Solve by Substitution 8x-y=4 5x+3y=17

$8 x - y = 4$ $5 x + 3 y = 17$

Step 1

Solve for $x$ in $8 x - y = 4$ .

Add $y$ to both sides of the equation.

$8 x = 4 + y$

$5 x + 3 y = 17$

Divide each term in $8 x = 4 + y$ by $8$ and simplify.

Divide each term in $8 x = 4 + y$ by $8$ .

$\frac{8 x}{8} = \frac{4}{8} + \frac{y}{8}$

$5 x + 3 y = 17$

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{8 x}{8} = \frac{4}{8} + \frac{y}{8}$

$5 x + 3 y = 17$

Divide $x$ by $1$ .

$x = \frac{4}{8} + \frac{y}{8}$

$5 x + 3 y = 17$

$x = \frac{4}{8} + \frac{y}{8}$

$5 x + 3 y = 17$

$x = \frac{4}{8} + \frac{y}{8}$

$5 x + 3 y = 17$

Simplify the right side.

Cancel the common factor of $4$ and $8$ .

Factor $4$ out of $4$ .

$x = \frac{4 (1)}{8} + \frac{y}{8}$

$5 x + 3 y = 17$

Cancel the common factors.

Factor $4$ out of $8$ .

$x = \frac{4 \cdot 1}{4 \cdot 2} + \frac{y}{8}$

$5 x + 3 y = 17$

Cancel the common factor.

$x = \frac{4 \cdot 1}{4 \cdot 2} + \frac{y}{8}$

$5 x + 3 y = 17$

Rewrite the expression.

$x = \frac{1}{2} + \frac{y}{8}$

$5 x + 3 y = 17$

$x = \frac{1}{2} + \frac{y}{8}$

$5 x + 3 y = 17$

$x = \frac{1}{2} + \frac{y}{8}$

$5 x + 3 y = 17$

$x = \frac{1}{2} + \frac{y}{8}$

$5 x + 3 y = 17$

$x = \frac{1}{2} + \frac{y}{8}$

$5 x + 3 y = 17$

$x = \frac{1}{2} + \frac{y}{8}$

$5 x + 3 y = 17$

Step 2

Replace all occurrences of $x$ with $\frac{1}{2} + \frac{y}{8}$ in each equation.

Replace all occurrences of $x$ in $5 x + 3 y = 17$ with $\frac{1}{2} + \frac{y}{8}$ .

$5 (\frac{1}{2} + \frac{y}{8}) + 3 y = 17$

$x = \frac{1}{2} + \frac{y}{8}$

Simplify the left side.

Simplify $5 (\frac{1}{2} + \frac{y}{8}) + 3 y$ .

Simplify each term.

Apply the distributive property.

$5 (\frac{1}{2}) + 5 (\frac{y}{8}) + 3 y = 17$

$x = \frac{1}{2} + \frac{y}{8}$

Combine $5$ and $\frac{1}{2}$ .

$\frac{5}{2} + 5 (\frac{y}{8}) + 3 y = 17$

$x = \frac{1}{2} + \frac{y}{8}$

Combine $5$ and $\frac{y}{8}$ .

$\frac{5}{2} + \frac{5 y}{8} + 3 y = 17$

$x = \frac{1}{2} + \frac{y}{8}$

$\frac{5}{2} + \frac{5 y}{8} + 3 y = 17$

$x = \frac{1}{2} + \frac{y}{8}$

To write $3 y$ as a fraction with a common denominator, multiply by $\frac{8}{8}$ .

$\frac{5}{2} + \frac{5 y}{8} + 3 y \cdot \frac{8}{8} = 17$

$x = \frac{1}{2} + \frac{y}{8}$

Simplify terms.

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

$\frac{5}{2} + \frac{5 y}{8} + \frac{3 y \cdot 8}{8} = 17$

$x = \frac{1}{2} + \frac{y}{8}$

Combine the numerators over the common denominator.

