III. Solve the following.1. Solve each system of linear equation by substitution...

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aishteru97 @aishteru97

October 2022 1 3 Report

III. Solve the following.
1. Solve each system of linear equation by substitution
2x + 2 = 2y
4x - 2y + 2 = 0

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UK2019

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✏️SUBSTITUTION

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Direction: Solve each system of linear equation by substitution.

$\begin{cases}2x + 2 = 2y & \green{(eq. \: 1)}\\ 4x - 2y + 2 = 0& \green{( eq. \: 2)}\end{cases}$

Solution: Find the value of x in the first equation.

$2x + \cancel{2 - 2 }= 2y - 2$

$2x = 2y - 2$

$\begin{gathered} \frac{ \cancel2x}{ \cancel2} = \frac{2y - 2}{2} \end{gathered}$

$x = y - 1$

- Substitute it to the second equation in terms of y.

$\begin{cases}x = y - 1&\\ 4(y - 1) - 2y + 2 = 0\end{cases}$

$\begin{cases}x = y - 1&\\ 4y - 4 - 2y + 2 = 0\end{cases}$

$\begin{cases}x = y - 1&\\ 2y - 2 = 0\end{cases}$

$\begin{cases}x = y - 1&\\ 2y \: \cancel{- \: 2 + 2 }= 0 + 2\end{cases}$

$\begin{cases}x = y - 1&\\ 2y = 2\end{cases}$

$\begin{cases}x = y - 1&\\ \begin{gathered} \frac{ \cancel2y}{ \cancel2} = \frac{2}{2} \end{gathered}\end{cases}$

$\begin{cases}x = y - 1&\\ y = 1\end{cases}$

- The y value is 1. Then, substitute to the first equation to find the value of x.

$\begin{cases}x = 1 - 1&\\ y = 1\end{cases}$

$\begin{cases}x = 0&\\ y = 1\end{cases}$

- Therefore, the solutions are:

$\large \boxed{ \rm \green{ \: (0, 1)\: }}$

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Verified answer

✏️SUBSTITUTION

===============================

===============================

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