Question
Solve: $6x + 3y = 7xy$ and $3x + 9y =11xy$.
✓
Answer
The given equations are as follow:
$6x + 3y = 7xy ...(i)$
$3x + 9y = 11 ....(ii)$
For equation (i), we have:
$\frac{6\text{x}+3\text{y}}{\text{xy}}=7$
$\frac{6\text{x}}{\text{xy}}+\frac{3\text{y}}{\text{xy}}=7$
$\frac{6\text{}}{\text{y}}+\frac{3\text{}}{\text{x}}=7...(\text{iii})$
For equation (ii), we have:
$\frac{3\text{x}+9\text{y}}{\text{xy}}=11$
$\frac{3\text{x}}{\text{xy}}+\frac{9\text{y}}{\text{xy}}=11$
$\frac{3\text{}}{\text{y}}+\frac{9\text{}}{\text{x}}=11...(\text{iv})$
On Substituting $\frac{1}{\text{y}}=\text{v}$ and $\frac{1}{\text{x}}=\text{u}$ in $(iii)$ and $(iv)$, we get:
$6v + 3u = 7 ...(v)$
$3v + 9u = 11 ....(vi)$
On multipiying $(v)$ by $3$, We get:
$18v + 9u = 21 ....(vii)$
On subtracting $(vi)$ from $(vii)$, we get:
$15v = 10$
$\Rightarrow\text{v}=\frac{10}{15}=\frac{2}{3}$
$\Rightarrow\frac{1}{\text{y}}=\frac{2}{3}$
$\Rightarrow\text{y}=\frac{3}{2}$
On subtracting $\text{y}=\frac{3}{2}$ in $(iii)$, we get:
$\frac{6}{\frac{3}{2}}+\frac{3}{\text{x}}=7$
$\Rightarrow4+\frac{3}{\text{x}}=7$
$\Rightarrow\frac{3}{\text{x}}=3$
$\Rightarrow3\text{x}=3$
$\Rightarrow\text{x}=1$
Hence, the required solution is $x = 1$ and $\text{y}=\frac{3}{2}$
$6x + 3y = 7xy ...(i)$
$3x + 9y = 11 ....(ii)$
For equation (i), we have:
$\frac{6\text{x}+3\text{y}}{\text{xy}}=7$
$\frac{6\text{x}}{\text{xy}}+\frac{3\text{y}}{\text{xy}}=7$
$\frac{6\text{}}{\text{y}}+\frac{3\text{}}{\text{x}}=7...(\text{iii})$
For equation (ii), we have:
$\frac{3\text{x}+9\text{y}}{\text{xy}}=11$
$\frac{3\text{x}}{\text{xy}}+\frac{9\text{y}}{\text{xy}}=11$
$\frac{3\text{}}{\text{y}}+\frac{9\text{}}{\text{x}}=11...(\text{iv})$
On Substituting $\frac{1}{\text{y}}=\text{v}$ and $\frac{1}{\text{x}}=\text{u}$ in $(iii)$ and $(iv)$, we get:
$6v + 3u = 7 ...(v)$
$3v + 9u = 11 ....(vi)$
On multipiying $(v)$ by $3$, We get:
$18v + 9u = 21 ....(vii)$
On subtracting $(vi)$ from $(vii)$, we get:
$15v = 10$
$\Rightarrow\text{v}=\frac{10}{15}=\frac{2}{3}$
$\Rightarrow\frac{1}{\text{y}}=\frac{2}{3}$
$\Rightarrow\text{y}=\frac{3}{2}$
On subtracting $\text{y}=\frac{3}{2}$ in $(iii)$, we get:
$\frac{6}{\frac{3}{2}}+\frac{3}{\text{x}}=7$
$\Rightarrow4+\frac{3}{\text{x}}=7$
$\Rightarrow\frac{3}{\text{x}}=3$
$\Rightarrow3\text{x}=3$
$\Rightarrow\text{x}=1$
Hence, the required solution is $x = 1$ and $\text{y}=\frac{3}{2}$
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