Question

Plzz help me........ ​

Answer 1

Answer:

1/2x + 1/3y = 2

multiply this equation by 2/3

2/3 × 1/2x + 2/3 × 1/3y = 4/3

1/3x + 2/9y = 4/3

now second equation 1/3x + 1/2y = 13/6

Subtract (2) from (1)

2/9y - 1/2y = 4/3 - 13/6

1/y. ( 2/9 - 1/2) = (8 -13)/6

( 36 - 9)/18 = -5y /6

27/18 = -5y /6

9/6 = -5y/6

9 = -5y

y = -9/5

Substitute y = -9/5 in (1) eq

1/2x + 1/3y = 2

1/2x + 1/3(-9/5) = 2

1/2x -5/27 = 2

1/2x = 2+ 5/27

1/2x = (54+5)/27

1/2x = 59/27

27 = 118 x

x = 27/118

#answerwithquality #BAL

Answer 2

Answer:

Given pair is:

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

$let \: \frac{1}{x} = a \: and \: \frac{1}{y} = b$

Therefore,

$\frac{1}{2} a + \frac{1}{3} b = 2 \\ \\ \frac{3a + 2b}{6} = 2 \\ \\ 3a + 2b = 12 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - - (1)$

and,

$\frac{1}{3} a + \frac{1}{2} b = \frac{13}{6} \\ \\ \frac{2a + 3b}{6} = \frac{13}{6} \\ \\2a + 3b = 13 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - - (2)$

eq(1) x 3 and eq(2) x 2

$9a + 6b = 36 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - - (3) \\ 4a + 6b = 26 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - - (4)$

By Elimination Method of eq(3) and eq(4)

$9a + 6b = 36 \: \: \: \: \: \: \: \: \: \\ 4a + 6b = 26 \\ - \: \: \: - \: \: \: \: \: \: \: \: \: \: \: - \\ 5a = 10 \\ a = 2$

Put a = 2 in eq(1)

$3(2) + 2b = 12 \\ 6 + 2b = 12 \\ 2b = 6\\ b = 3$

Therefore,

$\frac{1}{x} = a \\ \\ \frac{1}{x} = 5 \\ \\ x = \frac{1}{5}$

and,

$\frac{1}{y} = b \\ \\ \frac{1}{y} = 3 \\ \\ y = \frac{1}{3}$

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