How do you solve system of equations $y = x + 1$ $y = x+1$ and $y = 2 x - 1$ $y = 2x -1$ ?

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How do you solve system of equations $y = x + 1$ $y = x+1$ and $y = 2 x - 1$ $y = 2x -1$ ?

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Answer 1 · 2015-03-29T18:48:35

x=2; y=3

Substitute y= x+1 from the first equation into the second equation. it would become x+1= 2x-1. Now add 1 on both sides and also subtract x from both sides. It would become 2=x. that is x=2. Now in the 1st equation, plugin x=2, to get y= 2+1 =3.

Answer 2 · 2015-03-29T18:54:26

$y = x + 1$ $y = x + 1$ ...equation 1
$y = 2 x - 1$ $y = 2x - 1$ ...equation 2

subtract equation 1 from equation 2.

equation 2 - equation 1
$\Rightarrow y - y = [2 x - 1] - [x + 1]$ $=> y - y = [2x - 1] - [x + 1]$

$\Rightarrow 0 = 2 x - x - 1 - 1$ $=> 0 = 2x - x - 1 -1$

$\Rightarrow 0 = x - 2$ $=> 0 = x - 2$
$x - 2 = 0$ $x - 2 = 0$ Right ?!

$\Rightarrow x = 2$ $=> x = 2$

put $x = 2$ $x = 2$ in equation 1 OR 2

in equation 1:
$y = x + 1$ $y = x + 1$
$\Rightarrow y = 2 + 1$ $=> y = 2 + 1$
$\Rightarrow y = 3$ $=> y = 3$

OR
in equation 2:
$y = 2 x - 1$ $y = 2x - 1$
$\Rightarrow y = 2 \cdot (2) - 1$ $=> y = 2*(2) - 1$
$y = 4 - 1$ $y = 4 - 1$
$\Rightarrow y = 3$ $=> y = 3$ As above for equation 1

Hence $x = 2$ $x = 2$ and $y = 3$ $y = 3$

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How do you solve system of equations $y = x + 1$ $y = x+1$ and $y = 2 x - 1$ $y = 2x -1$ ?

2 Answers

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