Question

find the solution for the following

Answer 1

AnswEr:-

Your Answer is:-

• u = 5.
• v = 7.

ExplanaTion:-

Given equations:-

$\tt\longrightarrow \dfrac{45}{u} + \dfrac{7}{v} = 10... \: eq(1) \\ \\ \tt\longrightarrow \frac{15}{u} + \frac{14}{v} = 5... \: eq(2)$

To Find:-

• The value of u and v.

So,

$\\ \rm \bigstar \: \: Let \: \: \dfrac{1}{u} \: \: be \: \: x \: \: and \: \: \dfrac{1}{v} \: \: be \: \: y. \\ \\ \sf { \underline{ \underline{ \bullet \: \: So \: \: by \: \: putting \: \: the \: \: value \: \: of \: \: \frac{1}{u} \: \: and \: \: \frac{1}{v } \: \: we \: \: get:-}}}$

$\\ \tt\leadsto45x + 7y = 10... \: eq(3) \\ \\ \tt\leadsto15x + 14y = 5... \: eq(4)$

$\\ \\ \sf{ \underline{ \underline{ \bullet \: \: by \: \: multiplying \: \: eq(3) \: \: by \: \: 2 \: \: we \: \: get:-}}}$

$\\ \\ \tt\leadsto2(45x + 7y = 10.) \\ \\ \tt\leadsto90x + 14y = 20... \: eq(5)$

$\\ \\ \sf { \underline{ \underline{ \bullet\: \: by \: \: subtracting \: \: eq(4) \: \: from \: \: eq(5) \: \: we \: \: get:-}}}$

$\\ \tt\mapsto(90x + 14y) - (15x + 14y) = 20 - 5. \\ \\ \tt\mapsto90x \: \cancel{+ 14y } \: - 15x \: \cancel{ - 14y} = 15. \\ \\ \rm(by \: \: op e n ing \: \: brackets). \\ \\ \tt\mapsto90x - 15x = 15. \\ \\ \tt\mapsto75x = 15. \\ \\ \tt\mapsto x = \cancel \frac{15}{75} . \\ \\ \tt\mapsto{ \underline{ \underline{x = \frac{1}{5}.}}}$

$\\ \sf { \underline{ \underline{ \bullet \: \: So \: \: by \: \: putting \: \: the \: \: value \: \: of \: \: x \: \:in \: \: eq(4) \: \: we \: \: get:-}}}$

$\\ \tt\mapsto15x + 14y = 5. \\ \\ \tt\mapsto \cancel{15} \frac{1}{ \cancel{5}} + 14y = 5. \\ \\ \tt\mapsto3 + 14y = 5. \\ \\ \tt\mapsto14y = 5 - 3. \\ \\ \tt\mapsto14y = 2. \\ \\ \tt\mapsto y = \cancel \frac{2}{14} . \\ \\ \tt\mapsto{ \underline{ \underline{ y = \frac{1}{7} .}}}$

$\rm\therefore$ The Value Of x and y is $\tt\dfrac{1}{5}\:\:And\:\: \dfrac{1}{7}$ Respectively.

But,

$\\ \tt\longrightarrow x = \dfrac{1}{u}.\\ \\ \tt\longrightarrow \dfrac{\cancel{1}}{5} = \dfrac{\cancel1}{u}.\\ \\ \longrightarrow{\boxed{\underline {\tt u = 5.}}}\\ \\ \rm Also,\\ \\\tt\longrightarrow y = \dfrac{1}{v}\\ \\ \tt\longrightarrow \dfrac{\cancel1}{7} = \dfrac{\cancel1}{v}\\ \\ \longrightarrow{\boxed{\underline{\tt v = 7.}}}$