Linear Equations in Two Variables — Maharashtra (SSC) Class 10 Mathematics Solutions (Free)

Free step-by-step Maharashtra (SSC) Class 10 Mathematics solutions for "Linear Equations in Two Variables" — important questions with detailed answers, download PDF for board exam preparation.

Linear Equations in Two Variables — Maharashtra (SSC) Class 10 Mathematics Solutions (Free)

Free step-by-step Maharashtra (SSC) Class 10 Mathematics solutions for "Linear Equations in Two Variables" — important questions with detailed answers, download PDF for board exam preparation.

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Linear Equations in Two Variables — Textbook Questions Answered (8)

Q1: Solve the system: 2x + 3y = 8 and x - y = 1 using substitution method.

Step 1: Express one variable from second equation. From x - y = 1, we get x = y + 1. Step 2: Substitute x = y + 1 in first equation. 2(y + 1) + 3y = 8. Step 3: Expand and simplify. 2y + 2 + 3y = 8, so 5y + 2 = 8. Step 4: Solve for y. 5y = 6, therefore y = 6/5 = 1.2. Step 5: Find x using x = y + 1. x = 1.2 + 1 = 2.2. Step 6: Verify. 2(2.2) + 3(1.2) = 4.4 + 3.6 = 8. Check: 2.2 - 1.2 = 1. Both equations satisfied. Answer: x = 2.2, y = 1.2 or x = 11/5, y = 6/5.

Q2: Solve 3x + 4y = 7 and 2x - 5y = -8 using elimination method.

Step 1: Multiply first equation by 2 to match x coefficient. 2(3x + 4y) = 2(7) gives 6x + 8y = 14. Step 2: Multiply second equation by 3 to match x coefficient. 3(2x - 5y) = 3(-8) gives 6x - 15y = -24. Step 3: Subtract second from first to eliminate x. (6x + 8y) - (6x - 15y) = 14 - (-24). Step 4: Simplify. 8y + 15y = 38, so 23y = 38. Step 5: Solve for y. y = 38/23. Step 6: Substitute y in original first equation. 3x + 4(38/23) = 7. Step 7: Simplify. 3x + 152/23 = 7. Step 8: Solve for x. 3x = 7 - 152/23 = 161/23 - 152/23 = 9/23. So x = 3/23. Step 9: Verify both equations with these values. Answer: x = 3/23, y = 38/23.

Q3: Solve using cross-multiplication: 4x + 5y = 13 and 3x + 2y = 5.

Step 1: Write coefficients in cross-multiplication form. a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0. Step 2: Rewrite equations. 4x + 5y - 13 = 0 and 3x + 2y - 5 = 0. Step 3: Apply cross-multiplication. x/(b1c2 - b2c1) = y/(c1a2 - c2a1) = 1/(a1b2 - a2b1). Step 4: Calculate denominators. b1c2 - b2c1 = 5(-5) - 2(-13) = -25 + 26 = 1. Step 5: Calculate y denominator. c1a2 - c2a1 = (-13)(3) - (-5)(4) = -39 + 20 = -19. Step 6: Calculate z denominator. a1b2 - a2b1 = 4(2) - 3(5) = 8 - 15 = -7. Step 7: Find x. x/1 = 1/(-7), so x = -1/7. Step 8: Find y. y/(-19) = 1/(-7), so y = 19/7. Step 9: Verify. 4(-1/7) + 5(19/7) = -4/7 + 95/7 = 91/7 = 13. Check: 3(-1/7) + 2(19/7) = -3/7 + 38/7 = 35/7 = 5. Answer: x = -1/7, y = 19/7.

Q4: Determine if the system x + 2y = 5 and 2x + 4y = 12 has a solution.

Step 1: Write equations in standard form. x + 2y = 5 and 2x + 4y = 12. Step 2: Check if equations are dependent (same line) or inconsistent (parallel). Multiply first equation by 2. Step 3: Calculation. 2(x + 2y) = 2(5) gives 2x + 4y = 10. Step 4: Compare with second equation. We have 2x + 4y = 10 but second equation says 2x + 4y = 12. Step 5: Analyze. The coefficients of x and y are proportional (ratio 1:2) but the constants are different (10 and 12). Step 6: Conclusion. These are parallel lines that never intersect. Answer: The system has NO SOLUTION (inconsistent system). The lines are parallel.

