Question

Solve the following pair of equations 5/x+y + 2/x-y =3 , 15/x+y - 4/x-y = -1

Answer 1

$\begin{gathered}\frak{Given}\begin{cases}{ \sf \dfrac{5}{x + y} + \dfrac{2}{x - y} = 3 }\\\\\\{ \sf \dfrac{15}{x + y} - \dfrac{4}{x - y} = (- 1) }\end{cases}\end{gathered}$

To Find

• $\\ \sf \: \: \: x \: \: \: and \: \: \: y$

Solution

• $\\{ \sf \dfrac{5}{x + y} + \dfrac{2}{x - y} = 3 \: \: \: \: \: - - - - (1)}$
• ${ \sf \dfrac{15}{x + y} - \dfrac{4}{x - y} = -1 \: \: \: \: \: - - - - (2)}$

Let

$\\ \sf \dfrac{1}{x + y} = a \: \: \: \: \: \: \: , \: \: \: \: \: \dfrac{1}{x - y} = b$

• Putting a and b in both Equations 1 and 2 we get

$\\ \sf5a + 2b = 3 \: \: \: \: \: \: - - - - (3)$

$\\ \sf15a - 4b = - 1 \: \: \: \: \: \: - - - - (4)$

From Equation 3 we get

$\\\sf~~~~~~~:~~\implies 5a + 2b = 3$

• Subtracting Both sides by -5a

$\\\sf ~~~~~~~:~~\implies 5a - 5a + 2b = 3 - 5a$

$\sf ~~~~~~~:~~\implies 2b = 3 - 5a$

• Dividing Both sides by 2

$\\\sf ~~~~~~~:~~\implies 2b \div 2=( 3 - 5a )\div 2$

$\\\sf~~~~~~~:~~\implies \pink{b = \dfrac{( 3 - 5a ) }{2}}$

• Putting b in Equation 4

$\\\sf~~~~~~~:~~\implies 15a - 4 \left( \dfrac{3 - 5a}{2} \right) = - 1$

$\sf~~~~~~~:~~\implies 15a - \dfrac{12- 20a}{2} = - 1$

$\sf ~~~~~~~:~~\implies \dfrac{30a - 12 + 20a}{2} = - 1$

$\sf ~~~~~~~:~~\implies 50a - 12 = - 2$

$\sf ~~~~~~~:~~\implies 50a = - 2 + 12$

$\sf ~~~~~~~:~~\implies 50a = 10$

$\sf ~~~~~~~:~~\implies a = \dfrac{10}{50}$

$\sf ~~~~~~~:~~\implies \pink{a = \dfrac{1}{5}}$

Putting value of a in Equation 3

$\sf~~~~~~~:~~\implies 5 \times \dfrac{1}{5} + 2b = 3$

$\sf~~~~~~~:~~\implies 1 + 2b = 3$

$\sf~~~~~~~:~~\implies 2b = 3 - 1$

$\sf~~~~~~~:~~\implies 2b = 2$

$\sf~~~~~~~:~~\implies b = 1$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c | } \\ \qquad&\qquad \\ \sf a \implies \dfrac{1}{5} =\dfrac{1}{x + y} & \sf b \implies 1 =\dfrac{1}{x - y} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: x + y = 5& \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf x - y = 1\end{array}}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

Let

$\\ \sf \: x + y = 5 \: \: \: \: \: \: - - - - (5)$

$\sf \: x - y = 1 \: \: \: \: \: \: - - - - (6)$

• On Solving Equation 5

$\sf ~~~~~~~:~~\implies \: x + y = 5$

$\sf ~~~~~~~:~~\implies \: x = 5 - y$

• Putting x in Equation 6

$\sf~~~~~~~:~~\implies \: 5 - y - y = 1$

$\sf~~~~~~~:~~\implies \: - 2y = 1 - 5$

$\sf~~~~~~~:~~\implies \: - 2y = - 4$

$\sf ~~~~~~~:~~\implies \: y = 2$

• Putting y = 2 in Equation 6

$\sf ~~~~~~~:~~\implies \: x - y = 1$

$\sf ~~~~~~~:~~\implies \: x - 2= 1$

$\sf ~~~~~~~:~~\implies \: x = 2 + 1$

$\sf ~~~~~~~:~~\implies \: x = 3$

• Hance The value of x and y are 3 and 2 respectively.