CBSE Class 10 Mathematics Linear Equations Assignment Set 10

Linear Equations

Solve for \(x, y\) the following equations (Q1 to Q10)

Question 1. \(\frac{8}{x+2y} + \frac{3}{2x-y} = 3 \;;\; \frac{12}{x+2y} - \frac{6}{2x-y} = 1\)
Answer: Let \(u = \frac{1}{x+2y}\) and \(v = \frac{1}{2x-y}\).
The given equations become:
\(8u + 3v = 3\) --- (1)
\(12u - 6v = 1\) --- (2)
Multiply equation (1) by 2: \(16u + 6v = 6\) --- (3)
Add equation (2) and equation (3):
\((12u - 6v) + (16u + 6v) = 1 + 6\)
\(28u = 7 \implies u = \frac{7}{28} = \frac{1}{4}\)
Substitute \(u = \frac{1}{4}\) in equation (1):
\(8\left(\frac{1}{4}\right) + 3v = 3 \implies 2 + 3v = 3 \implies 3v = 1 \implies v = \frac{1}{3}\)
Now, revert back to original variables:
\(x + 2y = \frac{1}{u} = 4\) --- (4)
\(2x - y = \frac{1}{v} = 3\) --- (5)
Multiply equation (5) by 2: \(4x - 2y = 6\) --- (6)
Add equation (4) and equation (6):
\((x + 2y) + (4x - 2y) = 4 + 6\)
\(5x = 10 \implies x = 2\)
Substitute \(x = 2\) in equation (4):
\(2 + 2y = 4 \implies 2y = 2 \implies y = 1
Thus, the solution is \((2, 1)\).
In simple words: Substitute new variables to convert the non-linear equations into linear ones, solve for the substitutes, and then solve for \(x\) and \(y\).

Exam Tip: Always check if substitution simplifies fractions into standard linear forms before attempting cross-multiplication.

Question 2. \(\frac{x+y}{xy} = 2 \;;\; \frac{x-y}{xy} = 8\)
Answer: Split the fractions in both equations:
\(\frac{1}{y} + \frac{1}{x} = 2\) --- (1)
\(\frac{1}{y} - \frac{1}{x} = -8\) (Wait, looking at source: \(\frac{x-y}{xy} = 8 \implies \frac{1}{y} - \frac{1}{x} = 8\)) --- (2)
Let \(u = \frac{1}{x}\) and \(v = \frac{1}{y}\). The equations are:
\(v + u = 2\)
\(v - u = 8\)
Add both equations:
\(2v = 10 \implies v = 5 \implies \frac{1}{y} = 5 \implies y = \frac{1}{5}\)
Subtract the equations:\(-2u = -6 \implies u = 3 \implies \frac{1}{x} = 3 \implies x = \frac{1}{3}\)
Thus, the solution is \(\left(-\frac{1}{3}, \frac{1}{5}\right)\).
In simple words: Divide each term in the numerator by the denominator product \(xy\) to separate the variables into reciprocals.

Exam Tip: Separating terms of the numerator over a common denominator is a quick way to handle product terms in denominators.

Question 3. \(\frac{10}{x-y} + \frac{5}{x+y} = 6 \;;\; \frac{5}{x-y} + \frac{10}{x+y} = \frac{9}{2}\)
Answer: Let \(u = \frac{1}{x-y}\) and \(v = \frac{1}{x+y}\).
\(10u + 5v = 6\) --- (1)
\(5u + 10v = \frac{9}{2}\) --- (2)
Multiply equation (2) by 2: \(10u + 20v = 9\) --- (3)
Subtract equation (1) from equation (3):
\((10u + 20v) - (10u + 5v) = 9 - 6\)
\(15v = 3 \implies v = \frac{3}{15} = \frac{1}{5}\)
Substitute \(v = \frac{1}{5}\) in equation (1):
\(10u + 5\left(\frac{1}{5}\right) = 6 \implies 10u + 1 = 6 \implies 10u = 5 \implies u = \frac{1}{2}\)
Now find \(x\) and \(y\):
\(x - y = \frac{1}{u} = 2\) --- (4)
\(x + y = \frac{1}{v} = 5\) --- (5)
Add equation (4) and equation (5):
\(2x = 7 \implies x = \frac{7}{2}\)
Substitute \(x = \frac{7}{2}\) in equation (5):
\(\frac{7}{2} + y = 5 \implies y = 5 - \frac{7}{2} = \frac{3}{2}\)
Thus, the solution is \(\left(\frac{7}{2}, \frac{3}{2}\right)\).
In simple words: Use substitution for the binomial denominators, solve the simple linear system, and then solve for \(x\) and \(y\).

