Practice
Solving Systems of Linear Equations Practice
Fifty original graphing, substitution, elimination, classification, modeling, and parameter questions.
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Question 1
Explanation
Substitute \(y=2x+0\) into \(x+y=3\). Solving gives \(x=1\), then \(y=2\).
Question 2
Explanation
Substitute \(y=3x-5\) into \(x+y=3\). Solving gives \(x=2\), then \(y=1\).
Question 3
Explanation
Substitute \(y=4x-12\) into \(x+y=3\). Solving gives \(x=3\), then \(y=0\).
Question 4
Explanation
Substitute \(y=5x-21\) into \(x+y=3\). Solving gives \(x=4\), then \(y=-1\).
Question 5
Explanation
Substitute \(y=6x-32\) into \(x+y=3\). Solving gives \(x=5\), then \(y=-2\).
Question 6
Explanation
Multiply the second equation by \(3\) and add to eliminate \(y\); this gives \(14x=-14\), so \(x=-1\). Back-substitution gives \(y=2\).
Question 7
Explanation
Multiply the second equation by \(3\) and add to eliminate \(y\); this gives \(14x=0\), so \(x=0\). Back-substitution gives \(y=3\).
Question 8
Explanation
Multiply the second equation by \(3\) and add to eliminate \(y\); this gives \(14x=14\), so \(x=1\). Back-substitution gives \(y=4\).
Question 9
Explanation
Multiply the second equation by \(3\) and add to eliminate \(y\); this gives \(14x=28\), so \(x=2\). Back-substitution gives \(y=5\).
Question 10
Explanation
Multiply the second equation by \(3\) and add to eliminate \(y\); this gives \(14x=42\), so \(x=3\). Back-substitution gives \(y=6\).
Question 11
Explanation
Different slopes force one intersection.
Question 12
Explanation
Equal slopes with different intercepts produce parallel distinct lines.
Question 13
Explanation
Matching slopes and intercepts describe the same line.
Question 14
Explanation
Different slopes force one intersection.
Question 15
Explanation
Equal slopes with different intercepts produce parallel distinct lines.
Question 16
Explanation
The marked intersection satisfies both distinct lines.
Question 17
Explanation
The lines have equal slopes and different intercepts, so they never meet.
Question 18
Explanation
Both equations plot the same line, so every point on it is shared.
Question 19
Explanation
The marked intersection satisfies both distinct lines.
Question 20
Explanation
The lines have equal slopes and different intercepts, so they never meet.
Question 21
Explanation
The marked intersection satisfies both distinct lines.
Question 22
Explanation
The lines have equal slopes and different intercepts, so they never meet.
Question 23
Explanation
Both equations plot the same line, so every point on it is shared.
Question 24
Explanation
The marked intersection satisfies both distinct lines.
Question 25
Explanation
The lines have equal slopes and different intercepts, so they never meet.
Question 26
Explanation
Let \(A+S=32\) and \(8A+5S=196\). Substitution or elimination yields \(A=12\).
Question 27
Explanation
Let \(A+S=32\) and \(9A+6S=225\). Substitution or elimination yields \(A=11\).
Question 28
Explanation
Let \(A+S=32\) and \(10A+7S=254\). Substitution or elimination yields \(A=10\).
Question 29
Explanation
Let \(A+S=32\) and \(11A+8S=283\). Substitution or elimination yields \(A=9\).
Question 30
Explanation
Let \(A+S=32\) and \(12A+9S=312\). Substitution or elimination yields \(A=8\).
Question 31
Explanation
The second line is \(y=(k/2)x+4\). Match slopes: \(k/2=1\), so \(k=2\). Intercepts differ, confirming no solution.
Question 32
Explanation
The second line is \(y=(k/2)x+4\). Match slopes: \(k/2=2\), so \(k=4\). Intercepts differ, confirming no solution.
Question 33
Explanation
The second line is \(y=(k/2)x+4\). Match slopes: \(k/2=3\), so \(k=6\). Intercepts differ, confirming no solution.
Question 34
Explanation
The second line is \(y=(k/2)x+4\). Match slopes: \(k/2=4\), so \(k=8\). Intercepts differ, confirming no solution.
Question 35
Explanation
The second line is \(y=(k/2)x+4\). Match slopes: \(k/2=5\), so \(k=10\). Intercepts differ, confirming no solution.
