Solve by Addition/Elimination 1/5x+y=9/5 , 1/10x+1/3y=11/10

$\frac{1}{5} x + y = \frac{9}{5}$ , $\frac{1}{10} x + \frac{1}{3} y = \frac{11}{10}$

Multiply each term in $\frac{1}{5} x + y = \frac{9}{5}$ by $5$ to eliminate the fractions.

Multiply each term in $\frac{1}{5} x + y = \frac{9}{5}$ by $5$ .

$\frac{1}{5} x \cdot 5 + y \cdot 5 = \frac{9}{5} \cdot 5$

$\frac{1}{10} x + \frac{1}{3} y = \frac{11}{10}$

Simplify the left side.

Simplify each term.

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

$\frac{x}{5} \cdot 5 + y \cdot 5 = \frac{9}{5} \cdot 5$

$\frac{1}{10} x + \frac{1}{3} y = \frac{11}{10}$

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{x}{5} \cdot 5 + y \cdot 5 = \frac{9}{5} \cdot 5$

$\frac{1}{10} x + \frac{1}{3} y = \frac{11}{10}$

Rewrite the expression.

$x + y \cdot 5 = \frac{9}{5} \cdot 5$

$\frac{1}{10} x + \frac{1}{3} y = \frac{11}{10}$

$x + y \cdot 5 = \frac{9}{5} \cdot 5$

$\frac{1}{10} x + \frac{1}{3} y = \frac{11}{10}$

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

$x + 5 y = \frac{9}{5} \cdot 5$

$\frac{1}{10} x + \frac{1}{3} y = \frac{11}{10}$

$x + 5 y = \frac{9}{5} \cdot 5$

$\frac{1}{10} x + \frac{1}{3} y = \frac{11}{10}$

$x + 5 y = \frac{9}{5} \cdot 5$

$\frac{1}{10} x + \frac{1}{3} y = \frac{11}{10}$

Simplify the right side.

Cancel the common factor of $5$ .

Cancel the common factor.

$x + 5 y = \frac{9}{5} \cdot 5$

$\frac{1}{10} x + \frac{1}{3} y = \frac{11}{10}$

Rewrite the expression.

$x + 5 y = 9$

$\frac{1}{10} x + \frac{1}{3} y = \frac{11}{10}$

$x + 5 y = 9$

$\frac{1}{10} x + \frac{1}{3} y = \frac{11}{10}$

$x + 5 y = 9$

$\frac{1}{10} x + \frac{1}{3} y = \frac{11}{10}$

$x + 5 y = 9$

$\frac{1}{10} x + \frac{1}{3} y = \frac{11}{10}$

Multiply each term in $\frac{1}{10} x + \frac{1}{3} y = \frac{11}{10}$ by $30$ to eliminate the fractions.

Multiply each term in $\frac{1}{10} x + \frac{1}{3} y = \frac{11}{10}$ by $30$ .

$\frac{1}{10} x \cdot 30 + \frac{1}{3} y \cdot 30 = \frac{11}{10} \cdot 30$

$x + 5 y = 9$

Simplify the left side.

Simplify each term.

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

$\frac{x}{10} \cdot 30 + \frac{1}{3} y \cdot 30 = \frac{11}{10} \cdot 30$

$x + 5 y = 9$

Cancel the common factor of $10$ .

Factor $10$ out of $30$ .

$\frac{x}{10} \cdot (10 (3)) + \frac{1}{3} y \cdot 30 = \frac{11}{10} \cdot 30$

$x + 5 y = 9$

Cancel the common factor.

$\frac{x}{10} \cdot (10 \cdot 3) + \frac{1}{3} y \cdot 30 = \frac{11}{10} \cdot 30$

$x + 5 y = 9$

Rewrite the expression.

$x \cdot 3 + \frac{1}{3} y \cdot 30 = \frac{11}{10} \cdot 30$

$x + 5 y = 9$

$x \cdot 3 + \frac{1}{3} y \cdot 30 = \frac{11}{10} \cdot 30$

$x + 5 y = 9$

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

$3 \cdot x + \frac{1}{3} y \cdot 30 = \frac{11}{10} \cdot 30$

$x + 5 y = 9$

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

$3 x + \frac{y}{3} \cdot 30 = \frac{11}{10} \cdot 30$

$x + 5 y = 9$

Cancel the common factor of $3$ .

Factor $3$ out of $30$ .

$3 x + \frac{y}{3} \cdot (3 (10)) = \frac{11}{10} \cdot 30$

$x + 5 y = 9$

Cancel the common factor.

