➕ Algebra 1 · Systems of Equations
Solving Systems by Elimination: true or false
Decide fast, then read the reason. Wrong answers are the best teachers, so every statement explains itself.
All statements and answers
To eliminate y from x + 6y = 15 and 5x − 6y = 3, you add the equations.
✓ True. The y-coefficients 6 and −6 are opposites, so adding gives 6x = 18, which leaves x = 3 (and then y = 2).
To eliminate x from 5x + 2y = 9 and 5x − 3y = −1, you add the equations.
✗ False. The x-coefficients are equal, not opposite. Adding gives 10x, which does not eliminate x. Subtract instead: 5y = 10, so y = 2 (and then x = 1).
Multiplying 3x − y = 4 by 5 gives 15x − 5y = 4.
✗ False. The constant must be multiplied too: 15x − 5y = 20.
Subtracting x + y = 7 from 3x + y = 15 gives 2x = 8.
✓ True. (3x + y) − (x + y) = 15 − 7. The y-terms cancel and 3x − x = 2x, so 2x = 8.
Subtracting 2x − 3y = 9 from 2x + y = 13 gives −2y = 4.
✗ False. The minus must flip the sign of −3y: y − (−3y) = 4y. And 13 − 9 = 4. So the result is 4y = 4, which gives y = 1. The answer −2y comes from computing y − 3y.
If adding the two equations makes both variables vanish and leaves 0 = 7, the system has no solution.
✓ True. 0 = 7 is false for every x and y. A pair that solved both equations would also solve the sum, so no pair can solve both.
If elimination leaves 0 = 0, then the solution is x = 0 and y = 0.
✗ False. 0 = 0 is true no matter what x and y are. The two equations describe the same line, so there are infinitely many solutions.
A solution found by elimination is an ordered pair, so after you find x you still need y.
✓ True. A solution must make both equations true, and it needs both coordinates. Back-substitute to find the other one.
Subtracting an equation gives the same result as multiplying it by −1 (every term) and then adding.
✓ True. Subtracting 3x − 2y = 10 is the same as adding −3x + 2y = −10. Both flip every sign.
You can multiply one equation by 0 to help eliminate a variable.
✗ False. Multiplying by 0 turns the equation into 0 = 0, which throws away all its information. Multiply by a nonzero number.
If you back-substitute into one original equation and it checks, the solution is guaranteed to be correct.
✗ False. You found the second value from that equation, so it checks automatically. A mistake in the elimination step would only show up in the other equation.
The system x + 2y = 4 and x + 2y = 10 has no solution.
✓ True. x + 2y cannot equal 4 and 10 at the same time. Subtracting the equations gives 0 = −6, which is false.
To eliminate a variable from x + 5y = 12 and 3x + 2y = 10, adding or subtracting the equations as they are will cancel a variable.
✗ False. No variable has equal or opposite coefficients, so adding or subtracting does not cancel anything. Multiply first, for example the first equation by 3: 3x + 15y = 36. Subtracting the second then gives 13y = 26, so y = 2 and x = 2.
If A = B and C = D, then A + C = B + D.
✓ True. Adding equal amounts to equal amounts gives equal results. This is why adding two equations is allowed.