How do you solve systems of equations by solution 2x-3y=-1 and y=x-1?

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Answer 1 · 2018-03-21T04:27:58

$x = 1$ $x=1$ and $y = - 1$ $y=-1$

$2 x - 3 y = - 1 (1)$ $2x-3y=-1" " " "(1)$

$y = x - 1 (2)$ $y=x-1" " " "(2)$

$(2)$ $(2)$ can be written as

$x - y = 1$ $x-y= 1$

Therefore

$x - y = 1$ $x-y=1" "$ (multiply by $2$ $2$ )

$2 x - 2 y = 2 (3)$ $2x - 2y= 2" " " "(3)$

So, $(1) - (3)$ $(1)-(3)$

$2 x - 3 y = - 1$ $2x-3y= -1$
$- 2 x + 2 y = 2$ $-2x+2y= 2$

$- y = 1$ $-y= 1$

Therefore

$y = - 1$ $y=-1$

Now substitute the value of $y$ $y$ in $(1)$ $(1)$

$2 x - 3 (- 1) = 1$ $2x-3(-1)= 1$

$2 x + 3 = 1$ $2x+3=1$

$2 x = 3 - 1$ $2x=3-1$

$x = \frac{2}{2}$ $x= 2/2$

$x = 1$ $x= 1$

Therefore

$x = 1$ $x=1$ and $y = - 1$ $y=-1$