CBSE Class 10 Mathematics Pair of Linear Equation In Two Variables ... - Free Printable

To solve the given pairs of equations by reducing them to a pair of linear equations, we will use substitution and simplification techniques. Let's go through each problem step by step.

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(i)

\[
\frac{1}{2x} + \frac{1}{3y} = 2
\]
\[
\frac{1}{3x} + \frac{1}{2y} = \frac{13}{6}
\]

#### Step 1: Introduce substitutions
Let:
\[
u = \frac{1}{x} \quad \text{and} \quad v = \frac{1}{y}
\]
Then the equations become:
\[
\frac{u}{2} + \frac{v}{3} = 2
\]
\[
\frac{u}{3} + \frac{v}{2} = \frac{13}{6}
\]

#### Step 2: Eliminate fractions
Multiply the first equation by 6:
\[
6 \left( \frac{u}{2} + \frac{v}{3} \right) = 6 \cdot 2
\]
\[
3u + 2v = 12 \quad \text{(Equation 1)}
\]

Multiply the second equation by 6:
\[
6 \left( \frac{u}{3} + \frac{v}{2} \right) = 6 \cdot \frac{13}{6}
\]
\[
2u + 3v = 13 \quad \text{(Equation 2)}
\]

#### Step 3: Solve the system of linear equations
We have:
\[
3u + 2v = 12 \quad \text{(Equation 1)}
\]
\[
2u + 3v = 13 \quad \text{(Equation 2)}
\]

Multiply Equation 1 by 3 and Equation 2 by 2 to eliminate \( v \):
\[
9u + 6v = 36 \quad \text{(Equation 3)}
\]
\[
4u + 6v = 26 \quad \text{(Equation 4)}
\]

Subtract Equation 4 from Equation 3:
\[
(9u + 6v) - (4u + 6v) = 36 - 26
\]
\[
5u = 10
\]
\[
u = 2
\]

Substitute \( u = 2 \) back into Equation 1:
\[
3(2) + 2v = 12
\]
\[
6 + 2v = 12
\]
\[
2v = 6
\]
\[
v = 3
\]

#### Step 4: Back-substitute to find \( x \) and \( y \)
Recall that \( u = \frac{1}{x} \) and \( v = \frac{1}{y} \):
\[
u = 2 \implies \frac{1}{x} = 2 \implies x = \frac{1}{2}
\]
\[
v = 3 \implies \frac{1}{y} = 3 \implies y = \frac{1}{3}
\]

Thus, the solution is:
\[
\boxed{\left( \frac{1}{2}, \frac{1}{3} \right)}
\]

---

(ii)

\[
\frac{5}{x-1} + \frac{1}{y-2} = 2
\]
\[
\frac{6}{x-1} - \frac{3}{y-2} = 1
\]

#### Step 1: Introduce substitutions
Let:
\[
u = \frac{1}{x-1} \quad \text{and} \quad v = \frac{1}{y-2}
\]
Then the equations become:
\[
5u + v = 2 \quad \text{(Equation 1)}
\]
\[
6u - 3v = 1 \quad \text{(Equation 2)}
\]

#### Step 2: Solve the system of linear equations
We have:
\[
5u + v = 2 \quad \text{(Equation 1)}
\]
\[
6u - 3v = 1 \quad \text{(Equation 2)}
\]

From Equation 1, solve for \( v \):
\[
v = 2 - 5u \quad \text{(Equation 3)}
\]

Substitute \( v = 2 - 5u \) into Equation 2:
\[
6u - 3(2 - 5u) = 1
\]
\[
6u - 6 + 15u = 1
\]
\[
21u - 6 = 1
\]
\[
21u = 7
\]
\[
u = \frac{1}{3}
\]

Substitute \( u = \frac{1}{3} \) back into Equation 3:
\[
v = 2 - 5 \left( \frac{1}{3} \right)
\]
\[
v = 2 - \frac{5}{3}
\]
\[
v = \frac{6}{3} - \frac{5}{3}
\]
\[
v = \frac{1}{3}
\]