Exam Tip: Symmetric equations like this often simplify greatly if you add and subtract the two original equations.

Question 4. \(\frac{12}{2x-3y} + \frac{20}{3x+y} = \frac{31}{5} \;;\; \frac{10}{2x-3y} - \frac{16}{3x+y} = -3\)
Answer: Let \(u = \frac{1}{2x-3y}\) and \(v = \frac{1}{3x+y}\).
\(12u + 20v = \frac{31}{5}\) --- (1)
\(10u - 16v = -3\) --- (2)
Multiply equation (1) by 4: \(48u + 80v = \frac{124}{5}\)
Multiply equation (2) by 5: \(50u - 80v = -15\)
Add the two modified equations:
\((48u + 50u) = \frac{124}{5} - 15 \implies 98u = \frac{124 - 75}{5} = \frac{49}{5}\)
\(u = \frac{49}{5 \times 98} = \frac{1}{10}\)
Substitute \(u = \frac{1}{10}\) in equation (2):
\(10\left(\frac{1}{10}\right) - 16v = -3 \implies 1 - 16v = -3 \implies -16v = -4 \implies v = \frac{1}{4}\)
Now solve for \(x\) and \(y\):
\(2x - 3y = \frac{1}{u} = 10\) --- (3)
\(3x + y = \frac{1}{v} = 4\) --- (4)
Multiply equation (4) by 3: \(9x + 3y = 12\) --- (5)
Add equation (3) and equation (5):
\((2x - 3y) + (9x + 3y) = 10 + 12 \implies 11x = 22 \implies x = 2\)
Substitute \(x = 2\) in equation (4):
\(3(2) + y = 4 \implies 6 + y = 4 \implies y = -2\)
Thus, the solution is \((2, -2)\).
In simple words: Substitute variables to clear fractions, eliminate \(v\) using coefficients, and solve the resulting system for \(x\) and \(y\).

Exam Tip: Handle fractional constants carefully by multiplying through to clear denominators early.

Question 5. \(\frac{7}{x-2y} + \frac{9}{x+2y} = \frac{13}{2} \;;\; \frac{8}{x-2y} + \frac{15}{x+2y} = 9\)
Answer: Let \(u = \frac{1}{x-2y}\) and \(v = \frac{1}{x+2y}\).
\(7u + 9v = \frac{13}{2} \implies 14u + 18v = 13\) --- (1)
\(8u + 15v = 9 \implies 24u + 45v = 27\) --- (2)
Multiply equation (1) by 12: \(168u + 216v = 156\)
Multiply equation (2) by 7: \(168u + 315v = 189\)
Subtract the equations:
\((315 - 216)v = 189 - 156 \implies 99v = 33 \implies v = \frac{33}{99} = \frac{1}{3}\)
Substitute \(v = \frac{1}{3}\) in \(14u + 18v = 13\):
\(14u + 18\left(\frac{1}{3}\right) = 13 \implies 14u + 6 = 13 \implies 14u = 7 \implies u = \frac{1}{2}\)
Now solve for \(x\) and \(y\):
\(x - 2y = \frac{1}{u} = 2\) --- (3)
\(x + 2y = \frac{1}{v} = 3\) --- (4)
Add equation (3) and equation (4):
\(2x = 5 \implies x = \frac{5}{2}\)
Substitute \(x = \frac{5}{2}\) in equation (4):
\(\frac{5}{2} + 2y = 3 \implies 2y = 3 - \frac{5}{2} = \frac{1}{2} \implies y = \frac{1}{4}\)
Thus, the solution is \(\left(\frac{5}{2}, \frac{1}{4}\right)\).
In simple words: Substitute variables, solve for \(u\) and \(v\) using elimination, and then find \(x\) and \(y\).

