Question

solve this in substitutional method of linear equation in two variable
\begin{gathered}2x - y + 5 = 0 \\ x - 2y + 3 = 0\end{gathered}2x−y+5=0x−2y+3=0​
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Answer 1

Correct Question

Solve the given equations in substitutional method of linear equations in two variables.

2x - y + 5 = 0

x - 2y + 3 = 0

Answer

$\sf \dashrightarrow 2x - y + 5 = 0$

$\sf \dashrightarrow 2x - y = -5$

$\sf \dashrightarrow 2x = -5 + y$

$\sf \dashrightarrow x = \dfrac{-5 + y}{2}$

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

$\sf \dashrightarrow x - 2y + 3 = 0$

$\sf \dashrightarrow \dfrac{-5 + y}{2} - 2y + 3 = 0$

$\sf \dashrightarrow \dfrac{-5 + y - 4y}{2} + 3 = 0$

$\sf \dashrightarrow \dfrac{-5 - 3y}{2} + 3 = 0$

$\sf \dashrightarrow \dfrac{-5 - 3y + 6}{2} = 0$

$\sf \dashrightarrow \dfrac{1 - 3y}{2} = 0$

$\sf \dashrightarrow 1 - 3y = 0$

$\sf \dashrightarrow -3y = 0 - 1$

$\sf \dashrightarrow -3y = -1$

$\sf \dashrightarrow y = \dfrac{-1}{-3}$

$\sf \dashrightarrow y = \dfrac{1}{3}$

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

$\sf \dashrightarrow 2x - 5 + y = 0$

$\sf \dashrightarrow 2x - 5 + \dfrac{1}{3} = 0$

$\sf \dashrightarrow 2x - \dfrac{15 + 1}{3} = 0$

$\sf \dashrightarrow 2x - \dfrac{16}{3} = 0$

$\sf \dashrightarrow \dfrac{6x - 16}{3} = 0$

$\sf \dashrightarrow 6x - 16 = 0$

$\sf \dashrightarrow 6x = 16$

$\sf \dashrightarrow x = \dfrac{16}{6}$

$\sf \dashrightarrow x = \dfrac{8}{3}$

Hence, the values of x and y are 8/3 and 1/3 respectively.