# How do you solve the following linear system: y = 3x - 2 , 14x - 3y = 0?

##### 1 Answer
May 8, 2018

The solution is $\left(- \frac{6}{5} , - \frac{28}{5}\right)$ or $\left(- 1.2 , - 5.6\right)$.

#### Explanation:

Solve the linear system:

$\text{Equation 1} :$ $y = 3 x - 2$

$\text{Equation 2} :$ $14 x - 3 y = 0$

The solution is the point $\left(x , y\right)$ that the two lines have in common, which is the point of intersection. I'm going to use substitution to solve the system.

Equation 1 is already solved for $y$. Substitute $3 x - 2$ for $y$ in Equation 2 and solve for $x$.

$14 x - 3 \left(3 x - 2\right) = 0$

Expand.

$14 x - 9 x + 6 = 0$

Simplify.

$5 x + 6 = 0$

Subtract $6$ from both sides.

$5 x = - 6$

Divide both sides by $5$.

$x = - \frac{6}{5}$ or $- 1.2$

Substitute $- \frac{6}{5}$ for $x$ in Equation 1. Solve for $y$.

$y = 3 \left(- \frac{6}{5}\right) - 2$

Expand.

$y = - \frac{18}{5} - 2$

Multiply $2$ by $\frac{5}{5}$ to get an equivalent fraction with $5$ as the denominator.

$y = - \frac{18}{5} - 2 \times \frac{5}{5}$

$y = - \frac{18}{5} - \frac{10}{5}$

Simplify.

$y = - \frac{28}{5}$ or $- 5.6$

The solution is $\left(- \frac{6}{5} , - \frac{28}{5}\right)$ or $\left(- 1.2 , - 5.6\right)$.

graph{(y-3x+2)(14x-3y+0)=0 [-6.366, 4.73, -8.243, -2.696]}