Exam Tip: Multiplying equations to make coefficients equal avoids messy fraction arithmetic mid-way.

Question 6. \(4x + 3y = xy \;;\; \frac{9}{x} + \frac{8}{y} = 7\)
Answer: From the first equation, divide both sides by \(xy\):
\(\frac{4}{y} + \frac{3}{x} = 1\) --- (1)
The second equation is:
\(\frac{9}{x} + \frac{8}{y} = 7\) --- (2)
Let \(u = \frac{1}{x}\) and \(v = \frac{1}{y}\). Equations become:
\(3u + 4v = 1\) --- (3)
\(9u + 8v = 7\) --- (4)
Multiply equation (3) by 3: \(9u + 12v = 3\) --- (5)
Subtract equation (4) from equation (5):
\((12v - 8v) = 3 - 7 \implies 4v = -4 \implies v = -1\)
Substitute \(v = -1\) in equation (3):
\(3u + 4(-1) = 1 \implies 3u - 4 = 1 \implies 3u = 5 \implies u = \frac{5}{3}\)
Now find \(x\) and \(y\):
\(x = \frac{1}{u} = \frac{3}{5}\)
\(y = \frac{1}{v} = -1\)
Thus, the solution is \(\left(\frac{3}{5}, -1\right)\).
In simple words: Divide the product equation by \(xy\) to convert it into a standard reciprocal form.

Exam Tip: Equations containing \(xy\) terms can always be linearized by dividing through by \(xy\).

Question 7. \(\frac{5}{x-1} + \frac{1}{y-2} = 2 \;;\; \frac{6}{x-1} - \frac{3}{y-2} = 1\)
Answer: Let \(u = \frac{1}{x-1}\) and \(v = \frac{1}{y-2}\).
\(5u + v = 2\) --- (1)
\(6u - 3v = 1\) --- (2)
Multiply equation (1) by 3: \(15u + 3v = 6\) --- (3)
Add equation (2) and equation (3):
\((6u + 15u) = 1 + 6 \implies 21u = 7 \implies u = \frac{7}{21} = \frac{1}{3}\)
Substitute \(u = \frac{1}{3}\) in equation (1):
\(5\left(\frac{1}{3}\right) + v = 2 \implies \frac{5}{3} + v = 2 \implies v = 2 - \frac{5}{3} = \frac{1}{3}\)
Now solve for \(x\) and \(y\):
\(x - 1 = \frac{1}{u} = 3 \implies x = 4\)
\(y - 2 = \frac{1}{v} = 3 \implies y = 5\)
Thus, the solution is \((4, 5)\).
In simple words: Substitute \(u\) and \(v\) for the shifted terms, solve the linear system, and shift back to find \(x\) and \(y\).

Exam Tip: Do not open parentheses in denominators; treat expressions like \((x-1)\) as single units via substitution.

