# How do you find the x and y intercept of 0.2x+3.2y=12.8?

Sep 30, 2016

$x$ intercept: $\left(64 , 0\right)$
$y$ intercept: $\left(0 , 4\right)$

#### Explanation:

The axis intercepts are, as the name suggests, points on the given axis. So, the $x$ intercept must be a point on the $x$ axis, and the $y$ intercept must be a point on the $y$ axis.

By definition, a point lays on the $x$ axis if its $y$ coordinate equals zero, and vice versa. Let's use these conditions to find the results:

• For the $x$ intercept, we need to set $y = 0$:

$0.2 x + 3.2 y = 12.8 \setminus \to 0.2 x + 3.2 \left(0\right) = 12.8 \setminus \implies 0.2 x = 12.8$

From here, we can easily solve for $x$:

$x = \frac{12.8}{0.2} = 64$

And since we already knew that $y = 0$, if $x = 64$ our point will be $\left(64 , 0\right)$

• For the $y$ intercept, we do exactly the opposite: we know that our point will be of the form $\left(0 , y\right)$, and we need to find $y$, setting $x = 0$: after cancelling the $x$ coefficient, the equation becomes

$3.2 y = 12.8 \setminus \implies y = \frac{12.8}{3.2} = 4$