Feeling confident with single-variable problems but ready to conquer multiple equations at once? Dive into our free systems of equations quiz - designed for students, educators, and math enthusiasts looking to sharpen their approach to linear systems. In this linear systems quiz, you'll test linear equations skills, challenge yourself with tricky simultaneous equation pairs, and build the mastery needed for your next linear equations exam. Start with this engaging equation quiz, then level up with our comprehensive linear equations test . Ready to ace your systems of equations test? Get started now!
Solve the system: x + y = 10 and x - y = 4.
(7, 3)
(8, 2)
(3, 7)
(5, 5)
By adding the two equations, you get 2x = 14 so x = 7. Substitute back into x + y = 10 to find y = 3. This is a straightforward elimination method. For more, see Khan Academy: Systems of Equations.
Solve the system: x = 2y + 1 and x + y = 7.
(3, 1)
(6, 2)
(5, 2)
(7, 3)
Substitute x = 2y + 1 into x + y = 7 to get (2y + 1) + y = 7, so 3y = 6 and y = 2. Then x = 2(2) + 1 = 5. See Khan Academy for details.
Solve: 3x + y = 10 and x - y = 2.
(3, 2)
(3, 1)
(4, 1)
(2, 2)
From x - y = 2, we have x = y + 2. Substitute into 3(y + 2) + y = 10, giving 4y + 6 = 10 so y = 1 and x = 3. See Khan Academy for more examples.
Find the intersection of y = 2x + 1 and y = -x + 4.
(1, 3)
(1, -3)
(-1, 3)
(2, 5)
Set 2x + 1 = -x + 4, so 3x = 3 and x = 1, then y = 2(1) + 1 = 3. This is basic graph intersection. For more, see Khan Academy.
Solve the system: x + y = 7 and 2x + y = 10.
(2, 5)
(4, 3)
(5, 2)
(3, 4)
Subtract the first equation from the second to get x = 3, then substitute into x + y = 7 to get y = 4. More practice at Khan Academy.
Solve: -x + 2y = 4 and x - y = -1.
(1, 3)
(3, 2)
(2, 3)
(2, 1)
Add the equations to eliminate x: (-x + 2y) + (x - y) = 4 + (-1) gives y = 3, then x = y - 1 = 2. See Khan Academy.
Which describes the solutions of x + 2y = 4 and 2x + 4y = 8?
Infinite solutions
Unique solution
Exactly two solutions
No solution
The second equation is just twice the first, so the lines coincide and there are infinitely many solutions. See Khan Academy.
Which describes the solutions of x + y = 3 and 2x + 2y = 8?
Unique solution
Dependent solution
Infinite solutions
No solution
The second equation simplifies to x + y = 4, which contradicts x + y = 3, so the lines are parallel and there is no solution. See Khan Academy.
Solve the system: 0.5x + 0.25y = 3 and x - y = 2.
(2, 3)
(14/3, 8/3)
(8/3, 14/3)
(3, 2)
From x - y = 2, x = y + 2. Substituting gives 0.5(y+2) + 0.25y = 3 ? 0.75y = 2 ? y = 8/3, x = 14/3. For more practice, see Khan Academy.
Solve: 2x + 3y = 12 and 4x - y = 6.
(3, 1)
(2, 2)
(15/7, 18/7)
(18/7, 15/7)
Solve 4x - y = 6 for y = 4x - 6, substitute in 2x + 3(4x - 6) = 12 ? 14x = 30 ? x = 15/7, y = 18/7. See Khan Academy.
Solve the system: 3x + 2y = -1 and x + y = 4.
(4, -1)
(-1, 4)
(-9, 13)
(13, -9)
From x + y = 4, x = 4 - y. Substitute in 3(4 - y) + 2y = -1 ? 12 - 3y + 2y = -1 ? -y = -13 ? y = 13, x = -9. See Khan Academy.
Solve: 5x + 4y = 1 and 3x - 2y = 8.
(-17/11, 37/22)
(17/11, -37/22)
(1, 0)
(0, 1)
Multiply 3x - 2y = 8 by 2 to align y-terms, add with 5x + 4y = 1 to get 11x = 17 ? x = 17/11, then y = -37/22. More at Khan Academy.
Solve the system: 1.2x + 0.8y = 4 and x - y = 1.
(3, 1)
(1, 3)
(2.4, 1.4)
(1.4, 2.4)
x = y + 1 from the second equation, substitute into the first: 1.2(y+1) + 0.8y = 4 ? 2y + 1.2 = 4 ? y = 1.4, x = 2.4. See Khan Academy.
A $1000 investment yields $68 interest at rates of 5% and 8%. How much was invested at 8%?
(300, 700)
(500, 500)
(600, 400)
(400, 600)
Let x at 5% and y at 8%, x + y = 1000 and .05x + .08y = 68 ? solve to get y = 600 at 8% and x = 400 at 5%. More practice at Khan Academy.
Solve: 4y - 2x = 10 and x + 3y = 5.
(-2, 1)
(1, 2)
(-1, 2)
(2, -1)
Rewrite -2x + 4y = 10 as x - 2y = -5, then subtract from x + 3y = 5 to find y = 2 and x = -1. See Khan Academy.
Solve the system: 7x - y = 11 and 3x + 5y = 1.
(28/19, -13/19)
(11/7, -1/7)
(0, 1)
(1, 2)
From 7x - y = 11, y = 7x - 11, substitute into 3x + 5(7x - 11) = 1 ? 38x = 56 ? x = 28/19, y = -13/19. See Khan Academy.
Solve the system: x + y + z = 6, 2x - y + 3z = 14, and x + 4y - z = 2.