Q5: Write a system of linear equations for: 'A shop sells pens at Rs 5 and notebooks at Rs 8. If 10 items cost Rs 65, find quantities.'

Step 1: Define variables. Let x = number of pens, y = number of notebooks. Step 2: Write first equation from quantity condition. Total items = 10, so x + y = 10. Step 3: Write second equation from cost condition. Total cost = Rs 65, so 5x + 8y = 65. Step 4: Solve using substitution. From x + y = 10, get x = 10 - y. Step 5: Substitute in cost equation. 5(10 - y) + 8y = 65. Step 6: Expand. 50 - 5y + 8y = 65. Step 7: Simplify. 3y = 15, so y = 5. Step 8: Find x. x = 10 - 5 = 5. Step 9: Verify. Quantity: 5 + 5 = 10. Cost: 5(5) + 8(5) = 25 + 40 = 65. Answer: 5 pens and 5 notebooks.

Q6: Solve 5x - 3y = 4 and 7x + 2y = 20 using determinant method (Cramer's rule).

Step 1: Write in matrix form Ax = B where A is coefficient matrix, x is variable matrix, B is constant matrix. Step 2: Identify coefficients. a1 = 5, b1 = -3, c1 = 4; a2 = 7, b2 = 2, c2 = 20. Step 3: Calculate determinant D of coefficient matrix. D = a1*b2 - a2*b1 = 5(2) - 7(-3) = 10 + 21 = 31. Step 4: Calculate Dx (replace x coefficients with constants). Dx = c1*b2 - c2*b1 = 4(2) - 20(-3) = 8 + 60 = 68. Step 5: Calculate Dy (replace y coefficients with constants). Dy = a1*c2 - a2*c1 = 5(20) - 7(4) = 100 - 28 = 72. Step 6: Apply Cramer's rule. x = Dx/D = 68/31, y = Dy/D = 72/31. Step 7: Verify. 5(68/31) - 3(72/31) = 340/31 - 216/31 = 124/31 = 4. Check: 7(68/31) + 2(72/31) = 476/31 + 144/31 = 620/31 = 20. Answer: x = 68/31, y = 72/31.

Q7: For what value of k does the system kx + 3y = 8 and 2x + 6y = 7 have no solution?

Step 1: For no solution, lines must be parallel (same slope but different y-intercepts). Step 2: Rewrite equations in slope form. First: kx + 3y = 8, slope = -k/3. Second: 2x + 6y = 7, slope = -2/6 = -1/3. Step 3: For parallel lines, slopes must be equal. -k/3 = -1/3. Step 4: Solve for k. k/3 = 1/3, so k = 1. Step 5: Verify that constants are different when k = 1. First becomes x + 3y = 8, second is 2x + 6y = 7. Step 6: Multiply first by 2: 2x + 6y = 16 and 2x + 6y = 7. Same left side but different right side confirms no solution. Answer: k = 1.

Q8: Solve the system 2x + 3y = 11 and 5x - 4y = 2 and verify the solution.

Step 1: Use elimination method. Multiply first equation by 5 and second by 2 to eliminate x. Step 2: Calculate. 5(2x + 3y) = 5(11) gives 10x + 15y = 55. Step 3: Calculate. 2(5x - 4y) = 2(2) gives 10x - 8y = 4. Step 4: Subtract second from first. (10x + 15y) - (10x - 8y) = 55 - 4. Step 5: Simplify. 23y = 51. Step 6: Solve for y. y = 51/23. Step 7: Substitute in first equation. 2x + 3(51/23) = 11. Step 8: Simplify. 2x + 153/23 = 11. Step 9: Solve for x. 2x = 11 - 153/23 = 253/23 - 153/23 = 100/23. So x = 50/23. Step 10: Verify first equation. 2(50/23) + 3(51/23) = 100/23 + 153/23 = 253/23 = 11. Correct. Step 11: Verify second equation. 5(50/23) - 4(51/23) = 250/23 - 204/23 = 46/23 = 2. Correct. Answer: x = 50/23, y = 51/23.

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