Question

if 3x+y = 81 and 81x+y = 38, then the value of x and y are respectively.​

Answer 1

Answer

$\sf \dashrightarrow 3x + y = 81 \: \: --- (i)$

$\sf \dashrightarrow 81x + y = 38 \: \: --- (ii)$

By first equation,

$\sf \dashrightarrow 3x + y = 81$

$\sf \dashrightarrow 3x = 81 - y$

$\sf \dashrightarrow x = \dfrac{81 - y}{3}$

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

$\sf \dashrightarrow 81x + y = 38$

$\sf \dashrightarrow 81 \bigg( \dfrac{81 - y}{3} \bigg) + y = 38$

$\sf \dashrightarrow \dfrac{6561 - 81y}{3} + y = 38$

$\sf \dashrightarrow \dfrac{6561 - 81y + 3y}{3} = 38$

$\sf \dashrightarrow \dfrac{6561 - 78y}{3} = 38$

$\sf \dashrightarrow 6561 - 78y = 38 \times 3$

$\sf \dashrightarrow 6561 - 78y = 114$

$\sf \dashrightarrow -78y = 114 - 6561$

$\sf \dashrightarrow -78y = -6447$

$\sf \dashrightarrow y = \dfrac{-6447}{-78}$

$\sf \dashrightarrow y = \dfrac{2149}{26}$

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

$\sf \dashrightarrow 3x + y = 81$

$\sf \dashrightarrow 3x + \dfrac{2149}{26} = 81$

$\sf \dashrightarrow \dfrac{78x + 2149}{26} = 81$

$\sf \dashrightarrow 78x + 2149 = 81 \times 26$

$\sf \dashrightarrow 78x + 2149 = 2106$

$\sf \dashrightarrow 78x = -43$

$\sf \dashrightarrow x = \dfrac{-43}{78}$

Hence, the values of x and y are -43/78 and 2149/26 respectively.