Question

Find the solution of the pair of equations: x/10+y/5=1 and x/8+y/6=15. hence, find a, if y= ax+5​

Answer 1

Answer

$\sf \dashrightarrow \dfrac{x}{10} + \dfrac{y}{5} = 1 \: \: --- (i)$

$\sf \dashrightarrow \dfrac{x}{8} + \dfrac{y}{6} = 15 \: \: --- (ii)$

By first equation,

$\sf \dashrightarrow \dfrac{x}{10} + \dfrac{y}{5} = 1$

$\sf \dashrightarrow \dfrac{x + 2y}{10} = 1$

$\sf \dashrightarrow x + 2y = 10 \: \: --- (iii)$

By second equation,

$\sf \dashrightarrow \dfrac{x}{8} + \dfrac{y}{6} = 15$

$\sf \dashrightarrow \dfrac{3x + 4y}{24} = 15$

$\sf \dashrightarrow 3x + 4y = 360 \: \: --- (iv)$

By third equation,

$\sf \dashrightarrow x + 2y = 10$

$\sf \dashrightarrow x = 10 - 2y$

Now, let's find the value of y by fourth equation.

$\sf \dashrightarrow 3x + 4y = 360$

$\sf \dashrightarrow 3 (10 - 2y) + 4y = 360$

$\sf \dashrightarrow 30 - 6y + 4y = 360$

$\sf \dashrightarrow 30 - 2y = 360$

$\sf \dashrightarrow -2y = 360 - 30$

$\sf \dashrightarrow -2y = 330$

$\sf \dashrightarrow y = \dfrac{330}{-2}$

$\sf \dashrightarrow y = -165$

Now, let's find the value of x by third equation.

$\sf \dashrightarrow x + 2y = 10$

$\sf \dashrightarrow x + 2(-165) = 10$

$\sf \dashrightarrow x + (-330) = 10$

$\sf \dashrightarrow x - 330 = 10$

$\sf \dashrightarrow x = 10 + 330$

$\sf \dashrightarrow x = 340$

Now, let's find the value of a.

$\sf \dashrightarrow y = ax + 5$

$\sf \dashrightarrow -165 = a(340) + 5$

$\sf \dashrightarrow a(330) + 5 = -165$

$\sf \dashrightarrow a(330) = -165 - 5$

$\sf \dashrightarrow a(330) = -170$

$\sf \dashrightarrow a = \dfrac{-170}{330}$

$\sf \dashrightarrow a = \dfrac{-17}{33}$

Hence, the value of a is $\sf \dfrac{-17}{33}$.