Linear Equations in Two Variables — Important Questions

(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 ✗

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

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.

₹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