# How do you find the x and y intercepts for x+3y=-1?

Jan 18, 2016

The explanation is given below.

#### Explanation:

$x -$intercept and $y -$intercept occur where the line crosses the axes.

There are a couple of ways you can find the intercepts. Let us see both.

Method 1:
On the $x -$intercept the $y$ coordinate of the point is $0$. So just plug in $y = 0$ in the equation and solve for $x$.

$x + 3 \left(0\right) = - 1$
$x = - 1$

The $x -$intercept is $\left(- 1 , 0\right)$

On the $y$ intercept the $x$ coordinate of the point is $0$. So just plug in $x = 0$ in the equation and solve for $y$

$0 + 3 y = - 1$
$3 y = - 1$
$y = - \frac{1}{3}$

The $y -$intercept is $\left(0 , - \frac{1}{3}\right)$

Method 2:
Another method which is quite useful when the equation of line is given in the standard for such as our problem is to write the equation in the intercept form.

The intercept form is $\frac{x}{a} + \frac{y}{b} = 1$ where $x -$intercept is $\left(a , 0\right)$ and $y -$intercept is $\left(0 , b\right)$

$x + 3 y = - 1$

Divide by $- 1$ on both the sides, we get,

$\frac{x}{-} 1 + 3 \frac{y}{-} 1 = 1$

$\frac{x}{- 1} + \frac{y}{- \frac{1}{3}} = 1$ comparing with the intercept form we get

$a = - 1$ and $b = - \frac{1}{3}$

Our $x -$intercept is $\left(- 1 , 0\right)$
And $y -$intercept is $\left(0. - \frac{1}{3}\right)$

As you can see we get the same result as the first method.

The final choice of method is yours.