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

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

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

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

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

$y = x - 1$

Simplify the left side.

Simplify $2 x - 3 (x - 1)$ .

Simplify each term.

Apply the distributive property.

$2 x - 3 x - 3 \cdot - 1 = - 1$

$y = x - 1$

Multiply $- 3$ by $- 1$ .

$2 x - 3 x + 3 = - 1$

$y = x - 1$

$2 x - 3 x + 3 = - 1$

$y = x - 1$

Subtract $3 x$ from $2 x$ .

$- x + 3 = - 1$

$y = x - 1$

$- x + 3 = - 1$

$y = x - 1$

$- x + 3 = - 1$

$y = x - 1$

$- x + 3 = - 1$

$y = x - 1$

Solve for $x$ in $- x + 3 = - 1$ .

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

Subtract $3$ from both sides of the equation.

$- x = - 1 - 3$

$y = x - 1$

Subtract $3$ from $- 1$ .

$- x = - 4$

$y = x - 1$

$- x = - 4$

$y = x - 1$

Divide each term in $- x = - 4$ by $- 1$ and simplify.

Divide each term in $- x = - 4$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 4}{- 1}$

$y = x - 1$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 4}{- 1}$

$y = x - 1$

Divide $x$ by $1$ .

$x = \frac{- 4}{- 1}$

$y = x - 1$

$x = \frac{- 4}{- 1}$

$y = x - 1$

Simplify the right side.

Divide $- 4$ by $- 1$ .

$x = 4$

$y = x - 1$

$x = 4$

$y = x - 1$

$x = 4$

$y = x - 1$

$x = 4$

$y = x - 1$

Replace all occurrences of $x$ with $4$ in each equation.

Replace all occurrences of $x$ in $y = x - 1$ with $4$ .

$y = (4) - 1$

$x = 4$

Simplify the right side.

Subtract $1$ from $4$ .

$y = 3$

$x = 4$

$y = 3$

$x = 4$

$y = 3$

$x = 4$

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

$(4, 3)$

The result can be shown in multiple forms.

Point Form:

$(4, 3)$

Equation Form:

$x = 4, y = 3$

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

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