Solve by Substitution 6x-3y=5 , x+2y=0

$6 x - 3 y = 5$ , $x + 2 y = 0$

Step 1

Subtract $2 y$ from both sides of the equation.

$x = - 2 y$

$6 x - 3 y = 5$

Step 2

Replace all occurrences of $x$ with $- 2 y$ in each equation.

Replace all occurrences of $x$ in $6 x - 3 y = 5$ with $- 2 y$ .

$6 (- 2 y) - 3 y = 5$

$x = - 2 y$

Simplify the left side.

Simplify $6 (- 2 y) - 3 y$ .

Multiply $- 2$ by $6$ .

$- 12 y - 3 y = 5$

$x = - 2 y$

Subtract $3 y$ from $- 12 y$ .

$- 15 y = 5$

$x = - 2 y$

$- 15 y = 5$

$x = - 2 y$

$- 15 y = 5$

$x = - 2 y$

$- 15 y = 5$

$x = - 2 y$

Step 3

Divide each term in $- 15 y = 5$ by $- 15$ and simplify.

Divide each term in $- 15 y = 5$ by $- 15$ .

$\frac{- 15 y}{- 15} = \frac{5}{- 15}$

$x = - 2 y$

Simplify the left side.

Cancel the common factor of $- 15$ .

Cancel the common factor.

$\frac{- 15 y}{- 15} = \frac{5}{- 15}$

$x = - 2 y$

Divide $y$ by $1$ .

$y = \frac{5}{- 15}$

$x = - 2 y$

$y = \frac{5}{- 15}$

$x = - 2 y$

$y = \frac{5}{- 15}$

$x = - 2 y$

Simplify the right side.

Cancel the common factor of $5$ and $- 15$ .

Factor $5$ out of $5$ .

$y = \frac{5 (1)}{- 15}$

$x = - 2 y$

Cancel the common factors.

Factor $5$ out of $- 15$ .

$y = \frac{5 \cdot 1}{5 \cdot - 3}$

$x = - 2 y$

Cancel the common factor.

$y = \frac{5 \cdot 1}{5 \cdot - 3}$

$x = - 2 y$

Rewrite the expression.

$y = \frac{1}{- 3}$

$x = - 2 y$

$y = \frac{1}{- 3}$

$x = - 2 y$

$y = \frac{1}{- 3}$

$x = - 2 y$

Move the negative in front of the fraction.

$y = - \frac{1}{3}$

$x = - 2 y$

$y = - \frac{1}{3}$

$x = - 2 y$

$y = - \frac{1}{3}$

$x = - 2 y$

Step 4

Replace all occurrences of $y$ with $- \frac{1}{3}$ in each equation.

Replace all occurrences of $y$ in $x = - 2 y$ with $- \frac{1}{3}$ .

$x = - 2 (- \frac{1}{3})$

$y = - \frac{1}{3}$

Simplify the right side.

Multiply $- 2 (- \frac{1}{3})$ .

Multiply $- 1$ by $- 2$ .

$x = 2 (\frac{1}{3})$

$y = - \frac{1}{3}$

Combine $2$ and $\frac{1}{3}$ .

$x = \frac{2}{3}$

$y = - \frac{1}{3}$

$x = \frac{2}{3}$

$y = - \frac{1}{3}$

$x = \frac{2}{3}$

$y = - \frac{1}{3}$

$x = \frac{2}{3}$

$y = - \frac{1}{3}$

Step 5

The solution to the system is the complete set of ordered pairs that are valid solutions.

$(\frac{2}{3}, - \frac{1}{3})$

Step 6

The result can be shown in multiple forms.

Point Form:

$(\frac{2}{3}, - \frac{1}{3})$

Equation Form:

$x = \frac{2}{3}, y = - \frac{1}{3}$

Step 7

Solve by Substitution 6x-3y=5 , x+2y=0

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