Solve the following system of equations:
$\frac{x}{2}\ +\ y\ =\ 0.8$
$\frac{7}{\left( x\ +\ \frac{y}{2}\right)} \ =\ 10$

Given:

The given system of equations is:

$\frac{x}{2}\ +\ y\ =\ 0.8$

$\frac{7}{\left( x\ +\ \frac{y}{2}\right)} \ =\ 10$

To do:

We have to solve the given system of equations.

Solution:

The given system of equations can be written as,

$\frac{x}{2}+y=0.8$

Multiplying by $2$ on both sides, we get,

$2(\frac{x}{2})+2(y)=2(0.8)$

$x+2y=1.6$

Multiplying by $10$ on both sides, we get,

$10(x)+10(2y)=10(1.6)$

$10x+20y=16$---(i)

$\frac{7}{x+\frac{y}{2}}=10$

$\Rightarrow \frac{7}{\frac{2x+y}{2}}=10$

$\Rightarrow \frac{7\times2}{2x+y}=10$

$\Rightarrow \frac{14}{2x+y}=10$

On cross multiplication, we get,

$14=10(2x+y)$

$\Rightarrow 14=20x+10y$

$\Rightarrow 20x=14-10y$

$\Rightarrow x=\frac{14-10y}{20}$----(ii)

Substitute $x=\frac{14-10y}{20}$ in equation (i), we get,

$10(\frac{14-10y}{20})+20y=16$

$\frac{14-10y}{2}+20y=16$ 

Multiplying by $2$ on both sides, we get,

$2(\frac{14-10y}{2})+2(20y)=2(16)$

$14-10y+40y=32$

$30y=32-14$

$30y=18$

$y=\frac{18}{30}$

$y=\frac{3}{5}$

Substituting the value of $y=\frac{3}{5}$ in equation (ii), we get,

$x=\frac{14-10(\frac{3}{5})}{20}$

$x=\frac{14-6}{20}$

$x=\frac{8}{20}$

$x=\frac{2}{5}$

Therefore, the solution of the given system of equations is $x=\frac{2}{5}$ and $y=\frac{3}{5}$.

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Updated on: 10-Oct-2022

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