$\frac{5}{2} + \frac{5 y + 3 y \cdot 8}{8} = 17$

$x = \frac{1}{2} + \frac{y}{8}$

$\frac{5}{2} + \frac{5 y + 3 y \cdot 8}{8} = 17$

$x = \frac{1}{2} + \frac{y}{8}$

Simplify each term.

Simplify the numerator.

Factor $y$ out of $5 y + 3 y \cdot 8$ .

Factor $y$ out of $5 y$ .

$\frac{5}{2} + \frac{y \cdot 5 + 3 y \cdot 8}{8} = 17$

$x = \frac{1}{2} + \frac{y}{8}$

Factor $y$ out of $3 y \cdot 8$ .

$\frac{5}{2} + \frac{y \cdot 5 + y (3 \cdot 8)}{8} = 17$

$x = \frac{1}{2} + \frac{y}{8}$

Factor $y$ out of $y \cdot 5 + y (3 \cdot 8)$ .

$\frac{5}{2} + \frac{y (5 + 3 \cdot 8)}{8} = 17$

$x = \frac{1}{2} + \frac{y}{8}$

$\frac{5}{2} + \frac{y (5 + 3 \cdot 8)}{8} = 17$

$x = \frac{1}{2} + \frac{y}{8}$

Multiply $3$ by $8$ .

$\frac{5}{2} + \frac{y (5 + 24)}{8} = 17$

$x = \frac{1}{2} + \frac{y}{8}$

Add $5$ and $24$ .

$\frac{5}{2} + \frac{y \cdot 29}{8} = 17$

$x = \frac{1}{2} + \frac{y}{8}$

$\frac{5}{2} + \frac{y \cdot 29}{8} = 17$

$x = \frac{1}{2} + \frac{y}{8}$

Move $29$ to the left of $y$ .

$\frac{5}{2} + \frac{29 y}{8} = 17$

$x = \frac{1}{2} + \frac{y}{8}$

$\frac{5}{2} + \frac{29 y}{8} = 17$

$x = \frac{1}{2} + \frac{y}{8}$

$\frac{5}{2} + \frac{29 y}{8} = 17$

$x = \frac{1}{2} + \frac{y}{8}$

$\frac{5}{2} + \frac{29 y}{8} = 17$

$x = \frac{1}{2} + \frac{y}{8}$

$\frac{5}{2} + \frac{29 y}{8} = 17$

$x = \frac{1}{2} + \frac{y}{8}$

Step 3

Solve for $y$ in $\frac{5}{2} + \frac{29 y}{8} = 17$ .

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

Subtract $\frac{5}{2}$ from both sides of the equation.

$\frac{29 y}{8} = 17 - \frac{5}{2}$

$x = \frac{1}{2} + \frac{y}{8}$

To write $17$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{29 y}{8} = 17 \cdot \frac{2}{2} - \frac{5}{2}$

$x = \frac{1}{2} + \frac{y}{8}$

Combine $17$ and $\frac{2}{2}$ .

$\frac{29 y}{8} = \frac{17 \cdot 2}{2} - \frac{5}{2}$

$x = \frac{1}{2} + \frac{y}{8}$

Combine the numerators over the common denominator.

$\frac{29 y}{8} = \frac{17 \cdot 2 - 5}{2}$

$x = \frac{1}{2} + \frac{y}{8}$

Simplify the numerator.

Multiply $17$ by $2$ .

$\frac{29 y}{8} = \frac{34 - 5}{2}$

$x = \frac{1}{2} + \frac{y}{8}$

Subtract $5$ from $34$ .

$\frac{29 y}{8} = \frac{29}{2}$

$x = \frac{1}{2} + \frac{y}{8}$

$\frac{29 y}{8} = \frac{29}{2}$

$x = \frac{1}{2} + \frac{y}{8}$

$\frac{29 y}{8} = \frac{29}{2}$

$x = \frac{1}{2} + \frac{y}{8}$

Multiply both sides of the equation by $\frac{8}{29}$ .

