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(B) Given pair of equations is:$\frac{x}{10} + \frac{y}{5} = 1 \dots (i)$$\frac{x}{8} + \frac{y}{6} = 15 \dots (ii)$Multiplying equation $(i)$ by $10$

Vedclass · Accepted Answer

$x = 340, y = -165, \lambda = -\frac{1}{2}$

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(B) Given pair of equations is:$\frac{x}{10} + \frac{y}{5} = 1 \dots (i)$$\frac{x}{8} + \frac{y}{6} = 15 \dots (ii)$Multiplying equation $(i)$ by $10$,we get $x + 2y = 10 \dots (iii)$.Multiplying equation $(ii)$ by $24$,we get $3x + 4y = 360 \dots (iv)$.To eliminate $y$,multiply equation $(iii)$ by $2$:$2x + 4y = 20 \dots (v)$Subtracting equation $(v)$ from equation $(iv)$:$(3x + 4y) - (2x + 4y) = 360 - 20$$x = 340$Substituting $x = 340$ into equation $(iii)$:$340 + 2y = 10$$2y = -330$$y = -165$Given the relation $y = \lambda x + 5$,substitute $x = 340$ and $y = -165$:$-165 = \lambda(340) + 5$$-170 = 340\lambda$$\lambda = -\frac{170}{340} = -\frac{1}{2}$Thus,the solution is $x = 340, y = -165$ and $\lambda = -\frac{1}{2}$.

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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