Question

1. Solve the following pairs of equations by reducing them to a pair of linear equations:
7x-2y/xy=5
8x+7y/xy=15​

Answer 1

AnswEr:

Given Equation

$\bullet\:\sf\: \dfrac{7x \:-\:2y}{xy} = 5$

$\:\:\: \sf\dfrac{8x \:+\:7y}{xy} = 15$

$:\implies\sf\: \dfrac{7x}{xy} \:-\: \dfrac{2y}{xy} = 5$ ---------[Equation 1]

$:\implies\sf\: \dfrac{7}{y}\:-\:\dfrac{2}{x} = 5$

$:\implies\sf\: 7 \times \dfrac{1}{y}\:-\: 2 \times \dfrac{1}{x}=5$

Here, we get $\sf\dfrac{1}{y} = v \:\& \: \dfrac{1}{x} = u$

Similarly,

$:\implies\sf\: \dfrac{8x}{xy}\:+\:\dfrac{7y}{xy} = 15$ ---------[Equation 2]

$:\implies\sf\: \dfrac{8}{y} \:+\;\dfrac{7}{x} = 15$

$:\implies\sf\: 8 \times \dfrac{1}{y} \:+\: 7 \times \dfrac{1}{x} = 15$

Now, From Equation 1 & 2

$:\implies\sf\: 7v - 2u = 5$ ------[Equation 3]

$:\implies\sf\: 8v + 7u = 15$-----[Equation 4]

Multiplying Equation Equation (3) by 7 and Equation (4) by 2

$:\implies\sf\: 7 \times [7v \:-\:2u] = 5 \times 7$

$:\implies\sf\bold{49v\: -\: 14u\: =\: 35}$ -------[Equation 5]

$:\implies\sf\: 2 \times [8v \:+\:7u] = 15 \times 2$

$:\implies\sf\bold{16v\:+\:14u\:=\:30}$ --------[Equation 6]

Now, From Equations 5 & 6

$:\implies\sf\: 49v - 14u = 35$

$:\implies\sf\: 16v + 14u = 30$

$:\implies\sf\: 65v = 65$

$:\implies\sf\: v = \cancel\dfrac{65}{65}$

$:\implies\boxed{\sf{\pink{v\:=\:1}}}$

$\rule{200}2$

Substituting the Value of v in Equation 3

$:\implies\sf\: 7v - 2u = 5$

$:\implies\sf\: 7(1) - 2u = 5$

$:\implies\sf\: -2u = 5 - 7$

$:\implies\sf\: -2u = -2$

$:\implies\sf\: u = \cancel\dfrac{-2}{-2}$

$:\implies\boxed{\sf{\pink{u\:=\:1}}}$

$\rule{200}2$

$:\implies\sf\: \dfrac{1}{y} = v$

$:\implies\sf\: \dfrac{1}{y} = 1$

$:\implies\sf\: 1 = y \times 1$

$:\implies\boxed{\sf{\pink{y\:=\:1}}}$

$\rule{200}2$

$:\implies\sf\:\dfrac{1}{x}=u$

$:\implies\sf\: \dfrac{1}{x}=1$

$:\implies\sf\: 1 = x \times 1$

$:\implies\boxed{\sf{\pink{x\:=\:1}}}$

Hence, Value of x & y is 1.

Answer 2

Question:

1. Solve the following pairs of equations by reducing them to a pair of linear equations:

$\sf \star \: \frac{7x - 2y}{xy} = 5 \\ \\ \sf \: \frac{8x + 7y}{xy} = 15$

Solution:

$\sf\frac{7x - 2y}{xy} = 5 \\ \\ \sf\frac{7x}{xy} - \frac{2y}{xy} = 5 \\ \\ \sf \frac{7}{y} = \frac{2}{x} = 5 \\ \\ \sf \frac{ - 2}{x} + \frac{7}{y} = 5.....(1) \\ \\ \sf\frac{8x + 7y}{xy} = 15 \\ \\ \sf \frac{8}{y} + \frac{7}{x} = 15 \\ \\ \sf \frac{7}{x} + \frac{8}{y} = 15.....(2)$

$\sf \orange{required \: equations \: are : }$

$\sf \frac{ - 2}{x} + \frac{7}{y} = 5.....(1) \\ \\ \sf \frac{7}{x} + \frac{8}{y} = 15.....(2) \\ \\ \sf \: - 2u \: + 7v \: = 5....(3) \: and \\ \sf \: 7u \: + 8v \: = 15....(4) \\ \\ \sf \blue{ \:\: from \: equation \: (3)}$

$\sf- 2u \: + 7v \: = 5 \\ \\ \sf 7v \: = \: 5 + 2u \\ \\ \sf \: v \: = \frac{5 + 2u}{7} \\ \\ \sf \pink{ \underline{now \: , \: putting \: \: value \: of \: v \: in \: eq \: (4)}} \\ \\ \sf \: 7u \: + 8v \: = 15 \\ \\ \sf \: 7u \: + 8( \frac{5 + 2u}{7} ) = 15 \\ \\ \sf \green {\underline{ \: now, \: multiplying \: 7 \: both \: sides}} \\ \\ \sf \: 7 \times 7u \: + 7 \times 8( \frac{5 + 2u}{7} )= 7 \times 15 \\ \\ \sf49u \: + 8(5 + 2u) = 105 \: \\ \\ \sf{ \: 49u + 40 + 16u = 105} \\ \\ \sf \: 49u \: + 16u \: = 105 - 40 \\ \\ \sf \implies \: \: \cancel\frac{65u}{65} \\ \\ \: \: \: \: \: \: \: \sf {\boxed {\bold{u \: = 1}}}$

$\sf \blue{putting \: the \: value \: of \: u \: in \: eq(3)}$

$\sf- 2u \: + 7v \: = 5 \\ - 2 \: + 7v \: = 5 \\ \sf 7v \: = 5 + 2 \\ \sf7v \: = 7 \\ \sf v \: = \cancel\frac{7}{7} \\ \\ \: \ \sf {\boxed {\bold{v \: = 1}}} \: \\ \\ \sf \red {\underline{hence, \: value \: of \: u \: = 1 \: and \: v = 1}}$

$\sf \green{But \: in \: question \: it \: is \: asking \: to } \\ \sf \green{find \: the \: value \: of \: x \: and \: y.}$

$\sf \red{ \underline{finding \: the \: value \: of \: x}} \\ \\ \sf{ \: u \: = \frac{1}{x}} \\ \\ \sf{ \: 1 = \frac{1}{x} } \\ \\ \bold \blue { \underline{ \: x = 1}} \: \: \:$

$\sf \red{ \underline{finding \: the \: value \: of \: y}} \\ \\ \sf \: v \: = \frac{1}{y} \\ \\ \: \sf{1 = \frac{1}{y} } \\ \\ \: \: \bold \blue{ \underline{y = 1}}$

$\sf \underline \pink{hence \: , \: the \: value \: of \: x = 1 \: and \: y = 1}$