Solve by Addition/Elimination 1/5x+y=12/5 , 1/10x+1/3y=13/10

$\frac{1}{5} x + y = \frac{12}{5}$ , $\frac{1}{10} x + \frac{1}{3} y = \frac{13}{10}$

Step 1

Multiply each term in $\frac{1}{5} x + y = \frac{12}{5}$ by $5$ to eliminate the fractions.

Multiply each term in $\frac{1}{5} x + y = \frac{12}{5}$ by $5$ .

$\frac{1}{5} x \cdot 5 + y \cdot 5 = \frac{12}{5} \cdot 5$

$\frac{1}{10} x + \frac{1}{3} y = \frac{13}{10}$

Simplify the left side.

Simplify each term.

Combine $\frac{1}{5}$ and $x$ .

$\frac{x}{5} \cdot 5 + y \cdot 5 = \frac{12}{5} \cdot 5$

$\frac{1}{10} x + \frac{1}{3} y = \frac{13}{10}$

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{x}{5} \cdot 5 + y \cdot 5 = \frac{12}{5} \cdot 5$

$\frac{1}{10} x + \frac{1}{3} y = \frac{13}{10}$

Rewrite the expression.

$x + y \cdot 5 = \frac{12}{5} \cdot 5$

$\frac{1}{10} x + \frac{1}{3} y = \frac{13}{10}$

$x + y \cdot 5 = \frac{12}{5} \cdot 5$

$\frac{1}{10} x + \frac{1}{3} y = \frac{13}{10}$

Move $5$ to the left of $y$ .

$x + 5 y = \frac{12}{5} \cdot 5$

$\frac{1}{10} x + \frac{1}{3} y = \frac{13}{10}$

$x + 5 y = \frac{12}{5} \cdot 5$

$\frac{1}{10} x + \frac{1}{3} y = \frac{13}{10}$

$x + 5 y = \frac{12}{5} \cdot 5$

$\frac{1}{10} x + \frac{1}{3} y = \frac{13}{10}$

Simplify the right side.

Cancel the common factor of $5$ .

Cancel the common factor.

$x + 5 y = \frac{12}{5} \cdot 5$

$\frac{1}{10} x + \frac{1}{3} y = \frac{13}{10}$

Rewrite the expression.

$x + 5 y = 12$

$\frac{1}{10} x + \frac{1}{3} y = \frac{13}{10}$

$x + 5 y = 12$

$\frac{1}{10} x + \frac{1}{3} y = \frac{13}{10}$

$x + 5 y = 12$

$\frac{1}{10} x + \frac{1}{3} y = \frac{13}{10}$

$x + 5 y = 12$

$\frac{1}{10} x + \frac{1}{3} y = \frac{13}{10}$

Step 2

Multiply each term in $\frac{1}{10} x + \frac{1}{3} y = \frac{13}{10}$ by $30$ to eliminate the fractions.

Multiply each term in $\frac{1}{10} x + \frac{1}{3} y = \frac{13}{10}$ by $30$ .

$\frac{1}{10} x \cdot 30 + \frac{1}{3} y \cdot 30 = \frac{13}{10} \cdot 30$

$x + 5 y = 12$

Simplify the left side.

Simplify each term.

Combine $\frac{1}{10}$ and $x$ .

$\frac{x}{10} \cdot 30 + \frac{1}{3} y \cdot 30 = \frac{13}{10} \cdot 30$

$x + 5 y = 12$

Cancel the common factor of $10$ .

Factor $10$ out of $30$ .

$\frac{x}{10} \cdot (10 (3)) + \frac{1}{3} y \cdot 30 = \frac{13}{10} \cdot 30$

$x + 5 y = 12$

Cancel the common factor.

$\frac{x}{10} \cdot (10 \cdot 3) + \frac{1}{3} y \cdot 30 = \frac{13}{10} \cdot 30$

$x + 5 y = 12$

Rewrite the expression.

$x \cdot 3 + \frac{1}{3} y \cdot 30 = \frac{13}{10} \cdot 30$

$x + 5 y = 12$

$x \cdot 3 + \frac{1}{3} y \cdot 30 = \frac{13}{10} \cdot 30$

$x + 5 y = 12$

Move $3$ to the left of $x$ .

$3 \cdot x + \frac{1}{3} y \cdot 30 = \frac{13}{10} \cdot 30$

$x + 5 y = 12$

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

$3 x + \frac{y}{3} \cdot 30 = \frac{13}{10} \cdot 30$

$x + 5 y = 12$

Cancel the common factor of $3$ .

Factor $3$ out of $30$ .

$3 x + \frac{y}{3} \cdot (3 (10)) = \frac{13}{10} \cdot 30$

$x + 5 y = 12$

Cancel the common factor.

$3 x + \frac{y}{3} \cdot (3 \cdot 10) = \frac{13}{10} \cdot 30$

$x + 5 y = 12$

Rewrite the expression.