Question 36
Explanation
Multiply the first equation by \(3\): \(3y=3x-6\). Matching both coefficients makes the equations equivalent.
Question 37
Explanation
Multiply the first equation by \(3\): \(3y=6x-3\). Matching both coefficients makes the equations equivalent.
Question 38
Explanation
Multiply the first equation by \(3\): \(3y=9x+0\). Matching both coefficients makes the equations equivalent.
Question 39
Explanation
Multiply the first equation by \(3\): \(3y=12x+3\). Matching both coefficients makes the equations equivalent.
Question 40
Explanation
Multiply the first equation by \(3\): \(3y=15x+6\). Matching both coefficients makes the equations equivalent.
Question 41
Explanation
Substitution shows \((x,y)=(-2,1)\) makes the left sides \(-5\) and \(1\), respectively. The other choices fail at least one equation.
Question 42
Explanation
Substitution shows \((x,y)=(-1,2)\) makes the left sides \(-4\) and \(5\), respectively. The other choices fail at least one equation.
Question 43
Explanation
Substitution shows \((x,y)=(0,3)\) makes the left sides \(-3\) and \(9\), respectively. The other choices fail at least one equation.
Question 44
Explanation
Substitution shows \((x,y)=(1,4)\) makes the left sides \(-2\) and \(13\), respectively. The other choices fail at least one equation.
Question 45
Explanation
Substitution shows \((x,y)=(2,5)\) makes the left sides \(-1\) and \(17\), respectively. The other choices fail at least one equation.
Question 46
Explanation
Double the first equation and add to the second to eliminate \(y\): \(11x=11\), so \(x=1\). Then \(y=1\), and \(x+y=2\).
Question 47
Explanation
Double the first equation and add to the second to eliminate \(y\): \(11x=22\), so \(x=2\). Then \(y=0\), and \(x+y=2\).
Question 48
Explanation
Double the first equation and add to the second to eliminate \(y\): \(11x=33\), so \(x=3\). Then \(y=-1\), and \(x+y=2\).
Question 49
Explanation
Double the first equation and add to the second to eliminate \(y\): \(11x=44\), so \(x=4\). Then \(y=-2\), and \(x+y=2\).
Question 50
Explanation
Double the first equation and add to the second to eliminate \(y\): \(11x=55\), so \(x=5\). Then \(y=-3\), and \(x+y=2\).
Keyboard: use Tab to move, arrow keys to change answer choices, and Enter to check an answer.
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Questions to review
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- Question 1Solve by substitutionEasy
- Question 2Solve by substitutionEasy
- Question 3Solve by substitutionEasy
- Question 4Solve by substitutionEasy
- Question 5Solve by substitutionEasy
- Question 6Solve by eliminationEasy
- Question 7Solve by eliminationEasy
- Question 8Solve by eliminationEasy
- Question 9Solve by eliminationEasy
- Question 10Solve by eliminationEasy
- Question 11Classify a systemEasy
- Question 12Classify a systemEasy
- Question 13Classify a systemEasy
- Question 14Classify a systemEasy
- Question 15Classify a systemEasy
- Question 16Graph a systemMedium
- Question 17Graph a systemMedium
- Question 18Graph a systemMedium
- Question 19Graph a systemMedium
- Question 20Graph a systemMedium
- Question 21Graph a systemMedium
- Question 22Graph a systemMedium
- Question 23Graph a systemMedium
- Question 24Graph a systemMedium
- Question 25Graph a systemMedium
- Question 26Model a systemMedium
- Question 27Model a systemMedium
- Question 28Model a systemMedium
- Question 29Model a systemMedium
- Question 30Model a systemMedium
- Question 31System parameter no solutionMedium
- Question 32System parameter no solutionMedium
- Question 33System parameter no solutionMedium
- Question 34System parameter no solutionMedium
- Question 35System parameter no solutionMedium
- Question 36System parameter identityMedium
- Question 37System parameter identityMedium
- Question 38System parameter identityMedium
- Question 39System parameter identityMedium
- Question 40System parameter identityMedium
- Question 41Verify system solutionHard
- Question 42Verify system solutionHard
- Question 43Verify system solutionHard
- Question 44Verify system solutionHard
- Question 45Verify system solutionHard
- Question 46Multi-step system evaluationHard
- Question 47Multi-step system evaluationHard
- Question 48Multi-step system evaluationHard
- Question 49Multi-step system evaluationHard
- Question 50Multi-step system evaluationHard