How do you solve the system $y = x - 1$ $y=x-1$ and $x + y = 3$ $x+y=3$ using substitution?

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Answer 1 · 2017-01-20T18:34:33

$x = 2$ $x=2$ and $y = 1$ $y=1$

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

In the second equation, substitute $y$ $y$ with its value from the first equation, $(x - 1)$ $color(red)((x-1))$ .

$x + y = 3$ $x+y=3$

$x + (x - 1) = 3$ $x+color(red)((x-1))=3$

Open the brackets and simplify.

$x + x - 1 = 3$ $x+color(red)(x-1)=3$

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

Add $1$ $1$ to each side.

$2 x = 4$ $2x=4$

Divide both sides by $2$ $2$ .

$x = 2$ $x=2$

In the first equation, substitute $x$ $x$ with $2$ $color(blue)2$ .

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

$y = 2 - 1$ $y=color(blue)2-1$

$y = 1$ $y=1$

How do you solve the system y=x-1y=x−1y=x-1 and x+y=3x+y=3x+y=3 using substitution?