$3 x + y \cdot 10 = \frac{13}{10} \cdot 30$

$x + 5 y = 12$

$3 x + y \cdot 10 = \frac{13}{10} \cdot 30$

$x + 5 y = 12$

Move $10$ to the left of $y$ .

$3 x + 10 y = \frac{13}{10} \cdot 30$

$x + 5 y = 12$

$3 x + 10 y = \frac{13}{10} \cdot 30$

$x + 5 y = 12$

$3 x + 10 y = \frac{13}{10} \cdot 30$

$x + 5 y = 12$

Simplify the right side.

Cancel the common factor of $10$ .

Factor $10$ out of $30$ .

$3 x + 10 y = \frac{13}{10} \cdot (10 (3))$

$x + 5 y = 12$

Cancel the common factor.

$3 x + 10 y = \frac{13}{10} \cdot (10 \cdot 3)$

$x + 5 y = 12$

Rewrite the expression.

$3 x + 10 y = 13 \cdot 3$

$x + 5 y = 12$

$3 x + 10 y = 13 \cdot 3$

$x + 5 y = 12$

Multiply $13$ by $3$ .

$3 x + 10 y = 39$

$x + 5 y = 12$

$3 x + 10 y = 39$

$x + 5 y = 12$

$3 x + 10 y = 39$

$x + 5 y = 12$

Step 3

Multiply each equation by the value that makes the coefficients of $x$ opposite.

$(- 3) \cdot (x + 5 y) = (- 3) (12)$

$3 x + 10 y = 39$

Step 4

Simplify.

Simplify the left side.

Simplify $(- 3) \cdot (x + 5 y)$ .

Apply the distributive property.

$- 3 x - 3 (5 y) = (- 3) (12)$

$3 x + 10 y = 39$

Multiply $5$ by $- 3$ .

$- 3 x - 15 y = (- 3) (12)$

$3 x + 10 y = 39$

$- 3 x - 15 y = (- 3) (12)$

$3 x + 10 y = 39$

$- 3 x - 15 y = (- 3) (12)$

$3 x + 10 y = 39$

Simplify the right side.

Multiply $- 3$ by $12$ .

$- 3 x - 15 y = - 36$

$3 x + 10 y = 39$

$- 3 x - 15 y = - 36$

$3 x + 10 y = 39$

$- 3 x - 15 y = - 36$

$3 x + 10 y = 39$

Step 5

Add the two equations together to eliminate $x$ from the system.

			$-$	$3$	$x$	$-$	$1$	$5$	$y$	$=$	$-$	$3$	$6$
$+$				$3$	$x$	$+$	$1$	$0$	$y$	$=$		$3$	$9$
							$-$	$5$	$y$	$=$			$3$

Step 6

Divide each term in $- 5 y = 3$ by $- 5$ and simplify.

Divide each term in $- 5 y = 3$ by $- 5$ .

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

Simplify the left side.

Cancel the common factor of $- 5$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

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

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

Simplify the right side.

Move the negative in front of the fraction.

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

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

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

Step 7

Substitute the value found for $y$ into one of the original equations, then solve for $x$ .

Substitute the value found for $y$ into one of the original equations to solve for $x$ .

$- 3 x - 15 (- \frac{3}{5}) = - 36$

Simplify each term.

Cancel the common factor of $5$ .

Move the leading negative in $- \frac{3}{5}$ into the numerator.

$- 3 x - 15 (\frac{- 3}{5}) = - 36$

Factor $5$ out of $- 15$ .

$- 3 x + 5 (- 3) \frac{- 3}{5} = - 36$

Cancel the common factor.

$- 3 x + 5 \cdot - 3 \frac{- 3}{5} = - 36$

Rewrite the expression.

$- 3 x - 3 \cdot - 3 = - 36$

$- 3 x - 3 \cdot - 3 = - 36$

Multiply $- 3$ by $- 3$ .

$- 3 x + 9 = - 36$

$- 3 x + 9 = - 36$

Move all terms not containing $x$ to the right side of the equation.

Subtract $9$ from both sides of the equation.

$- 3 x = - 36 - 9$

Subtract $9$ from $- 36$ .

$- 3 x = - 45$

$- 3 x = - 45$

Divide each term in $- 3 x = - 45$ by $- 3$ and simplify.

Divide each term in $- 3 x = - 45$ by $- 3$ .

$\frac{- 3 x}{- 3} = \frac{- 45}{- 3}$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 x}{- 3} = \frac{- 45}{- 3}$

Divide $x$ by $1$ .

$x = \frac{- 45}{- 3}$

$x = \frac{- 45}{- 3}$

$x = \frac{- 45}{- 3}$

Simplify the right side.

Divide $- 45$ by $- 3$ .

$x = 15$

$x = 15$

$x = 15$

$x = 15$

Step 8

The solution to the independent system of equations can be represented as a point.

$(15, - \frac{3}{5})$

Step 9

The result can be shown in multiple forms.

Point Form:

$(15, - \frac{3}{5})$

Equation Form:

$x = 15, y = - \frac{3}{5}$

Step 10

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