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
$\Rightarrow\frac{\text{x}}{2}+\text{y}=0.8$
⇒ x + 2y = 1.6 ......(i)
And, $\frac{7}{\text{x}+\frac{\text{y}}{2}}=10$
$\Rightarrow10\Big(\text{x}+\frac{\text{y}}{2}\Big)=7$
⇒ 20x + 10y = 14
⇒ 10x + 5y = 7 .......(ii)
Multiplying (i) by 10, we get
⇒ 10x + 20y = 16 ......(iii)
Subtracting (ii) from (iii), we get
⇒ 15y = 9
$\Rightarrow\text{y}=\frac{9}{15}=0.6$
Putting y = 0.6 in (iii), we get
⇒ 10x + 20 × 0.6 = 16
⇒ 10x = 16 - 12
$\Rightarrow\text{x}=\frac{4}{10}$
$\Rightarrow\text{x}=0.4$
Thus, x = 0.4 and y = 0.6
$\Rightarrow\frac{\text{x}}{2}+\text{y}=0.8$
⇒ x + 2y = 1.6 ......(i)
And, $\frac{7}{\text{x}+\frac{\text{y}}{2}}=10$
$\Rightarrow10\Big(\text{x}+\frac{\text{y}}{2}\Big)=7$
⇒ 20x + 10y = 14
⇒ 10x + 5y = 7 .......(ii)
Multiplying (i) by 10, we get
⇒ 10x + 20y = 16 ......(iii)
Subtracting (ii) from (iii), we get
⇒ 15y = 9
$\Rightarrow\text{y}=\frac{9}{15}=0.6$
Putting y = 0.6 in (iii), we get
⇒ 10x + 20 × 0.6 = 16
⇒ 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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