#### Step 3: Back-substitute to find \( x \) and \( y \)
Recall that \( u = \frac{1}{x-1} \) and \( v = \frac{1}{y-2} \):
\[
u = \frac{1}{3} \implies \frac{1}{x-1} = \frac{1}{3} \implies x - 1 = 3 \implies x = 4
\]
\[
v = \frac{1}{3} \implies \frac{1}{y-2} = \frac{1}{3} \implies y - 2 = 3 \implies y = 5
\]

Thus, the solution is:
\[
\boxed{(4, 5)}
\]

---

(iii)

\[
\frac{10}{x+y} + \frac{2}{x-y} = 4
\]
\[
\frac{15}{x+y} - \frac{5}{x-y} = -2
\]

#### Step 1: Introduce substitutions
Let:
\[
u = \frac{1}{x+y} \quad \text{and} \quad v = \frac{1}{x-y}
\]
Then the equations become:
\[
10u + 2v = 4 \quad \text{(Equation 1)}
\]
\[
15u - 5v = -2 \quad \text{(Equation 2)}
\]

#### Step 2: Simplify the equations
Divide Equation 1 by 2:
\[
5u + v = 2 \quad \text{(Equation 3)}
\]

Divide Equation 2 by 5:
\[
3u - v = -\frac{2}{5} \quad \text{(Equation 4)}
\]

#### Step 3: Solve the system of linear equations
Add Equation 3 and Equation 4:
\[
(5u + v) + (3u - v) = 2 - \frac{2}{5}
\]
\[
8u = \frac{10}{5} - \frac{2}{5}
\]
\[
8u = \frac{8}{5}
\]
\[
u = \frac{1}{5}
\]

Substitute \( u = \frac{1}{5} \) back into Equation 3:
\[
5 \left( \frac{1}{5} \right) + v = 2
\]
\[
1 + v = 2
\]
\[
v = 1
\]

#### Step 4: Back-substitute to find \( x \) and \( y \)
Recall that \( u = \frac{1}{x+y} \) and \( v = \frac{1}{x-y} \):
\[
u = \frac{1}{5} \implies \frac{1}{x+y} = \frac{1}{5} \implies x + y = 5
\]
\[
v = 1 \implies \frac{1}{x-y} = 1 \implies x - y = 1
\]

Solve the system:
\[
x + y = 5
\]
\[
x - y = 1
\]

Add the equations:
\[
(x + y) + (x - y) = 5 + 1
\]
\[
2x = 6
\]
\[
x = 3
\]

Substitute \( x = 3 \) into \( x + y = 5 \):
\[
3 + y = 5
\]
\[
y = 2
\]

Thus, the solution is:
\[
\boxed{(3, 2)}
\]

---

(iv)

\[
\frac{1}{3x+y} + \frac{1}{3x-y} = \frac{3}{4}
\]
\[
\frac{1}{2(3x+y)} - \frac{1}{2(3x-y)} = \frac{-1}{8}
\]

#### Step 1: Introduce substitutions
Let:
\[
u = \frac{1}{3x+y} \quad \text{and} \quad v = \frac{1}{3x-y}
\]
Then the equations become:
\[
u + v = \frac{3}{4} \quad \text{(Equation 1)}
\]
\[
\frac{u}{2} - \frac{v}{2} = \frac{-1}{8}
\]

Multiply the second equation by 2:
\[
u - v = \frac{-1}{4} \quad \text{(Equation 2)}
\]

#### Step 2: Solve the system of linear equations
We have:
\[
u + v = \frac{3}{4} \quad \text{(Equation 1)}
\]
\[
u - v = \frac{-1}{4} \quad \text{(Equation 2)}
\]

Add Equation 1 and Equation 2:
\[
(u + v) + (u - v) = \frac{3}{4} + \frac{-1}{4}
\]
\[
2u = \frac{2}{4}
\]
\[
2u = \frac{1}{2}
\]
\[
u = \frac{1}{4}
\]