$3 x + \frac{y}{3} \cdot (3 \cdot 10) = \frac{11}{10} \cdot 30$

$x + 5 y = 9$

Rewrite the expression.

$3 x + y \cdot 10 = \frac{11}{10} \cdot 30$

$x + 5 y = 9$

$3 x + y \cdot 10 = \frac{11}{10} \cdot 30$

$x + 5 y = 9$

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

$3 x + 10 y = \frac{11}{10} \cdot 30$

$x + 5 y = 9$

$3 x + 10 y = \frac{11}{10} \cdot 30$

$x + 5 y = 9$

$3 x + 10 y = \frac{11}{10} \cdot 30$

$x + 5 y = 9$

Simplify the right side.

Cancel the common factor of $10$ .

Factor $10$ out of $30$ .

$3 x + 10 y = \frac{11}{10} \cdot (10 (3))$

$x + 5 y = 9$

Cancel the common factor.

$3 x + 10 y = \frac{11}{10} \cdot (10 \cdot 3)$

$x + 5 y = 9$

Rewrite the expression.

$3 x + 10 y = 11 \cdot 3$

$x + 5 y = 9$

$3 x + 10 y = 11 \cdot 3$

$x + 5 y = 9$

Multiply $11$ by $3$ .

$3 x + 10 y = 33$

$x + 5 y = 9$

$3 x + 10 y = 33$

$x + 5 y = 9$

$3 x + 10 y = 33$

$x + 5 y = 9$

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

$(- 3) \cdot (x + 5 y) = (- 3) (9)$

$3 x + 10 y = 33$

Simplify.

Simplify the left side.

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

Apply the distributive property.

$- 3 x - 3 (5 y) = (- 3) (9)$

$3 x + 10 y = 33$

Multiply $5$ by $- 3$ .

$- 3 x - 15 y = (- 3) (9)$

$3 x + 10 y = 33$

$- 3 x - 15 y = (- 3) (9)$

$3 x + 10 y = 33$

$- 3 x - 15 y = (- 3) (9)$

$3 x + 10 y = 33$

Simplify the right side.

Multiply $- 3$ by $9$ .

$- 3 x - 15 y = - 27$

$3 x + 10 y = 33$

$- 3 x - 15 y = - 27$

$3 x + 10 y = 33$

$- 3 x - 15 y = - 27$

$3 x + 10 y = 33$

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

			$-$	$3$	$x$	$-$	$1$	$5$	$y$	$=$	$-$	$2$	$7$
$+$				$3$	$x$	$+$	$1$	$0$	$y$	$=$		$3$	$3$
							$-$	$5$	$y$	$=$			$6$

Divide each term in $- 5 y = 6$ by $- 5$ and simplify.

Divide each term in $- 5 y = 6$ by $- 5$ .

$\frac{- 5 y}{- 5} = \frac{6}{- 5}$

Simplify the left side.

Cancel the common factor of $- 5$ .

Cancel the common factor.

$\frac{- 5 y}{- 5} = \frac{6}{- 5}$

Divide $y$ by $1$ .

$y = \frac{6}{- 5}$

$y = \frac{6}{- 5}$

$y = \frac{6}{- 5}$

Simplify the right side.

Move the negative in front of the fraction.

$y = - \frac{6}{5}$

$y = - \frac{6}{5}$

$y = - \frac{6}{5}$

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 $x$ .

$- 3 x - 15 (- \frac{6}{5}) = - 27$

Simplify each term.

Cancel the common factor of $5$ .

Move the leading negative in $- \frac{6}{5}$ into the numerator.

$- 3 x - 15 (\frac{- 6}{5}) = - 27$

Factor $5$ out of $- 15$ .

$- 3 x + 5 (- 3) \frac{- 6}{5} = - 27$

Cancel the common factor.

$- 3 x + 5 \cdot - 3 \frac{- 6}{5} = - 27$

Rewrite the expression.

$- 3 x - 3 \cdot - 6 = - 27$

$- 3 x - 3 \cdot - 6 = - 27$

Multiply $- 3$ by $- 6$ .

$- 3 x + 18 = - 27$

$- 3 x + 18 = - 27$

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

Subtract $18$ from both sides of the equation.

$- 3 x = - 27 - 18$

Subtract $18$ from $- 27$ .

$- 3 x = - 45$

$- 3 x = - 45$

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 $- 3$ .

$x = 15$

$x = 15$

$x = 15$

$x = 15$

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

$(15, - \frac{6}{5})$

The result can be shown in multiple forms.

Point Form:

$(15, - \frac{6}{5})$

Equation Form:

$x = 15, y = - \frac{6}{5}$

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