Question

solve the system of equations ,,,2/x+3/y=2 and 4/x-9/y=-1​

Answer 1

Answer

$\sf \dashrightarrow \dfrac{2}{x} + \dfrac{3}{y} = 2 \: \: --- (i)$

$\sf \dashrightarrow \dfrac{4}{x} - \dfrac{9}{y} = -1 \: \: --- (ii)$

Let $\sf \dfrac{1}{x}$ be u.

Let $\sf \dfrac{1}{y}$ be v.

So, the equations become

$\sf \dashrightarrow 2u + 3v = 2$

$\sf \dashrightarrow 4u - 9v = -1$

By first equation,

$\sf \dashrightarrow 2u + 3v = 2$

$\sf \dashrightarrow 2u = 2 - 3v$

$\sf \dashrightarrow u = \dfrac{2 - 3v}{2}$

Now, let's find the value of v by second equation.

$\sf \dashrightarrow 4u - 9v = -1$

$\sf \dashrightarrow 4 \bigg( \dfrac{2 - 3v}{2}\bigg) - 9v = -1$

$\sf \dashrightarrow \dfrac{8 - 12v}{2} - 9v = -1$

$\sf \dashrightarrow \dfrac{8 - 12v - 18v}{2} = -1$

$\sf \dashrightarrow \dfrac{8 - 20v}{2} = -1$

$\sf \dashrightarrow 8 - 20v = -1 \times 2$

$\sf \dashrightarrow 8 - 20v = -2$

$\sf \dashrightarrow -20v = -2 - 8$

$\sf \dashrightarrow -20v = -10$

$\sf \dashrightarrow v = \dfrac{-10}{-20}$

$\sf \dashrightarrow v = \dfrac{1}{2}$

Now, we can find the value of u by first equation.

$\sf \dashrightarrow 2u + 3v = 2$

$\sf \dashrightarrow 2u + 3 \bigg( \dfrac{1}{2} \bigg) = 2$

$\sf \dashrightarrow 2u + \dfrac{3}{2} = 2$

$\sf \dashrightarrow \dfrac{4u + 3}{2} = 2$

$\sf \dashrightarrow 4u + 3 = 2 \times 2$

$\sf \dashrightarrow 4u + 3 = 4$

$\sf \dashrightarrow 4u = 4 - 3$

$\sf \dashrightarrow 4u = 1$

$\sf \dashrightarrow u = \dfrac{1}{4}$

We know that,

$\sf \dashrightarrow \dfrac{1}{x} = u$

$\sf \dashrightarrow \dfrac{1}{x} = \dfrac{1}{4}$

$\sf \dashrightarrow x = 4$

We also know that,

$\sf \dashrightarrow \dfrac{1}{y} = v$

$\sf \dashrightarrow \dfrac{1}{y} = \dfrac{1}{2}$

$\sf \dashrightarrow y = 2$

Hence, the values of x and y are 4 and 2 respectively.