(1, 2, 3)
(4, 0, 2)
(3, 1, 2)
(2, 4, 0)
Use substitution or elimination: from x + y + z = 6 get x = 6 - y - z, substitute into the other two, solve for y, z then x. The result is (4, 0, 2). For method details see Khan Academy.
Solve: x + y - 2z = 1, 2x - y + z = 3, and -x + 3y - 3z = -4.
(-1, 2, 1)
(1, -2, -1)
(2, -1, 1)
(0, 0, 0)
Substitute from the first equation into the others to eliminate x, then solve the resulting two-variable system for y and z, then back-substitute for x. The solution is (1, -2, -1). See Khan Academy.
A chemist needs 200 ml of a 40% acid solution using 20% and 70% concentrations. How much of the 70% solution is needed?
(100, 100)
(140, 60)
(80, 120)
(120, 80)
Let x be amount of 20% and y of 70%, x + y = 200 and 0.20x + 0.70y = 80. Solving gives y = 80 ml at 70%. See Khan Academy.
For what value of k does the system 2x + ky = 4 and 3x + 6y = 9 have no solution?
6
4
-1
3
No solution arises if slopes match but intercepts differ: - 2/k = - 1/2 ? k = 4. Then the equations become inconsistent. See Purplemath: Parallel Lines.
Solve the system: x + 2y + 3z = 1, 2x - y + z = 2, and x + y - z = 0.
(-2/3, 1/3, -1/3)
(1, 0, 0)
(0, 1, 1)
(2/3, -1/3, 1/3)
Use substitution or Cramer's rule: solving gives x = 2/3, y = - 1/3, z = 1/3. For detailed steps see Khan Academy.
Determine the nature of solutions for 2x - 3y + z = 7, x + y - 2z = -1, and 3x - 2y - z = 4.
Dependent solution
Unique solution
No solution
Infinite solutions
Attempting elimination leads to a contradiction (parallel planes) so there is no solution. See Khan Academy discussion on parallel planes.
Which describes the solutions of 2x + 4y = 6 and x + 2y = 3?
No solution
One solution
Unique solution
Infinite solutions
The second equation is exactly half the first, so they represent the same line and there are infinitely many solutions. For more see Khan Academy.
Solve: x - y + 2z = 4, 2x + y - z = 1, and 3x - y + z = 7.
(2, -1, 0)
(8/5, -2, 1/5)
(-1, 2, -1)
(1, 0, 1)
Combine equations to eliminate variables in steps (for instance sum and subtract) to find y = -2, then back-substitute to get x = 8/5, z = 1/5. See Khan Academy.
Use matrix methods or elimination to solve: 4x - y + 2z = 9, -x + 3y - z = -6, 2x + y + z = 7.
(1, 2, 3)
(0, 0, 0)
(-17/3, 5/3, 50/3)
(-3, 2, 1)
Solve via substitution or invert the coefficient matrix to get x = -17/3, y = 5/3, z = 50/3. This requires careful elimination or Cramer's rule. See Khan Academy: Solving with Matrices.
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AI Study Notes
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Study Outcomes
Solve systems using substitution -
Apply the substitution method to find precise solutions to pairs of linear equations and verify results efficiently.
Solve systems using elimination -
Use the elimination method to combine linear equations, eliminate variables, and accurately determine solution pairs.
Interpret graphical solutions -
Plot linear equations on a coordinate plane to visualize the point of intersection representing the system's solution.
Analyze system consistency -
Differentiate between consistent, inconsistent, and dependent systems of linear equations based on their algebraic and graphical properties.
Verify solution accuracy -
Check solutions by substituting them back into the original equations to ensure correctness and deepen understanding.
Build test-taking confidence -
Receive instant feedback on your performance to refine strategies, improve speed, and boost readiness for linear equations exams.
Cheat Sheet
Substitution Method Mastery -
When preparing for a systems of equations quiz, substitution is a go-to technique recommended by MIT OpenCourseWare: you solve one equation for a variable (e.g., y = 2x + 3) and substitute it into the other. This transforms a 2×2 system into a single-variable equation like 3x + (2x + 3) = 11, making it easier to solve. Consistent practice isolating variables can dramatically improve your score on any linear systems quiz.
Elimination Method Efficiency -
Khan Academy highlights elimination as an efficient strategy for a systems of equations test and to test linear equations skills: multiply equations to align coefficients, then add or subtract to eliminate one variable. For instance, multiplying 2x + y = 7 by 2 gives 4x + 2y = 14, which you can subtract from 4x − 3y = 8 to eliminate x immediately. A mnemonic trick is 'Add to Eliminate' when coefficients match in sign and magnitude.
Graphing Linear Systems -
The National Council of Teachers of Mathematics recommends graphing both equations on the same axes to visually identify intersection points. Use slope-intercept form (y = mx + b) to draw lines; where the lines meet is the solution, whether it's one point (unique solution), none (parallel lines), or infinitely many (coincident lines). This hands-on approach reinforces understanding, especially before a linear systems quiz or linear equations exam.
Cramer's Rule for 2×2 Systems -
For a quick, formula-based solution, many university courses introduce Cramer's Rule: x = Dx/D and y = Dy/D, where D is the determinant of the coefficient matrix. For example, with system ax + by = e and cx + dy = f, D = ad − bc, Dx = ed − bf, Dy = af − ec, giving direct values for x and y. This method is especially useful on a linear systems quiz requiring precise, algebraic accuracy.
Identifying Solution Types -
Recognizing whether a system has one solution, no solution, or infinitely many is a crucial skill emphasized by Purplemath. Parallel lines (same slope, different intercepts) yield no solution, while identical equations indicate infinite solutions. Remember the phrase 'Parallel = No Place to Meet, Same Line = All the Time' to lock in this concept before any systems of equations test.
