Solve by Substitution y=3x+5 and 3y-2x=1

$y = 3 x + 5$ and $3 y - 2 x = 1$

Replace all occurrences of $y$ with $3 x + 5$ in each equation.

Replace all occurrences of $y$ in $3 y - 2 x = 1$ with $3 x + 5$ .

$3 (3 x + 5) - 2 x = 1$

$y = 3 x + 5$

Simplify the left side.

Simplify $3 (3 x + 5) - 2 x$ .

Simplify each term.

Apply the distributive property.

$3 (3 x) + 3 \cdot 5 - 2 x = 1$

$y = 3 x + 5$

Multiply $3$ by $3$ .

$9 x + 3 \cdot 5 - 2 x = 1$

$y = 3 x + 5$

Multiply $3$ by $5$ .

$9 x + 15 - 2 x = 1$

$y = 3 x + 5$

$9 x + 15 - 2 x = 1$

$y = 3 x + 5$

Subtract $2 x$ from $9 x$ .

$7 x + 15 = 1$

$y = 3 x + 5$

$7 x + 15 = 1$

$y = 3 x + 5$

$7 x + 15 = 1$

$y = 3 x + 5$

$7 x + 15 = 1$

$y = 3 x + 5$

Solve for $x$ in $7 x + 15 = 1$ .

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

Subtract $15$ from both sides of the equation.

$7 x = 1 - 15$

$y = 3 x + 5$

Subtract $15$ from $1$ .

$7 x = - 14$

$y = 3 x + 5$

$7 x = - 14$

$y = 3 x + 5$

Divide each term in $7 x = - 14$ by $7$ and simplify.

Divide each term in $7 x = - 14$ by $7$ .

$\frac{7 x}{7} = \frac{- 14}{7}$

$y = 3 x + 5$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 x}{7} = \frac{- 14}{7}$

$y = 3 x + 5$

Divide $x$ by $1$ .

$x = \frac{- 14}{7}$

$y = 3 x + 5$

$x = \frac{- 14}{7}$

$y = 3 x + 5$

$x = \frac{- 14}{7}$

$y = 3 x + 5$

Simplify the right side.

Divide $- 14$ by $7$ .

$x = - 2$

$y = 3 x + 5$

$x = - 2$

$y = 3 x + 5$

$x = - 2$

$y = 3 x + 5$

$x = - 2$

$y = 3 x + 5$

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

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

$y = 3 (- 2) + 5$

$x = - 2$

Simplify the right side.

Simplify $3 (- 2) + 5$ .

Multiply $3$ by $- 2$ .

$y = - 6 + 5$

$x = - 2$

Add $- 6$ and $5$ .

$y = - 1$

$x = - 2$

$y = - 1$

$x = - 2$

$y = - 1$

$x = - 2$

$y = - 1$

$x = - 2$

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

$(- 2, - 1)$

The result can be shown in multiple forms.

Point Form:

$(- 2, - 1)$

Equation Form:

$x = - 2, y = - 1$

Solve by Substitution y=3x+5 and 3y-2x=1

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