Question

Solve the following equation by elimination method.

3x+5y=7 and 11x=9 +13y.

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Answer 1

$3x + 5y = 7....(i)$

$11x - 13y = 9....(ii)$

Multiplying eqn. $(i)$ by $13$ and eqn. $(ii)$ by $5$, we have

$\longrightarrow{\green{}}$ $39x + 65y = 91$

$\longrightarrow{\green{}}$ $55x - 65y = 45$

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$\longrightarrow{\green{}}$ $94x = 136$

$\longrightarrow{\green{}}$ $x = \frac{136}{94}$

$\longrightarrow{\green{}}$ $x = \frac{68}{47}$

Plugging the value of $x$ in eqn. $(i)$ gives us

$\longrightarrow{\green{}}$ $3 \times \frac{68}{47} + 5y = 7$

$\longrightarrow{\green{}}$ $\frac{204}{47} + 5y = 7$

$\longrightarrow{\green{}}$ $5y = 7 - \frac{204}{47}$

$\longrightarrow{\green{}}$ $5y = \frac{7 \times 47}{1 \times 47} - \frac{204}{47}$

$\longrightarrow{\green{}}$ $5y = \frac{329 - 204}{47}$

$\longrightarrow{\green{}}$ $5y = \frac{125}{47}$

$\longrightarrow{\green{}}$ $y = \frac{125}{5 \times 47}$

$\longrightarrow{\green{}}$ $y = \frac{25}{47}$

Hence, $\boxed{ x = \frac{68}{47} }$ and $\boxed{ y = \frac{25}{47} }$.