Substitute \( u = \frac{1}{4} \) back into Equation 1:
\[
\frac{1}{4} + v = \frac{3}{4}
\]
\[
v = \frac{3}{4} - \frac{1}{4}
\]
\[
v = \frac{2}{4}
\]
\[
v = \frac{1}{2}
\]

#### Step 3: Back-substitute to find \( x \) and \( y \)
Recall that \( u = \frac{1}{3x+y} \) and \( v = \frac{1}{3x-y} \):
\[
u = \frac{1}{4} \implies \frac{1}{3x+y} = \frac{1}{4} \implies 3x + y = 4
\]
\[
v = \frac{1}{2} \implies \frac{1}{3x-y} = \frac{1}{2} \implies 3x - y = 2
\]

Solve the system:
\[
3x + y = 4
\]
\[
3x - y = 2
\]

Add the equations:
\[
(3x + y) + (3x - y) = 4 + 2
\]
\[
6x = 6
\]
\[
x = 1
\]

Substitute \( x = 1 \) into \( 3x + y = 4 \):
\[
3(1) + y = 4
\]
\[
3 + y = 4
\]
\[
y = 1
\]

Thus, the solution is:
\[
\boxed{(1, 1)}
\]

---

(v)

\[
6x + 3y = 6xy
\]
\[
2x + 4y = 5xy
\]

#### Step 1: Simplify the equations
Divide the first equation by 3:
\[
2x + y = 2xy \quad \text{(Equation 1)}
\]

Divide the second equation by 2:
\[
x + 2y = \frac{5}{2}xy \quad \text{(Equation 2)}
\]

#### Step 2: Introduce substitutions
Let:
\[
u = \frac{1}{x} \quad \text{and} \quad v = \frac{1}{y}
\]
Then:
\[
x = \frac{1}{u} \quad \text{and} \quad y = \frac{1}{v}
\]

Substitute into Equation 1:
\[
2 \left( \frac{1}{u} \right) + \frac{1}{v} = 2 \left( \frac{1}{u} \right) \left( \frac{1}{v} \right)
\]
\[
\frac{2}{u} + \frac{1}{v} = \frac{2}{uv}
\]
Multiply through by \( uv \):
\[
2v + u = 2 \quad \text{(Equation 3)}
\]

Substitute into Equation 2:
\[
\frac{1}{u} + 2 \left( \frac{1}{v} \right) = \frac{5}{2} \left( \frac{1}{u} \right) \left( \frac{1}{v} \right)
\]
\[
\frac{1}{u} + \frac{2}{v} = \frac{5}{2uv}
\]
Multiply through by \( 2uv \):
\[
2v + 4u = 5 \quad \text{(Equation 4)}
\]

#### Step 3: Solve the system of linear equations
We have:
\[
2v + u = 2 \quad \text{(Equation 3)}
\]
\[
4u + 2v = 5 \quad \text{(Equation 4)}
\]

Subtract Equation 3 from Equation 4:
\[
(4u + 2v) - (2v + u) = 5 - 2
\]
\[
3u = 3
\]
\[
u = 1
\]

Substitute \( u = 1 \) back into Equation 3:
\[
2v + 1 = 2
\]
\[
2v = 1
\]
\[
v = \frac{1}{2}
\]

#### Step 4: Back-substitute to find \( x \) and \( y \)
Recall that \( u = \frac{1}{x} \) and \( v = \frac{1}{y} \):
\[
u = 1 \implies \frac{1}{x} = 1 \implies x = 1
\]
\[
v = \frac{1}{2} \implies \frac{1}{y} = \frac{1}{2} \implies y = 2
\]

Thus, the solution is:
\[
\boxed{(1, 2)}
\]

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Final Answers:

\[
\boxed{\left( \frac{1}{2}, \frac{1}{3} \right), (4, 5), (3, 2), (1, 1), (1, 2)}
\]