Question

SOLVE IT BY ELEMINATION OR SUBSTITUTION PROCESS​

Answer 1

Answer

We'll solve this by substitution method.

$\sf \dashrightarrow \dfrac{10}{x + y} + \dfrac{2}{x - y} = 4 \: \: --- (i)$

$\sf \dashrightarrow \dfrac{15}{x + y} - \dfrac{9}{x - y} = -2$

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

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

So, the equations become

$\sf \dashrightarrow 10u + 2v = 4 \: \: --- (iii)$

$\sf \dashrightarrow 15u - 9v = -2 \: \: --- (iv)$

By third equation,

$\sf \dashrightarrow 10u + 2v = 4$

$\sf \dashrightarrow 10u = 4 - 2v$

$\sf \dashrightarrow u = \dfrac{4 - 2v}{10}$

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

$\sf \dashrightarrow 15u - 9v = -2$

$\sf \dashrightarrow 15 \bigg( \dfrac{4 - 2v}{10} \bigg) - 9v = -2$

$\sf \dashrightarrow \dfrac{60 - 30v}{10} - 9v = -2$

$\sf \dashrightarrow \dfrac{60 - 30v - 90v}{10} = -2$

$\sf \dashrightarrow \dfrac{60 - 120v}{10} = -2$

$\sf \dashrightarrow 60 - 120v = -2 \times 10$

$\sf \dashrightarrow 60 - 120v = -20$

$\sf \dashrightarrow -120v = -20 - 60$

$\sf \dashrightarrow -120v = -80$

$\sf \dashrightarrow v = \dfrac{-80}{-120}$

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

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

$\sf \dashrightarrow 10u + 2v = 4$

$\sf \dashrightarrow 10u + 2 \bigg( \dfrac{2}{3} \bigg) = 4$

$\sf \dashrightarrow 10u + \dfrac{4}{3} = 4$

$\sf \dashrightarrow \dfrac{30u + 4}{3} = 4$

$\sf \dashrightarrow 30u + 4 = 4 \times 3$

$\sf \dashrightarrow 30u + 4 = 12$

$\sf \dashrightarrow 30u = 12 - 4$

$\sf \dashrightarrow 30u = 8$

$\sf \dashrightarrow u = \dfrac{8}{30}$

$\sf \dashrightarrow u = \dfrac{4}{15}$

We know that,

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

$\sf \dashrightarrow \dfrac{1}{x + y} = \dfrac{4}{15}$

$\sf \dashrightarrow 4x + 4y = 15 \: \: --- (v)$

We also know that,

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

$\sf \dashrightarrow \dfrac{1}{x - y} = \dfrac{2}{3}$

$\sf \dashrightarrow 2x - 2y = 3 \: \: --- (vi)$

Now, by fifth equation.

$\sf \dashrightarrow 4x + 4y = 15$

$\sf \dashrightarrow 4x = 15 - 4y$

$\sf \dashrightarrow x = \dfrac{15 - 4y}{4}$

Now, we should find the value of y by sixth equation.

$\sf \dashrightarrow 2x - 2y = 3$

$\sf \dashrightarrow 2 \bigg( \dfrac{15 - 4y}{4} \bigg) - 2y = 3$

$\sf \dashrightarrow \dfrac{30 - 8y}{4} - 2y = 3$

$\sf \dashrightarrow \dfrac{30 - 8y - 8y}{4} = 3$

$\sf \dashrightarrow \dfrac{30 - 16y}{4} = 3$

$\sf \dashrightarrow 30 - 16y = 3 \times 4$

$\sf \dashrightarrow 30 - 16y = 12$

$\sf \dashrightarrow -16y = 12 - 30$

$\sf \dashrightarrow -16y = -18$

$\sf \dashrightarrow y = \dfrac{-18}{-16}$

$\sf \dashrightarrow y = \dfrac{9}{8}$

Now, we can find the value of x by fifth equation.

$\sf \dashrightarrow 4x + 4y = 15$

$\sf \dashrightarrow 4x + 4 \bigg( \dfrac{9}{8} \bigg) = 15$

$\sf \dashrightarrow 4x + \dfrac{36}{8} = 15$

$\sf \dashrightarrow \dfrac{32x + 36}{8} = 15$

$\sf \dashrightarrow 32x + 36 = 15 \times 8$

$\sf \dashrightarrow 32x + 16 = 120$

$\sf \dashrightarrow 32x = 120 - 16$

$\sf \dashrightarrow 32x = 104$

$\sf \dashrightarrow x = \dfrac{104}{32}$

$\sf \dashrightarrow x = \dfrac{13}{4}$

Hence, the values of x and y are $\sf \dfrac{13}{4}$ and $\sf \dfrac{9}{8}$ respectively.