Solve by Substitution y=-1+3/8x 2x-5y=6

$y = - 1 + \frac{3}{8} x$ $2 x - 5 y = 6$

Simplify $- 1 + \frac{3}{8} x$ .

Combine $\frac{3}{8}$ and $x$ .

$y = - 1 + \frac{3 x}{8}$

$2 x - 5 y = 6$

Reorder $- 1$ and $\frac{3 x}{8}$ .

$y = \frac{3 x}{8} - 1$

$2 x - 5 y = 6$

$y = \frac{3 x}{8} - 1$

$2 x - 5 y = 6$

Replace all occurrences of $y$ with $\frac{3 x}{8} - 1$ in each equation.

Replace all occurrences of $y$ in $2 x - 5 y = 6$ with $\frac{3 x}{8} - 1$ .

$2 x - 5 (\frac{3 x}{8} - 1) = 6$

$y = \frac{3 x}{8} - 1$

Simplify the left side.

Simplify $2 x - 5 (\frac{3 x}{8} - 1)$ .

Simplify each term.

Apply the distributive property.

$2 x - 5 \frac{3 x}{8} - 5 \cdot - 1 = 6$

$y = \frac{3 x}{8} - 1$

Multiply $- 5 \frac{3 x}{8}$ .

Combine $- 5$ and $\frac{3 x}{8}$ .

$2 x + \frac{- 5 (3 x)}{8} - 5 \cdot - 1 = 6$

$y = \frac{3 x}{8} - 1$

Multiply $3$ by $- 5$ .

$2 x + \frac{- 15 x}{8} - 5 \cdot - 1 = 6$

$y = \frac{3 x}{8} - 1$

$2 x + \frac{- 15 x}{8} - 5 \cdot - 1 = 6$

$y = \frac{3 x}{8} - 1$

Multiply $- 5$ by $- 1$ .

$2 x + \frac{- 15 x}{8} + 5 = 6$

$y = \frac{3 x}{8} - 1$

Move the negative in front of the fraction.

$2 x - \frac{15 x}{8} + 5 = 6$

$y = \frac{3 x}{8} - 1$

$2 x - \frac{15 x}{8} + 5 = 6$

$y = \frac{3 x}{8} - 1$

To write $2 x$ as a fraction with a common denominator, multiply by $\frac{8}{8}$ .

$2 x \cdot \frac{8}{8} - \frac{15 x}{8} + 5 = 6$

$y = \frac{3 x}{8} - 1$

Simplify terms.

Combine $2 x$ and $\frac{8}{8}$ .

$\frac{2 x \cdot 8}{8} - \frac{15 x}{8} + 5 = 6$

$y = \frac{3 x}{8} - 1$

Combine the numerators over the common denominator.

$\frac{2 x \cdot 8 - 15 x}{8} + 5 = 6$

$y = \frac{3 x}{8} - 1$

$\frac{2 x \cdot 8 - 15 x}{8} + 5 = 6$

$y = \frac{3 x}{8} - 1$

Simplify each term.

Simplify the numerator.

Factor $x$ out of $2 x \cdot 8 - 15 x$ .

Factor $x$ out of $2 x \cdot 8$ .

$\frac{x (2 \cdot 8) - 15 x}{8} + 5 = 6$

$y = \frac{3 x}{8} - 1$

Factor $x$ out of $- 15 x$ .

$\frac{x (2 \cdot 8) + x \cdot - 15}{8} + 5 = 6$

$y = \frac{3 x}{8} - 1$

Factor $x$ out of $x (2 \cdot 8) + x \cdot - 15$ .

$\frac{x (2 \cdot 8 - 15)}{8} + 5 = 6$

$y = \frac{3 x}{8} - 1$

$\frac{x (2 \cdot 8 - 15)}{8} + 5 = 6$

$y = \frac{3 x}{8} - 1$

Multiply $2$ by $8$ .

$\frac{x (16 - 15)}{8} + 5 = 6$

$y = \frac{3 x}{8} - 1$

Subtract $15$ from $16$ .

$\frac{x \cdot 1}{8} + 5 = 6$

$y = \frac{3 x}{8} - 1$

$\frac{x \cdot 1}{8} + 5 = 6$

$y = \frac{3 x}{8} - 1$

Multiply $x$ by $1$ .

$\frac{x}{8} + 5 = 6$

$y = \frac{3 x}{8} - 1$

$\frac{x}{8} + 5 = 6$

$y = \frac{3 x}{8} - 1$

$\frac{x}{8} + 5 = 6$

$y = \frac{3 x}{8} - 1$

$\frac{x}{8} + 5 = 6$

$y = \frac{3 x}{8} - 1$

$\frac{x}{8} + 5 = 6$

$y = \frac{3 x}{8} - 1$

Solve for $x$ in $\frac{x}{8} + 5 = 6$ .

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

Subtract $5$ from both sides of the equation.

$\frac{x}{8} = 6 - 5$

$y = \frac{3 x}{8} - 1$

Subtract $5$ from $6$ .

$\frac{x}{8} = 1$

$y = \frac{3 x}{8} - 1$

$\frac{x}{8} = 1$

$y = \frac{3 x}{8} - 1$

Multiply both sides of the equation by $8$ .

$8 (\frac{x}{8}) = 8 \cdot 1$

$y = \frac{3 x}{8} - 1$

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$8 (\frac{x}{8}) = 8 \cdot 1$

$y = \frac{3 x}{8} - 1$

Rewrite the expression.

$x = 8 \cdot 1$

$y = \frac{3 x}{8} - 1$

$x = 8 \cdot 1$

$y = \frac{3 x}{8} - 1$

$x = 8 \cdot 1$

$y = \frac{3 x}{8} - 1$

Simplify the right side.

Multiply $8$ by $1$ .

$x = 8$

$y = \frac{3 x}{8} - 1$

$x = 8$

$y = \frac{3 x}{8} - 1$

$x = 8$

$y = \frac{3 x}{8} - 1$

$x = 8$

$y = \frac{3 x}{8} - 1$

Replace all occurrences of $x$ with $8$ in each equation.

Replace all occurrences of $x$ in $y = \frac{3 x}{8} - 1$ with $8$ .

$y = \frac{3 (8)}{8} - 1$

$x = 8$

Simplify the right side.

Simplify $\frac{3 (8)}{8} - 1$ .

Multiply $3$ by $8$ .

$y = \frac{24}{8} - 1$

$x = 8$

Divide $24$ by $8$ .

$y = 3 - 1$

$x = 8$

Subtract $1$ from $3$ .

$y = 2$

$x = 8$

$y = 2$

$x = 8$

$y = 2$

$x = 8$

$y = 2$

$x = 8$

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

$(8, 2)$

The result can be shown in multiple forms.

Point Form:

$(8, 2)$

Equation Form:

$x = 8, y = 2$

Solve by Substitution y=-1+3/8x 2x-5y=6

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