Question

solve the following simultaneous equation
1/3x-1/4y+1=0 1/5x+1/2y=4/15
​

Answer 1

Answer

$\sf \dashrightarrow \dfrac{1}{3x} - \dfrac{1}{4y} + 1 = 0 \: \: --- (i)$

$\sf \dashrightarrow \dfrac{1}{5x} + \dfrac{1}{2y} = \dfrac{4}{15} \: \: --- (ii)$

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

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

So, the equations become

$\sf \dashrightarrow \dfrac{u}{3} + \dfrac{v}{4} = -1$

$\sf \dashrightarrow \dfrac{u}{5} + \dfrac{v}{2} = \dfrac{4}{15}$

By first equation,

$\sf \dashrightarrow \dfrac{u}{3} + \dfrac{v}{4} = -1$

$\sf \dashrightarrow \dfrac{4u + 3v}{12} = -1$

$\sf \dashrightarrow 4u + 3v = -12$

By second equation,

$\sf \dashrightarrow \dfrac{u}{5} + \dfrac{v}{2} = \dfrac{4}{15}$

$\sf \dashrightarrow \dfrac{2u + 5v}{10} = \dfrac{4}{15}$

$\sf \dashrightarrow 2u + 5v = \dfrac{4}{15} \times 10$

$\sf \dashrightarrow 2u + 5v = \dfrac{40}{15}$

By first equation,

$\sf \dashrightarrow 4u + 3v = -12$

$\sf \dashrightarrow 4u = -12 - 3v$

$\sf \dashrightarrow u = \dfrac{-12 - 3v}{4}$

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

$\sf \dashrightarrow 2u + 5v = \dfrac{40}{15}$

$\sf \dashrightarrow 2 \bigg( \dfrac{-12 - 3v}{4} \bigg) + 5v = \dfrac{40}{15}$

$\sf \dashrightarrow \dfrac{-24 - 6v}{4} + 5v = \dfrac{40}{15}$

$\sf \dashrightarrow \dfrac{-24 - 6v + 20v}{4} = \dfrac{40}{15}$

$\sf \dashrightarrow \dfrac{-24 + 14v}{4} = \dfrac{40}{15}$

$\sf \dashrightarrow -24 + 14v = 4 \bigg( \dfrac{40}{15} \bigg)$

$\sf \dashrightarrow -24 + 14v = \dfrac{160}{15}$

$\sf \dashrightarrow -24 + 14v = \dfrac{32}{3}$

$\sf \dashrightarrow 14v = \dfrac{32}{3} + 24$

$\sf \dashrightarrow 14v = \dfrac{32 + 72}{3}$

$\sf \dashrightarrow 14v = \dfrac{104}{3}$

$\sf \dashrightarrow v = \dfrac{104}{3} \div 14$

$\sf \dashrightarrow v = \dfrac{104}{3} \times \dfrac{1}{14}$

$\sf \dashrightarrow v = \dfrac{104}{42}$

Now, let's find the value of u by first equation.

$\sf \dashrightarrow 4u + 3v = -12$

$\sf \dashrightarrow 4u + 3 \bigg( \dfrac{104}{42} \bigg) = -12$

$\sf \dashrightarrow 4u + \dfrac{312}{42} = -12$

$\sf \dashrightarrow 4u + \dfrac{104}{13} = -12$

$\sf \dashrightarrow \dfrac{52u + 104}{13} = -12$

$\sf \dashrightarrow 52u + 104 = -12 \times 13$

$\sf \dashrightarrow 52u + 104 = -156$

$\sf \dashrightarrow 52u = -156 - 104$

$\sf \dashrightarrow 52u = -260$

$\sf \dashrightarrow u = \dfrac{-260}{52}$

$\sf \dashrightarrow u = -5$

As we know that,

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

$\sf \dashrightarrow \dfrac{1}{x} = -5$

$\sf \dashrightarrow x = -5$

We also know that,

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

$\sf \dashrightarrow \dfrac{1}{y} = \dfrac{104}{42}$

$\sf \dashrightarrow y = \dfrac{42}{104}$

$\sf \dashrightarrow y = \dfrac{21}{52}$

Hence, the values of x and y are -5 and $\bf \dfrac{21}{52}$ respectively.