Question
Solve for $x$ and $y$:
$\frac{5}{\text{x}}+\text{6y}=13,$
$\frac{3}{\text{x}}+\text{4y}=7\ (\text{x}\neq0).$
$\frac{5}{\text{x}}+\text{6y}=13,$
$\frac{3}{\text{x}}+\text{4y}=7\ (\text{x}\neq0).$
✓
Answer
Putting $\frac{1}{\text{x}}=\text{u}$ the given equations become
$5u + 6y = 13 ...(1)$
$3u + 4y = 7 ...(2)$
Multiplying $(1)$ by $4$ and $(2)$ by $6$, we get
$20u + 24y = 52 ...(3)$
$18u + 24y = 42 ...(4)$
Subtracting $(4)$ from $(3),$ we get
$2u = 10$
$\Rightarrow x = 5$
Substituting $u = 5$ in $(1),$ we get
$5 \times 5 + 6y = 13$
$\Rightarrow 6y = 13 - 25$
$\Rightarrow 6y = -12$
$\Rightarrow y = -2$
$u = 5$
$\Rightarrow\frac{1}{\text{x}}=5$
$\Rightarrow\text{5x}=1$
$\Rightarrow\text{x}=\frac{1}{5}$
$\therefore$ The solution is $\text{x}=\frac{1}{5}$ and $y = -2$
$5u + 6y = 13 ...(1)$
$3u + 4y = 7 ...(2)$
Multiplying $(1)$ by $4$ and $(2)$ by $6$, we get
$20u + 24y = 52 ...(3)$
$18u + 24y = 42 ...(4)$
Subtracting $(4)$ from $(3),$ we get
$2u = 10$
$\Rightarrow x = 5$
Substituting $u = 5$ in $(1),$ we get
$5 \times 5 + 6y = 13$
$\Rightarrow 6y = 13 - 25$
$\Rightarrow 6y = -12$
$\Rightarrow y = -2$
$u = 5$
$\Rightarrow\frac{1}{\text{x}}=5$
$\Rightarrow\text{5x}=1$
$\Rightarrow\text{x}=\frac{1}{5}$
$\therefore$ The solution is $\text{x}=\frac{1}{5}$ and $y = -2$
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