and the second by 5 so that the x-coefficients match, then subtract to isolate y. The system has unique solution x = -6, y = -5, so x + y = -11.
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If 4x + 1y = 26 and 1x + 5y = 35, what is the value of x + y?
- A 30
- B 11
- C -1
- D -11
Worked solution
Answer: B — 11
Solve by elimination. Multiply the first equation by 1 and the second by 4 so that the x-coefficients match, then subtract to isolate y. The system has unique solution x = 5, y = 6, so x + y = 11.
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If 3x + 1y = 16 and 5x + 3y = 24, what is the value of x + y?
- A -4
- B -12
- C 4
- D 8
Worked solution
Answer: C — 4
Solve by elimination. Multiply the first equation by 5 and the second by 3 so that the x-coefficients match, then subtract to isolate y. The system has unique solution x = 6, y = -2, so x + y = 4.
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If 3x + 2y = 1 and 3x + 1y = 2, what is the value of x + y?
- A -1
- B 1
- C 2
- D 0
Worked solution
Answer: D — 0
Solve by elimination. Multiply the first equation by 3 and the second by 3 so that the x-coefficients match, then subtract to isolate y. The system has unique solution x = 1, y = -1, so x + y = 0.
Why this band assignment matters
Every question in this drill is tagged with a target score band — 400–500, 500–600, 600–700, or 700–800 — based on its difficulty and the patterns the College Board uses for questions at each level. If you are aiming to break out of a 580 plateau, the 600–700 questions in this drill are your highest-leverage practice. If you are chasing 750+, the 700–800 questions here are the ones that separate the top 10% of test takers from everyone else.
Use the band tags to filter your work. If you can confidently solve every 400–500 and 500–600 question without notes, move to the 600–700 set. If those land cleanly, the 700–800 set is your final boss. The worked solutions in this drill are written so that even the hardest questions become learnable patterns once you have seen the structure of the solve a few times.