Pair of Linear Equations in Two Variables — Important Questions

x = 2

Step 1: The equations are 3x + y = 10 ... (1) and x + y = 6 ... (2). Step 2: Subtract equation (2) from equation (1) to eliminate y: (3x + y) - (x + y) = 10 - 6. Step 3: 2x = 4, so x = 2. Step 4: Substitute into equation (2): 2 + y = 6, so y = 4. Step 5: Verify: 3(2) + 4 = 10 ✓ and 2 + 4 = 6 ✓. A common mistake is subtracting in the wrong order, giving -2x = -4, but x = 2 is the same.

₹5

Step 1: Let orange cost = x and apple cost = y. Equations: 5x + 3y = 35 ...(1) and 2x + 4y = 28 ...(2). Step 2: Multiply (1) by 2 and (2) by 5: 10x + 6y = 70 and 10x + 20y = 140. Step 3: Subtract first from second: 14y = 70, so y = 5. Step 4: Substitute y = 5 into (2): 2x + 20 = 28, so x = 4. Step 5: Verify: 5(4) + 3(5) = 20 + 15 = 35 ✓. Each apple costs ₹5.

a1/a2 = b1/b2 ≠ c1/c2

Step 1: When a1/a2 = b1/b2 ≠ c1/c2, the lines are parallel to each other. Step 2: Parallel lines never intersect, meaning there is no common point. Step 3: No common point means no solution exists — this is an inconsistent pair. Step 4: Contrast: a1/a2 = b1/b2 = c1/c2 gives coincident lines (infinite solutions). Step 5: And a1/a2 ≠ b1/b2 gives intersecting lines (unique solution). Memorize all three conditions carefully.

y = 3

Step 1: Equations: 2x + 3y = 13 ...(1) and 5x - 2y = 4 ...(2). Step 2: Multiply (1) by 2 and (2) by 3: 4x + 6y = 26 and 15x - 6y = 12. Step 3: Add both equations: 19x = 38, so x = 2. Step 4: Substitute x = 2 into (1): 4 + 3y = 13, so 3y = 9, giving y = 3. Step 5: Verify with (2): 5(2) - 2(3) = 10 - 6 = 4 ✓. A common mistake is subtracting instead of adding in Step 3, which does not eliminate a variable.