# How do you solve the system by graphing 3x + y = 3 and x - 4y = 1?

$x = 1 , \setminus \setminus y = 0$

#### Explanation:

Given equations

$3 x + y = 3 \setminus \ldots \ldots . \left(1\right)$

$x - 4 y = 1 \setminus \ldots \ldots \left(2\right)$

Multiplying (2) by $3$ & subtracting from (1) as follows

$3 x + y - 3 \left(x - 4 y\right) = 3 - 3 \setminus \cdot 1$

$13 y = 0$

$y = 0$

setting $y = 0$ in (1), we get

$3 x + 0 = 3$

$x = 1$