SAT Math: Systems of linear equations
Systems of two linear equations.
- Algebra · Systems of two linear equations in two variables
- Both modules
- 13 practice questions
- Calculator allowed (Desmos)
What the test covers
- Elimination and substitution; Desmos intersection
- No / one / infinitely many solutions from coefficient ratios
Key ideas
- Solve by elimination (add or subtract multiples) or substitution.
- One solution: different slopes. No solution: same slope, different intercepts (parallel). Infinitely many: the same line.
- For 'no solution' with a constant \(k\), make the \(x\) and \(y\) coefficients proportional and the constants not.
Common mistakes
- A question asking for \(x+y\) may be answerable by adding the equations directly.
- Proportional coefficients alone are not enough for infinitely many solutions: the constants must match too.
Do it in Desmos
Graph both lines; the intersection is the solution. Parallel lines mean no solution.
Worked example
Worked example
If \(2x+3y=12\) and \(x-y=1\), what is the value of \(x+y\)?
- \(3\)
- \(5\)
- \(1\)
- \(7\)
Answer: B: \(5\)
From the second equation \(x=y+1\). Then \(2y+2+3y=12\), \(y=2\), \(x=3\), and \(x+y=5\).
Practice questions
Try each question before opening the solution. Desmos or your own calculator is allowed on every SAT Math question. Every answer here was re-checked independently by computer before it was published.
Question 1
In the system \(3x+ky=7\) and \(6x-4y=10\), \(k\) is a constant. The system has no solution. What is the value of \(k\)?
Show the answer and solution
Answer: -2
No solution means parallel lines: the coefficients are proportional but the constants are not. \(\tfrac36=\tfrac{k}{-4}\) gives \(k=-2\); then \(\tfrac{7}{10}\ne\tfrac12\), so the lines are distinct.
Question 2
The system \(ax+2y=8\) and \(3x+by=12\), where \(a\) and \(b\) are constants, has infinitely many solutions. What is the value of \(a+b\)?
Show the answer and solution
Answer: 5
The two equations must be multiples of each other: \(\tfrac a3=\tfrac2b=\tfrac8{12}=\tfrac23\). So \(a=2\), \(b=3\), \(a+b=5\).
Question 3
If \(2x+5y=25\) and \(x-2y=-1\), what is the value of \(x+y\)?
- \(8\)
- \(3\)
- \(2\)
- \(5\)
Show the answer and solution
Answer: A: \(8\)
Solving the system (by elimination or with Desmos) gives \(x=5\) and \(y=3\), so \(x+y=8\).
Question 4
If \(4x+y=18\) and \(x-2y=9\), what is the value of \(x+y\)?
- \(-2\)
- \(7\)
- \(3\)
- \(5\)
Show the answer and solution
Answer: C: \(3\)
Solving the system (by elimination or with Desmos) gives \(x=5\) and \(y=-2\), so \(x+y=3\).
Question 5
If \(4x+3y=5\) and \(x-y=-4\), what is the value of \(x+y\)?
- \(3\)
- \(2\)
- \(-4\)
- \(-1\)
Show the answer and solution
Answer: B: \(2\)
Solving the system (by elimination or with Desmos) gives \(x=-1\) and \(y=3\), so \(x+y=2\).
Question 6
If \(2x+5y=16\) and \(2x+y=0\), what is the value of \(x+y\)?
- \(2\)
- \(-6\)
- \(4\)
- \(-2\)
Show the answer and solution
Answer: A: \(2\)
Solving the system (by elimination or with Desmos) gives \(x=-2\) and \(y=4\), so \(x+y=2\).
Question 7
If \(2x+5y=37\) and \(2x+y=17\), what is the value of \(x+y\)?
- \(6\)
- \(5\)
- \(11\)
- \(1\)
Show the answer and solution
Answer: C: \(11\)
Solving the system (by elimination or with Desmos) gives \(x=6\) and \(y=5\), so \(x+y=11\).
Question 8
In the system \(3x+ky=6\) and \(6x+4y=13\), \(k\) is a constant. The system has no solution. What is the value of \(k\)?
Show the answer and solution
Answer: 2
No solution means parallel, distinct lines: the \(x\)- and \(y\)-coefficients are in the same ratio (\(3:6=k:4\), so \(k=2\)) but the constants are not (\(6\times2\ne13\)).
Question 9
In the system \(3x+ky=8\) and \(6x-4y=18\), \(k\) is a constant. The system has no solution. What is the value of \(k\)?
Show the answer and solution
Answer: -2
No solution means parallel, distinct lines: the \(x\)- and \(y\)-coefficients are in the same ratio (\(3:6=k:-4\), so \(k=-2\)) but the constants are not (\(8\times2\ne18\)).
Question 10
In the system \(5x+ky=2\) and \(10x-4y=5\), \(k\) is a constant. The system has no solution. What is the value of \(k\)?
Show the answer and solution
Answer: -2
No solution means parallel, distinct lines: the \(x\)- and \(y\)-coefficients are in the same ratio (\(5:10=k:-4\), so \(k=-2\)) but the constants are not (\(2\times2\ne5\)).
Question 11
In the system \(5x+ky=6\) and \(10x+4y=9\), \(k\) is a constant. The system has no solution. What is the value of \(k\)?
Show the answer and solution
Answer: 2
No solution means parallel, distinct lines: the \(x\)- and \(y\)-coefficients are in the same ratio (\(5:10=k:4\), so \(k=2\)) but the constants are not (\(6\times2\ne9\)).
Question 12
In the system \(2x+ky=3\) and \(4x-4y=7\), \(k\) is a constant. The system has no solution. What is the value of \(k\)?
Show the answer and solution
Answer: -2
No solution means parallel, distinct lines: the \(x\)- and \(y\)-coefficients are in the same ratio (\(2:4=k:-4\), so \(k=-2\)) but the constants are not (\(3\times2\ne7\)).
Keep going
- Previous topic: Linear equations in two variables
- Next topic: Linear inequalities
- All SAT Math skills · Adaptive SAT Math module simulator · Official SAT practice (Bluebook, Khan Academy)
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