# Find the solution of the pair of equations $\frac{x}{10}+\frac{y}{5}-1=0$ and $\frac{x}{8}+\frac{y}{6}=15$. Hence, find $\lambda$, if $y=\lambda x+5$.

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Given:

The given pair of equations is $\frac{x}{10}+\frac{y}{5}-1=0$ and $\frac{x}{8}+\frac{y}{6}=15$ and $y = \lambda x + 5$.

To do:

We have to solve the given system of equations and the value of $\lambda$.

Solution:

The given system of equations can be written as,

$\frac{x}{10}+\frac{y}{5}=1$

$\Rightarrow \frac{1(x)+2(y)}{10}=1$

$\Rightarrow x+2y=1(10)$   (On cross  multiplication)

$\Rightarrow x+2y=10$---(i)

$\frac{x}{8}+\frac{y}{6}=15$

$\Rightarrow \frac{3(x)+4(y)}{24}=15$   (LCM of 8 and 6 is 24)

$\Rightarrow 3x+4y=15(24)$   (On cross multiplication)

$\Rightarrow 3x=360-4y$

$\Rightarrow x=\frac{360-4y}{3}$----(ii)

Substitute $x=\frac{360-4y}{3}$ in equation (i), we get,

$\frac{360-4y}{3}+2y=10$

Multiplying by $3$ on both sides, we get,

$3(\frac{360-4y}{3})+3(2y)=3(10)$

$360-4y+6y=30$

$2y=30-360$

$2y=-330$

$y=\frac{-330}{2}$

$y=-165$

Substituting the value of $y=-165$ in equation (ii), we get,

$x=\frac{360-4(-165)}{3}$

$x=\frac{360+660}{3}$

$x=\frac{1020}{3}$

$x=340$

$y = \lambda x + 5$   (Given)

$-165=\lambda (340)+5$

$340\lambda=-165-5$

$\lambda=\frac{-170}{340}$

$\lambda=\frac{-1}{2}$

Therefore, the solution of the given system of equations is $x=340$, $y=-165$ and the value of $\lambda$ is $\frac{-1}{2}$.

Updated on 10-Oct-2022 13:27:15