How do you solve the system using the elimination method for x – 3y = 0 and 3y – 6 = 2x?

Jul 3, 2015

$\left\{\begin{matrix}x = - 6 \\ y = - 2\end{matrix}\right.$

Explanation:

To solve by elimination, let say

$\text{Equation 1}$ is $\text{ } x - 3 y = 0$
and
$\text{Equation 2}$ is $\text{ } 3 y - 6 = 2 x$

Now , to eliminate $y$ you would wanna add Equation 1and Equation 2.

To do that you have to add the Left hand Side($\text{LHS}$)of each equation.

Then you equate that to the sum of the Right Hand Sides($\text{RHS}$) of the two equations.

If you do that correctly then,

$\text{LHS} = x - 3 y + 3 y - 6 = x - 6$

Now, that's how you eliminated $y$

$\text{RHS} = 0 + 2 x = 2 x$

Now, do $\text{LHS"="RHS}$

$\implies x - 6 = 2 x$

$\implies - 2 x + x - 6 = 2 x - 2 x$

$\implies - x - 6 = 0$

$\implies - x - 6 + 6 = 6$

$\implies - x = 6$

$- 1 \times - x = - 1 \times 6$

$\implies \textcolor{b l u e}{x = - 6}$

Now, to obtain $y$ we want to eliminate $x$

$\text{Equation 1}$ is $\text{ } x - 3 y = 0$

$\text{Equation 2}$ is $\text{ } 3 y - 6 = 2 x$

Multiply both side of $\text{Equation 1}$ by $2$ then add the resulting equation with $\text{Equation 2}$

$\text{Equation 1}$ becomes $2 x - 6 y = 0$
Then with $\text{Equation 2}$

$\implies \text{LHS} = 2 x - 6 y + 3 y - 6 = 2 x - 3 y - 6$

$\implies \text{RHS} = 0 + 2 x = 2 x$

Now , $\text{RHS"="LHS}$

$\implies 2 x - 3 y - 6 = 2 x$

$\implies - 2 x + 2 x - 3 y - 6 = 2 x - 2 x$

$\implies - 3 y - 6 = 0$

$\implies - 3 y - 6 + 6 = 0 + 6$

$\implies \frac{- 3 y}{- 3} = \frac{6}{-} 3$

$\implies \textcolor{b l u e}{y = - 2}$