How do you solve $x + 3 y = - 1$ $x+3y=-1$ , $y = x + 1$ $y=x+1$ ?

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Answer 1 · 2018-03-03T07:23:45

For a problem like this you would use substitution method.

Since the problem already gives you the value of $y$ $y$ (which is $x + 1$ $x + 1$ ), you would use that information and apply it to the equation

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

How you would do this is by replacing $x + 1$ $x + 1$ (the value of $y$ $y$ ) into the original equation wherever you see $y$ $y$ .

So:

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

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

(Solve algebraically)

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

$4 x + 3 = - 1$ $4x + 3 = -1$

$4 x = - 4$ $4x = -4$

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

Now to find $y$ $y$ as a number, you would plug in $- 1$ $-1$ as $x$ $x$ and solve:

$y = x + 1$ $y = x + 1$

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

$y = 0$ $y = 0$

How do you solve x+3y=−1x+3y=-1, y=x+1y=x+1?