Answer

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Hint: Equate the coefficient of one of the variables of the second equation with the first equation and subtract it from the first equation to get the value of another variable. Then put that value in any of the equations to get the value of the first variable.

To solve the given system of equations firstly, we will write them below,

$\dfrac{15}{u}+\dfrac{2}{v}=17$

$\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{36}{5}$

To analyze above equations in a simple way we will just write them by separating variables from constants,

$15\left( \dfrac{1}{u} \right)+2\left( \dfrac{1}{v} \right)=17$……………………………………. (1)

$\left( \dfrac{1}{u} \right)+\left( \dfrac{1}{v} \right)=\dfrac{36}{5}$ .………………………………….. (2)

Now, we have to remove one of the variables by doing operations on the above equations to find the value of another variable. Therefore we will eliminate ‘u’ from above equations and for that we will multiply equation (2) by 15 therefore we will get,

$\therefore 15\left( \dfrac{1}{u} \right)+15\left( \dfrac{1}{v} \right)=15\times \dfrac{36}{5}$

Dividing 15 by 5 we will get,

$\therefore 15\left( \dfrac{1}{u} \right)+15\left( \dfrac{1}{v} \right)=3\times 36$

Multiplying ‘36’ by ‘3’ we will get,

$\therefore 15\left( \dfrac{1}{u} \right)+15\left( \dfrac{1}{v} \right)=108$ ……………………………….. (3)

To eliminate ‘u’ we will subtract equation (1) from equation (3) therefore we will get,

$\begin{align}

& 15\left( \dfrac{1}{u} \right)+15\left( \dfrac{1}{v} \right)=108 \\

& - \\

& 15\left( \dfrac{1}{u} \right)+2\left( \dfrac{1}{v} \right)=17 \\

\end{align}$

$0$ $+13\left( \dfrac{1}{v} \right)$$=91$

Dividing by ‘13’ on both sides we will get,

$\therefore \left( \dfrac{1}{v} \right)=\dfrac{91}{13}$

Dividing ‘92’ by ‘7’ we will get,

\[\therefore \left( \dfrac{1}{v} \right)=7\]

By simply inverting the above equation we will get,

\[\therefore v=\dfrac{1}{7}\] ………………………………………….. (4)

To find the value of ‘u’ we will put the value of ‘v’ in equation (2) therefore we will get,

$\therefore \dfrac{1}{u}+\dfrac{1}{\left( \dfrac{1}{7} \right)}=\dfrac{36}{5}$

Division can be written in the form of multiplication by simply inverting the value therefore we can write $\dfrac{1}{7}$ as shown below,

$\therefore \dfrac{1}{u}+1\times \left( \dfrac{7}{1} \right)=\dfrac{36}{5}$

$\therefore \dfrac{1}{u}+7=\dfrac{36}{5}$

By shifting ‘7’ on the opposite side of equation we will get,

$\therefore \dfrac{1}{u}=\dfrac{36}{5}-7$

As the denominators both terms on the right hand side of equation are not equal we will equate them by cross multiplication as shown below,

$\therefore \dfrac{1}{u}=\dfrac{36-5\left( 7 \right)}{5}$

By multiplying ‘7’ by ‘5’ we will get,

$\therefore \dfrac{1}{u}=\dfrac{36-35}{5}$

By subtracting ‘35’ from ‘36’ we will get,

$\therefore \dfrac{1}{u}=\dfrac{1}{5}$

By simply inverting the above equation we will get,

u = 5

Therefore the solution of the given system of equations can be written as ‘u=5’ and ‘v= $\dfrac{1}{7}$’.

Note: Do not cross multiply both the equations directly for vanishing its denominator as it will become lengthy,

$\dfrac{15}{u}+\dfrac{2}{v}=17$

$15v+2u=17uv$

Like you have to remove the ‘uv’ from the equations and then you have to solve it.

To solve the given system of equations firstly, we will write them below,

$\dfrac{15}{u}+\dfrac{2}{v}=17$

$\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{36}{5}$

To analyze above equations in a simple way we will just write them by separating variables from constants,

$15\left( \dfrac{1}{u} \right)+2\left( \dfrac{1}{v} \right)=17$……………………………………. (1)

$\left( \dfrac{1}{u} \right)+\left( \dfrac{1}{v} \right)=\dfrac{36}{5}$ .………………………………….. (2)

Now, we have to remove one of the variables by doing operations on the above equations to find the value of another variable. Therefore we will eliminate ‘u’ from above equations and for that we will multiply equation (2) by 15 therefore we will get,

$\therefore 15\left( \dfrac{1}{u} \right)+15\left( \dfrac{1}{v} \right)=15\times \dfrac{36}{5}$

Dividing 15 by 5 we will get,

$\therefore 15\left( \dfrac{1}{u} \right)+15\left( \dfrac{1}{v} \right)=3\times 36$

Multiplying ‘36’ by ‘3’ we will get,

$\therefore 15\left( \dfrac{1}{u} \right)+15\left( \dfrac{1}{v} \right)=108$ ……………………………….. (3)

To eliminate ‘u’ we will subtract equation (1) from equation (3) therefore we will get,

$\begin{align}

& 15\left( \dfrac{1}{u} \right)+15\left( \dfrac{1}{v} \right)=108 \\

& - \\

& 15\left( \dfrac{1}{u} \right)+2\left( \dfrac{1}{v} \right)=17 \\

\end{align}$

$0$ $+13\left( \dfrac{1}{v} \right)$$=91$

Dividing by ‘13’ on both sides we will get,

$\therefore \left( \dfrac{1}{v} \right)=\dfrac{91}{13}$

Dividing ‘92’ by ‘7’ we will get,

\[\therefore \left( \dfrac{1}{v} \right)=7\]

By simply inverting the above equation we will get,

\[\therefore v=\dfrac{1}{7}\] ………………………………………….. (4)

To find the value of ‘u’ we will put the value of ‘v’ in equation (2) therefore we will get,

$\therefore \dfrac{1}{u}+\dfrac{1}{\left( \dfrac{1}{7} \right)}=\dfrac{36}{5}$

Division can be written in the form of multiplication by simply inverting the value therefore we can write $\dfrac{1}{7}$ as shown below,

$\therefore \dfrac{1}{u}+1\times \left( \dfrac{7}{1} \right)=\dfrac{36}{5}$

$\therefore \dfrac{1}{u}+7=\dfrac{36}{5}$

By shifting ‘7’ on the opposite side of equation we will get,

$\therefore \dfrac{1}{u}=\dfrac{36}{5}-7$

As the denominators both terms on the right hand side of equation are not equal we will equate them by cross multiplication as shown below,

$\therefore \dfrac{1}{u}=\dfrac{36-5\left( 7 \right)}{5}$

By multiplying ‘7’ by ‘5’ we will get,

$\therefore \dfrac{1}{u}=\dfrac{36-35}{5}$

By subtracting ‘35’ from ‘36’ we will get,

$\therefore \dfrac{1}{u}=\dfrac{1}{5}$

By simply inverting the above equation we will get,

u = 5

Therefore the solution of the given system of equations can be written as ‘u=5’ and ‘v= $\dfrac{1}{7}$’.

Note: Do not cross multiply both the equations directly for vanishing its denominator as it will become lengthy,

$\dfrac{15}{u}+\dfrac{2}{v}=17$

$15v+2u=17uv$

Like you have to remove the ‘uv’ from the equations and then you have to solve it.

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