How do you find the slope and y-intercept for the line $5x-y+2=0$ ?

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Answer 1 · 2018-03-05T14:41:59

Just arrange the equation in $y=mx+c$ form,where $m$ is the slope and $c$ is the $Y$ intercept.

So,given equation in $5x-y+2=0$

or, $y=5x+2$

so,comparing with $y=mx+c$ we get, $m=5$ and $c=2$

Now,see the graph below to match the result obtained. graph{5x-y+2=0 [-10, 10, -5, 5]}

Answer 2 · 2018-03-05T14:43:41

See a solution process below:

We can put the equation for the line in slope-intercept form. The slope-intercept form of a linear equation is: $y = color(red)(m)x + color(blue)(b)$

Where $color(red)(m)$ is the slope and $color(blue)(b)$ is the y-intercept value.

$5x - y + 2 = 0$

$5x - y + color(red)(y) + 2 = 0 + color(red)(y)$

$5x - 0 + 2 = y$

$5x + 2 = y$

$y = color(red)(5)x + color(blue)(2)$

Therefore:

How do you find the slope and y-intercept for the line 5x-y+2=05x-y+2=0?