Question

2 upon x + 6 upon y =13 ; 3 upon x + 4 upon y = 12

Answer 1

Answer:

$\large\bold\red{x=\frac{5}{13}\:,\:y=\frac{10}{13}}$

Step-by-step explanation:

Given,

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

And,

$\frac{3}{x} + \frac{4}{y} = 12 \: \: \: ........(2)$

Now,

Let's assume that,

• $\frac{1}{x} = a$
• $\frac{1}{y} = b$

Therefore,

We have,

$= > 2a + 6b = 13 \: \: \: .......(3)$

And,

$= > 3a + 4b = 12\: \: .......(4)$

Multiplying eqn (3) with 3 and (4) with 2 , and subtracting ,

We get,

$= >( 6a + 18b) -( 6a + 8b) = 39 - 26 \\ \\ = >10b = 13 \\ \\ = > b = \frac{13}{10} \\ \\ = > \frac{1}{y} = \frac{13}{10} \\ \\ = > y = \frac{10}{13}$

Putting the value of b in eqn (3),

We get,

$= > 2a + 6 \times \frac{13}{10} = 13 \\ \\ = > 2a = 13 - \frac{78}{10} \\ \\ = > 2a = \frac{130 - 78}{10} \\ \\ = > 2a = \frac{52}{10} \\ \\ = > a = \frac{26}{10} \\ \\ = > a = \frac{13}{5} \\ \\ = > \frac{1}{x} = \frac{13}{5} \\ \\ = > x = \frac{5}{13}$

Answer 2

Answer:

x=5/13 is the correct answer ❤️❤️❤️