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.
TL;DR: Free step-by-step Maharashtra (SSC) Class 10 Mathematics solutions for "Linear Equations in Two Variables" — important questions with detailed answers…
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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 -…
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: Fin…
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 line…
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. Quant…
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