Solve by Substitution 3+2x-y=0 , -3-7y=10x

$3 + 2 x - y = 0$ , $- 3 - 7 y = 10 x$

Step 1

Solve for $x$ in $3 + 2 x - y = 0$ .

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

Subtract $3$ from both sides of the equation.

$2 x - y = - 3$

$- 3 - 7 y = 10 x$

Add $y$ to both sides of the equation.

$2 x = - 3 + y$

$- 3 - 7 y = 10 x$

$2 x = - 3 + y$

$- 3 - 7 y = 10 x$

Divide each term in $2 x = - 3 + y$ by $2$ and simplify.

Divide each term in $2 x = - 3 + y$ by $2$ .

$\frac{2 x}{2} = \frac{- 3}{2} + \frac{y}{2}$

$- 3 - 7 y = 10 x$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 3}{2} + \frac{y}{2}$

$- 3 - 7 y = 10 x$

Divide $x$ by $1$ .

$x = \frac{- 3}{2} + \frac{y}{2}$

$- 3 - 7 y = 10 x$

$x = \frac{- 3}{2} + \frac{y}{2}$

$- 3 - 7 y = 10 x$

$x = \frac{- 3}{2} + \frac{y}{2}$

$- 3 - 7 y = 10 x$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{3}{2} + \frac{y}{2}$

$- 3 - 7 y = 10 x$

$x = - \frac{3}{2} + \frac{y}{2}$

$- 3 - 7 y = 10 x$

$x = - \frac{3}{2} + \frac{y}{2}$

$- 3 - 7 y = 10 x$

$x = - \frac{3}{2} + \frac{y}{2}$

$- 3 - 7 y = 10 x$

Step 2

Replace all occurrences of $x$ with $- \frac{3}{2} + \frac{y}{2}$ in each equation.

Replace all occurrences of $x$ in $- 3 - 7 y = 10 x$ with $- \frac{3}{2} + \frac{y}{2}$ .

$- 3 - 7 y = 10 (- \frac{3}{2} + \frac{y}{2})$

$x = - \frac{3}{2} + \frac{y}{2}$

Simplify the right side.

Simplify $10 (- \frac{3}{2} + \frac{y}{2})$ .

Apply the distributive property.

$- 3 - 7 y = 10 (- \frac{3}{2}) + 10 (\frac{y}{2})$

$x = - \frac{3}{2} + \frac{y}{2}$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{3}{2}$ into the numerator.

$- 3 - 7 y = 10 (\frac{- 3}{2}) + 10 (\frac{y}{2})$

$x = - \frac{3}{2} + \frac{y}{2}$

Factor $2$ out of $10$ .

$- 3 - 7 y = 2 (5) (\frac{- 3}{2}) + 10 (\frac{y}{2})$

$x = - \frac{3}{2} + \frac{y}{2}$

Cancel the common factor.

$- 3 - 7 y = 2 \cdot (5 (\frac{- 3}{2})) + 10 (\frac{y}{2})$

$x = - \frac{3}{2} + \frac{y}{2}$

Rewrite the expression.

$- 3 - 7 y = 5 \cdot - 3 + 10 (\frac{y}{2})$

$x = - \frac{3}{2} + \frac{y}{2}$

$- 3 - 7 y = 5 \cdot - 3 + 10 (\frac{y}{2})$

$x = - \frac{3}{2} + \frac{y}{2}$

Multiply $5$ by $- 3$ .

$- 3 - 7 y = - 15 + 10 (\frac{y}{2})$

$x = - \frac{3}{2} + \frac{y}{2}$

Cancel the common factor of $2$ .

Factor $2$ out of $10$ .

$- 3 - 7 y = - 15 + 2 (5) (\frac{y}{2})$

$x = - \frac{3}{2} + \frac{y}{2}$

Cancel the common factor.

$- 3 - 7 y = - 15 + 2 \cdot (5 (\frac{y}{2}))$

$x = - \frac{3}{2} + \frac{y}{2}$

Rewrite the expression.

$- 3 - 7 y = - 15 + 5 y$

$x = - \frac{3}{2} + \frac{y}{2}$

$- 3 - 7 y = - 15 + 5 y$

$x = - \frac{3}{2} + \frac{y}{2}$

$- 3 - 7 y = - 15 + 5 y$

$x = - \frac{3}{2} + \frac{y}{2}$

$- 3 - 7 y = - 15 + 5 y$

$x = - \frac{3}{2} + \frac{y}{2}$

$- 3 - 7 y = - 15 + 5 y$

$x = - \frac{3}{2} + \frac{y}{2}$

Step 3

Solve for $y$ in $- 3 - 7 y = - 15 + 5 y$ .

Move all terms containing $y$ to the left side of the equation.

Subtract $5 y$ from both sides of the equation.

$- 3 - 7 y - 5 y = - 15$

$x = - \frac{3}{2} + \frac{y}{2}$

Subtract $5 y$ from $- 7 y$ .

$- 3 - 12 y = - 15$

$x = - \frac{3}{2} + \frac{y}{2}$

$- 3 - 12 y = - 15$

$x = - \frac{3}{2} + \frac{y}{2}$

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

Add $3$ to both sides of the equation.

$- 12 y = - 15 + 3$

$x = - \frac{3}{2} + \frac{y}{2}$

Add $- 15$ and $3$ .

$- 12 y = - 12$

$x = - \frac{3}{2} + \frac{y}{2}$

$- 12 y = - 12$

$x = - \frac{3}{2} + \frac{y}{2}$

Divide each term in $- 12 y = - 12$ by $- 12$ and simplify.

Divide each term in $- 12 y = - 12$ by $- 12$ .

$\frac{- 12 y}{- 12} = \frac{- 12}{- 12}$

$x = - \frac{3}{2} + \frac{y}{2}$

Simplify the left side.

Cancel the common factor of $- 12$ .

Cancel the common factor.

$\frac{- 12 y}{- 12} = \frac{- 12}{- 12}$

$x = - \frac{3}{2} + \frac{y}{2}$

Divide $y$ by $1$ .

$y = \frac{- 12}{- 12}$

$x = - \frac{3}{2} + \frac{y}{2}$

$y = \frac{- 12}{- 12}$

$x = - \frac{3}{2} + \frac{y}{2}$

$y = \frac{- 12}{- 12}$

$x = - \frac{3}{2} + \frac{y}{2}$

Simplify the right side.

Divide $- 12$ by $- 12$ .

$y = 1$

$x = - \frac{3}{2} + \frac{y}{2}$

$y = 1$

$x = - \frac{3}{2} + \frac{y}{2}$

$y = 1$

$x = - \frac{3}{2} + \frac{y}{2}$

$y = 1$

$x = - \frac{3}{2} + \frac{y}{2}$

Step 4

Replace all occurrences of $y$ with $1$ in each equation.

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

$x = - \frac{3}{2} + \frac{1}{2}$

$y = 1$

Simplify the right side.

Simplify $- \frac{3}{2} + \frac{1}{2}$ .

Combine the numerators over the common denominator.

$x = \frac{- 3 + 1}{2}$

$y = 1$

Simplify the expression.

Add $- 3$ and $1$ .

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

$y = 1$

Divide $- 2$ by $2$ .

$x = - 1$

$y = 1$

$x = - 1$

$y = 1$

$x = - 1$

$y = 1$

$x = - 1$

$y = 1$

$x = - 1$

$y = 1$

Step 5

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

$(- 1, 1)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(- 1, 1)$

Equation Form:

$x = - 1, y = 1$

Step 7

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