# How do you graph 2x + y = 3 by plotting points?

Mar 25, 2017

Find a few $\left(x , y\right)$ pairs that satisfy the equation; plot these points; connect them.

#### Explanation:

Pick some values for $x$, and use the equation to find what $y$ must be for each of those $x$'s.

For example, if $x$ is 0, then we have:

$\text{ } 2 x + y = 3$
$\implies 2 \left(0\right) + y = 3$
$\implies \text{ } 0 + y = 3$
$\implies \text{ } y = 3$

So when $x = 0$, we have $y = 3$. Meaning the point $\left(x , y\right) = \left(0 , 3\right)$ is a solution to $2 x + y = 3$, and so our graph of the equation will pass through $\left(0 , 3\right)$.

We can get a few more points:

If $x = 1$, then

$\text{ } 2 x + y = 3$
$\implies 2 \left(1\right) + y = 3$
$\implies \text{ } 2 + y = 3$
$\implies \text{ } y = 1$

So $\left(x , y\right) = \left(1 , 1\right)$ is also on our graph.

Similarly, if $x = 2$, then we get $y = \text{–} 1$, giving us the point $\left(x , y\right) = \left(2 , \text{–} 1\right)$.

We then take the points we've computed, $\left(0 , 3\right) , \left(1 , 1\right) , \left(2 , \text{–} 1\right)$, and plot them on a graph:

graph{((x)^2+(y-3)^2-.02)((x-1)^2+(y-1)^2-.02)((x-2)^2+(y+1)^2-.02)=0 [-10, 10, -5, 5]}

Last, since we know this equation is linear, we just need to connect the dots with a line:

graph{((x)^2+(y-3)^2-.02)((x-1)^2+(y-1)^2-.02)((x-2)^2+(y+1)^2-.02)(2x+y-3)=0 [-10, 10, -5, 5]}

And we're done!