Question 8. \(\frac{30}{x-y} + \frac{44}{x+y} = 10 \;;\; \frac{40}{x-y} + \frac{55}{x+y} = 13\)
Answer: Let \(u = \frac{1}{x-y}\) and \(v = \frac{1}{x+y}\).
\(30u + 44v = 10 \implies 15u + 22v = 5\) --- (1)
\(40u + 55v = 13 \implies \frac{40}{5}u + \frac{55}{5}v = \frac{13}{5} \implies 8u + 11v = \frac{13}{5}\) --- (2)
Multiply equation (2) by 2: \(16u + 22v = \frac{26}{5}\) --- (3)
Subtract equation (1) from equation (3):
\(16u - 15u = \frac{26}{5} - 5 \implies u = \frac{26 - 25}{5} = \frac{1}{5}\)
Substitute \(u = \frac{1}{5}\) in equation (1):
\(15\left(\frac{1}{5}\right) + 22v = 5 \implies 3 + 22v = 5 \implies 22v = 2 \implies v = \frac{1}{11}\)
Now solve for \(x\) and \(y\):
\(x - y = \frac{1}{u} = 5\) --- (4)
\(x + y = \frac{1}{v} = 11\) --- (5)
Add equation (4) and equation (5):
\(2x = 16 \implies x = 8\)
Substitute \(x = 8\) in equation (5):
\(8 + y = 11 \implies y = 3\)
Thus, the solution is \((8, 3)\).
In simple words: Simplify equations by dividing common factors if possible, then solve using substitution.

Exam Tip: Always reduce equations by common numerical factors before applying elimination.

Question 9. \(\frac{x}{a} - \frac{y}{b} = 0 \;;\; ax + by = a^2 + b^2\)
Answer: From the first equation:
\(\frac{x}{a} = \frac{y}{b} \implies y = \frac{b}{a}x\)
Substitute \(y = \frac{b}{a}x\) into the second equation:\;
\(ax + b\left(\frac{b}{a}x\right) = a^2 + b^2\)
\(ax + \frac{b^2}{a}x = a^2 + b^2\)
Take \(x\) common:
\(x\left(a + \frac{b^2}{a}\right) = a^2 + b^2\)
\(x\left(\frac{a^2 + b^2}{a}\right) = a^2 + b^2\)
Divide both sides by \((a^2 + b^2)\):
\(\frac{x}{a} = 1 \implies x = a\)
Substitute \(x = a\) into \(y = \frac{b}{a}x\):
\(y = \frac{b}{a}(a) = b\)
Thus, the solution is \((a, b)\).
In simple words: Express one variable in terms of the other using the first equation and substitute it into the second.

Exam Tip: Literal coefficients (\(a, b\)) require careful algebraic grouping instead of numerical arithmetic.

Question 10. \(\frac{1}{3x+y} + \frac{1}{3x-y} = \frac{3}{4} \;;\; \frac{1}{2(3x+y)} - \frac{1}{2(3x-y)} = \frac{-1}{8}\)
Answer: Let \(u = \frac{1}{3x+y}\) and \(v = \frac{1}{3x-y}\).
\(u + v = \frac{3}{4}\) --- (1)
\(\frac{1}{2}u - \frac{1}{2}v = -\frac{1}{8} \implies u - v = -2\left(\frac{1}{8}\right) = -\frac{1}{4}\) --- (2)
Add equation (1) and equation (2):
\((u + v) + (u - v) = \frac{3}{4} - \frac{1}{4}\)
\(2u = \frac{2}{4} = \frac{1}{2} \implies u = \frac{1}{4}\)
Substitute \(u = \frac{1}{4}\) in equation (1):
\(\frac{1}{4} + v = \frac{3}{4} \implies v = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2}\)
Now solve for \(x\) and \(y\):
\(3x + y = \frac{1}{u} = 4\) --- (3)
\(3x - y = \frac{1}{v} = 2\) --- (4)
Add equation (3) and equation (4):
\(6x = 6 \implies x = 1\)
Substitute \(x = 1\) in equation (3):
\(3(1) + y = 4 \implies 3 + y = 4 \implies y = 1\)
Thus, the solution is \((1, 1)\).
In simple words: Substitute variables, simplify the second equation by multiplying by 2, and solve the system easily.

Exam Tip: Multiplying through by denominators like 2 helps eliminate fractions before adding or subtracting equations.