How do you solve the system of equations $y = x + 3$ $y=x+3$ and $y = 3 x + 5$ $y=3x+5$ ?

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Answer 1 · 2016-11-23T14:02:08

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

Because we are given $y = x + 3$ $y = x + 3$ we can substitute $x + 3$ $x + 3$ for $y$ $y$ in the second equation and solve for $x$ $x$ :

$x + 3 = 3 x + 5$ $x + 3 = 3x + 5$

$x + 3 - x - 5 = 3 x + 5 - x - 5$ $x + 3 - x - 5 = 3x + 5 - x - 5$

$- 2 = 2 x$ $-2 = 2x$

$\frac{- 2}{2} = \frac{2 x}{2}$ $(-2)/2 = (2x)/2$

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

We can now substitute $- 1$ $-1$ for $x$ $x$ in the first equation to find $y$ $y$ :

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

$y = 2$ $y = 2$

How do you solve the system of equations y=x+3y=x+3 and y=3x+5y=3x+5?