MCQ
If $\frac{2\text{x}+\text{y}+2}{5}=\frac{3\text{x}-\text{y}+1}{3}=\frac{3\text{x}+2\text{y}+1}{6}$ then :
- ✓$x = 1, y =1$
- B$x = -1, y = -1$
- C$x = 1, y = 2$
- D$x = 2, y = 1$
✓
Answer
Correct option: A.
$x = 1, y =1$
$\frac{2\text{x}+\text{y}+2}{5}=\frac{3\text{x}-\text{y}+1}{3}=\frac{3\text{x}+2\text{y}+1}{6}$
$\Rightarrow\frac{2\text{x}+\text{y}+2}{5}=\frac{3\text{x}-\text{y}+1}{3}$ and $\frac{3\text{x}-\text{y}+1}{3}=\frac{3\text{x}+2\text{y}+1}{6}$
On solving we can obtain two equations in $x$ and $y$.
Multiplying each of the equation by the $\text{LCM}$ of the denominators, we get
$\Rightarrow 3(2x + y + 2) = 5(3x - y + 1)$ and $2(3x - y + 1) = 3x + 2y + 1$
$\Rightarrow 6x + 3y + 6 = 15x - 5y + 5$ and $6x - 2y + 2 = 3x + 2y + 1$
$\Rightarrow 9x - 8y = 1 ...(i)$
$\Rightarrow 3x - 4y = -1 ...(ii)$
Multiply $(ii)$ by $3,$ and subtract from $(i).$
$\Rightarrow 9x - 8y = 1$ and $9x - 12y = -3$
$\Rightarrow 4y = 4$
$\Rightarrow y = 1$
Substituting $y = 1$ in $(i),$ we get $x = 1$.
$\Rightarrow\frac{2\text{x}+\text{y}+2}{5}=\frac{3\text{x}-\text{y}+1}{3}$ and $\frac{3\text{x}-\text{y}+1}{3}=\frac{3\text{x}+2\text{y}+1}{6}$
On solving we can obtain two equations in $x$ and $y$.
Multiplying each of the equation by the $\text{LCM}$ of the denominators, we get
$\Rightarrow 3(2x + y + 2) = 5(3x - y + 1)$ and $2(3x - y + 1) = 3x + 2y + 1$
$\Rightarrow 6x + 3y + 6 = 15x - 5y + 5$ and $6x - 2y + 2 = 3x + 2y + 1$
$\Rightarrow 9x - 8y = 1 ...(i)$
$\Rightarrow 3x - 4y = -1 ...(ii)$
Multiply $(ii)$ by $3,$ and subtract from $(i).$
$\Rightarrow 9x - 8y = 1$ and $9x - 12y = -3$
$\Rightarrow 4y = 4$
$\Rightarrow y = 1$
Substituting $y = 1$ in $(i),$ we get $x = 1$.
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