Question
Solve the following systems of equations:
$\frac{\text{x}}{2}+\text{y}=0.8,$
$\frac{7}{\text{x}+\frac{\text{y}}{2}}=10.$
$\frac{\text{x}}{2}+\text{y}=0.8,$
$\frac{7}{\text{x}+\frac{\text{y}}{2}}=10.$
✓
Answer
The given equations are
$\Rightarrow\frac{\text{x}}{2}+\text{y}=0.8$
$\Rightarrow x + 2y = 1.6 ......(i)$
And, $\frac{7}{\text{x}+\frac{\text{y}}{2}}=10$
$\Rightarrow10\Big(\text{x}+\frac{\text{y}}{2}\Big)=7$
$\Rightarrow 20x + 10y = 14$
$\Rightarrow 10x + 5y = 7 .......(ii)$
Multiplying $(i)$ by $10$, we get
$\Rightarrow 10x + 20y = 16 ......(iii)$
Subtracting $(ii)$ from $(iii),$ we get
$\Rightarrow 15y = 9$
$\Rightarrow\text{y}=\frac{9}{15}=0.6$
Putting $y = 0.6$ in $(iii)$, we get
$\Rightarrow 10x + 20 \times 0.6 = 16$
$\Rightarrow 10x = 16 - 12$
$\Rightarrow\text{x}=\frac{4}{10}$
$\Rightarrow\text{x}=0.4$
Thus, $x = 0.4$ and $y = 0.6$
$\Rightarrow\frac{\text{x}}{2}+\text{y}=0.8$
$\Rightarrow x + 2y = 1.6 ......(i)$
And, $\frac{7}{\text{x}+\frac{\text{y}}{2}}=10$
$\Rightarrow10\Big(\text{x}+\frac{\text{y}}{2}\Big)=7$
$\Rightarrow 20x + 10y = 14$
$\Rightarrow 10x + 5y = 7 .......(ii)$
Multiplying $(i)$ by $10$, we get
$\Rightarrow 10x + 20y = 16 ......(iii)$
Subtracting $(ii)$ from $(iii),$ we get
$\Rightarrow 15y = 9$
$\Rightarrow\text{y}=\frac{9}{15}=0.6$
Putting $y = 0.6$ in $(iii)$, we get
$\Rightarrow 10x + 20 \times 0.6 = 16$
$\Rightarrow 10x = 16 - 12$
$\Rightarrow\text{x}=\frac{4}{10}$
$\Rightarrow\text{x}=0.4$
Thus, $x = 0.4$ and $y = 0.6$
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