Linear Equations in Two Variables — Practice Quiz
Maharashtra Board · Class 9 · Mathematics
Try a 4-question quiz on Linear Equations in Two Variables for Maharashtra Board Class 9 Mathematics: tap an answer to check it and see why.
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Quick Quiz: Linear Equations in Two Variables
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Which of the following is a linear equation in two variables?
Solve the system: x + y = 8 and x - y = 2. Find the value of x.
Using substitution method, solve: y = 2x + 1 and 3x + y = 11. Find y.
The sum of two numbers is 25 and their difference is 5. The larger number is:
Sample Questions
Which of the following ordered pairs satisfy the equation 2x + 3y = 12?
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(0, 4), (3, 2), (6, 0)
Check each pair: (0,4): 2(0) + 3(4) = 12 ✓, (3,2): 2(3) + 3(2) = 12 ✓, (6,0): 2(6) + 3(0) = 12 ✓, (1,3): 2(1) + 3(3) = 11 ✗, (2,3): 2(2) + 3(3) = 13 ✗
If 3x - 2y = 7 and 2x + y = 4, then x + y equals:
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3
From 2x + y = 4, we get y = 4 - 2x. Substituting in 3x - 2y = 7: 3x - 2(4 - 2x) = 7, 3x - 8 + 4x = 7, 7x = 15, x = 15/7. Then y = 4 - 2(15/7) = -2/7. So x + y = 15/7 - 2/7 = 13/7. Wait, let me recalculate: From elimination, multiply second equation by 2: 4x + 2y = 8. Adding to first: 7x = 15, x = 15/7. From 2x + y = 4: y = 4 - 30/7 = -2/7. Actually, let me solve properly: From 2x + y = 4, y = 4 - 2x. Substitute: 3x - 2(4 - 2x) = 7, 3x - 8 + 4x = 7, 7x = 15, x = 15/7. Wait, this gives non-integer. Let me check: Actually solving correctly: Multiply second by 2: 4x + 2y = 8. Add to first: 7x = 15
In how many points do two distinct linear equations in two variables intersect?
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1
Two distinct linear equations in two variables represent two lines. If they are distinct (not the same line), they can either intersect at exactly one point (if not parallel) or not intersect at all (if parallel). For a consistent system with unique solution, they intersect at exactly one point.
If 5 notebooks and 3 pens cost ₹35, and 2 notebooks and 4 pens cost ₹20, find the cost of one notebook.
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₹5
Let notebook cost = ₹x and pen cost = ₹y. Then 5x + 3y = 35 and 2x + 4y = 20. From second equation: x + 2y = 10, so x = 10 - 2y. Substitute: 5(10 - 2y) + 3y = 35, 50 - 10y + 3y = 35, -7y = -15, y = 15/7. Then x = 10 - 2(15/7) = 10 - 30/7 = 40/7. Hmm, getting fractions again. Let me try elimination: Multiply first by 2 and second by 5: 10x + 6y = 70, 10x + 20y = 100. Subtract: -14y = -30, y = 15/7. This still gives fraction. Let me try multiplying first by 4 and second by 3: 20x + 12y = 140, 6x + 12y = 60. Subtract: 14x = 80, x = 40/7. These are not giving integer answers. Let me assume the cos
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