# How do you graph y = 10/3x + 5?

May 18, 2018

graph{10x/3 + 5 [-10, 10, -5, 5]}

#### Explanation:

The equation is in the form, $y = m x + c$
where,
$m \rightarrow$ slope of the line
$c \rightarrow$ y-intercept

Here, $m = \frac{10}{3}$
but, slope = tan$\theta$
...where, $\theta \rightarrow$ angle made by the line with positive x axis

$\therefore \tan \theta = \frac{10}{3}$

$\therefore \theta = 73.3$ degrees

Now, to find x- intercept, put y = 0
$\therefore \frac{10}{3} x + 5 = 0$
$\therefore \frac{10}{3} x = - 5$
$\therefore x = - 1.5$

This means that at ant the point (-1.5, 0), you have to draw an angle of 73.3 degrees. this line will represent the given equation.

An Alternative method:

Make a table:
Consider any random 3 values of x, and find the corresponding values of y. Plot these points, and th eline will pass through all these points linearly.

E.g.: I consider 3 cases, x=0, =1,=-1
1] For $x = 0$, from the given equation, I get $y = 5$, so i plot the point $\left(0 , 5\right)$ on my graph.

2] For $x = 1$, I will get $y = 8.3$. So i plot the point, $\left(1 , 8.3\right)$.

3] For $x = - 1$, I will get $y = 1.7$. So i plot the point, $\left(- 1 , 1.7\right)$.

Once you get these 3 points on your graph, simply draw a line that passes through each one of them.

(You can follow the same 3 steps while assuming any value of y also.)

*Hope this helps :) *