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

I'll solve each pair of equations by reducing them to linear equations. Let me work through each one carefully.

Problem (i):
$$\frac{1}{2x} + \frac{1}{3y} = 2$$
$$\frac{1}{3x} + \frac{1}{2y} = \frac{13}{6}$$

Let me substitute: $u = \frac{1}{x}$ and $v = \frac{1}{y}$

The equations become:
$$\frac{u}{2} + \frac{v}{3} = 2$$ ... (1)
$$\frac{u}{3} + \frac{v}{2} = \frac{13}{6}$$ ... (2)

Multiply equation (1) by 6:
$$3u + 2v = 12$$ ... (3)

Multiply equation (2) by 6:
$$2u + 3v = 13$$ ... (4)

Now solve the system:
From (3): $3u + 2v = 12$
From (4): $2u + 3v = 13$

Multiply (3) by 3: $9u + 6v = 36$
Multiply (4) by 2: $4u + 6v = 26$

Subtract: $(9u + 6v) - (4u + 6v) = 36 - 26$
$5u = 10$
$u = 2$

Substitute into (3): $3(2) + 2v = 12$
$6 + 2v = 12$
$2v = 6$
$v = 3$

Since $u = \frac{1}{x} = 2$, we get $x = \frac{1}{2}$
Since $v = \frac{1}{y} = 3$, we get $y = \frac{1}{3}$

Answer (i): $x = \frac{1}{2}, y = \frac{1}{3}$

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Problem (ii):
$$\frac{5}{x-1} + \frac{1}{y-2} = 2$$
$$\frac{6}{x-1} - \frac{3}{y-2} = 1$$

Let me substitute: $u = \frac{1}{x-1}$ and $v = \frac{1}{y-2}$

The equations become:
$$5u + v = 2$$ ... (1)
$$6u - 3v = 1$$ ... (2)

From (1): $v = 2 - 5u$

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

Substitute back: $v = 2 - 5(\frac{1}{3}) = 2 - \frac{5}{3} = \frac{6-5}{3} = \frac{1}{3}$

Since $u = \frac{1}{x-1} = \frac{1}{3}$, we get $x - 1 = 3$, so $x = 4$
Since $v = \frac{1}{y-2} = \frac{1}{3}$, we get $y - 2 = 3$, so $y = 5$

Answer (ii): $x = 4, y = 5$

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Problem (iii):
$$\frac{10}{x+y} + \frac{2}{x-y} = 4$$
$$\frac{15}{x+y} - \frac{5}{x-y} = -2$$

Let me substitute: $u = \frac{1}{x+y}$ and $v = \frac{1}{x-y}$

The equations become:
$$10u + 2v = 4$$ ... (1)
$$15u - 5v = -2$$ ... (2)

Simplify (1) by dividing by 2:
$$5u + v = 2$$ ... (3)

From (3): $v = 2 - 5u$

Substitute into (2):
$15u - 5(2 - 5u) = -2$
$15u - 10 + 25u = -2$
$40u = 8$
$u = \frac{8}{40} = \frac{1}{5}$

Substitute back: $v = 2 - 5(\frac{1}{5}) = 2 - 1 = 1$

Since $u = \frac{1}{x+y} = \frac{1}{5}$, we get $x + y = 5$ ... (4)
Since $v = \frac{1}{x-y} = 1$, we get $x - y = 1$ ... (5)

Add (4) and (5): $2x = 6$, so $x = 3$
Subtract (5) from (4): $2y = 4$, so $y = 2$

Answer (iii): $x = 3, y = 2$

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Problem (iv):
$$\frac{1}{3x+y} + \frac{1}{3x-y} = \frac{3}{4}$$
$$\frac{1}{2(3x+y)} - \frac{1}{2(3x-y)} = \frac{-1}{8}$$

Let me substitute: $u = \frac{1}{3x+y}$ and $v = \frac{1}{3x-y}$

The equations become:
$$u + v = \frac{3}{4}$$ ... (1)
$$\frac{u}{2} - \frac{v}{2} = \frac{-1}{8}$$ ... (2)

Multiply (2) by 2:
$$u - v = \frac{-1}{4}$$ ... (3)

Add (1) and (3): $2u = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$
So $u = \frac{1}{4}$

Substitute into (1): $\frac{1}{4} + v = \frac{3}{4}$
$v = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$

Since $u = \frac{1}{3x+y} = \frac{1}{4}$, we get $3x + y = 4$ ... (4)
Since $v = \frac{1}{3x-y} = \frac{1}{2}$, we get $3x - y = 2$ ... (5)

Add (4) and (5): $6x = 6$, so $x = 1$
Substitute into (4): $3(1) + y = 4$, so $y = 1$

Answer (iv): $x = 1, y = 1$

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Problem (v):
$$6x + 3y = 6xy$$
$$2x + 4y = 5xy$$

Divide both equations by $xy$ (assuming $x \neq 0$ and $y \neq 0$):

First equation: $\frac{6x}{xy} + \frac{3y}{xy} = 6$
$\frac{6}{y} + \frac{3}{x} = 6$ ... (1)

Second equation: $\frac{2x}{xy} + \frac{4y}{xy} = 5$
$\frac{2}{y} + \frac{4}{x} = 5$ ... (2)

Let me substitute: $u = \frac{1}{x}$ and $v = \frac{1}{y}$

The equations become:
$$3u + 6v = 6$$ ... (3)
$$4u + 2v = 5$$ ... (4)

Simplify (3) by dividing by 3:
$$u + 2v = 2$$ ... (5)

From (5): $u = 2 - 2v$

Substitute into (4):
$4(2 - 2v) + 2v = 5$
$8 - 8v + 2v = 5$
$8 - 6v = 5$
$-6v = -3$
$v = \frac{1}{2}$

Substitute back: $u = 2 - 2(\frac{1}{2}) = 2 - 1 = 1$

Since $u = \frac{1}{x} = 1$, we get $x = 1$
Since $v = \frac{1}{y} = \frac{1}{2}$, we get $y = 2$

Let me verify:
First equation: $6(1) + 3(2) = 6 + 6 = 12$ and $6(1)(2) = 12$ ✓
Second equation: $2(1) + 4(2) = 2 + 8 = 10$ and $5(1)(2) = 10$ ✓

Answer (v): $x = 1, y = 2$

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Final Answer:
(i) $x = \frac{1}{2}, y = \frac{1}{3}$
(ii) $x = 4, y = 5$
(iii) $x = 3, y = 2$
(iv) $x = 1, y = 1$
(v) $x = 1, y = 2$