$\frac{8}{29} \cdot \frac{29 y}{8} = \frac{8}{29} \cdot \frac{29}{2}$

$x = \frac{1}{2} + \frac{y}{8}$

Simplify both sides of the equation.

Simplify the left side.

Simplify $\frac{8}{29} \cdot \frac{29 y}{8}$ .

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{8}{29} \cdot \frac{29 y}{8} = \frac{8}{29} \cdot \frac{29}{2}$

$x = \frac{1}{2} + \frac{y}{8}$

Rewrite the expression.

$\frac{1}{29} \cdot (29 y) = \frac{8}{29} \cdot \frac{29}{2}$

$x = \frac{1}{2} + \frac{y}{8}$

$\frac{1}{29} \cdot (29 y) = \frac{8}{29} \cdot \frac{29}{2}$

$x = \frac{1}{2} + \frac{y}{8}$

Cancel the common factor of $29$ .

Factor $29$ out of $29 y$ .

$\frac{1}{29} \cdot (29 (y)) = \frac{8}{29} \cdot \frac{29}{2}$

$x = \frac{1}{2} + \frac{y}{8}$

Cancel the common factor.

$\frac{1}{29} \cdot (29 y) = \frac{8}{29} \cdot \frac{29}{2}$

$x = \frac{1}{2} + \frac{y}{8}$

Rewrite the expression.

$y = \frac{8}{29} \cdot \frac{29}{2}$

$x = \frac{1}{2} + \frac{y}{8}$

$y = \frac{8}{29} \cdot \frac{29}{2}$

$x = \frac{1}{2} + \frac{y}{8}$

$y = \frac{8}{29} \cdot \frac{29}{2}$

$x = \frac{1}{2} + \frac{y}{8}$

$y = \frac{8}{29} \cdot \frac{29}{2}$

$x = \frac{1}{2} + \frac{y}{8}$

Simplify the right side.

Simplify $\frac{8}{29} \cdot \frac{29}{2}$ .

Cancel the common factor of $2$ .

Factor $2$ out of $8$ .

$y = \frac{2 (4)}{29} \cdot \frac{29}{2}$

$x = \frac{1}{2} + \frac{y}{8}$

Cancel the common factor.

$y = \frac{2 \cdot 4}{29} \cdot \frac{29}{2}$

$x = \frac{1}{2} + \frac{y}{8}$

Rewrite the expression.

$y = \frac{4}{29} \cdot 29$

$x = \frac{1}{2} + \frac{y}{8}$

$y = \frac{4}{29} \cdot 29$

$x = \frac{1}{2} + \frac{y}{8}$

Cancel the common factor of $29$ .

Cancel the common factor.

$y = \frac{4}{29} \cdot 29$

$x = \frac{1}{2} + \frac{y}{8}$

Rewrite the expression.

$y = 4$

$x = \frac{1}{2} + \frac{y}{8}$

$y = 4$

$x = \frac{1}{2} + \frac{y}{8}$

$y = 4$

$x = \frac{1}{2} + \frac{y}{8}$

$y = 4$

$x = \frac{1}{2} + \frac{y}{8}$

$y = 4$

$x = \frac{1}{2} + \frac{y}{8}$

$y = 4$

$x = \frac{1}{2} + \frac{y}{8}$

Step 4

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

Replace all occurrences of $y$ in $x = \frac{1}{2} + \frac{y}{8}$ with $4$ .

$x = \frac{1}{2} + \frac{4}{8}$

$y = 4$

Simplify the right side.

Simplify $\frac{1}{2} + \frac{4}{8}$ .

Cancel the common factor of $4$ and $8$ .

Factor $4$ out of $4$ .

$x = \frac{1}{2} + \frac{4 (1)}{8}$

$y = 4$

Cancel the common factors.

Factor $4$ out of $8$ .

$x = \frac{1}{2} + \frac{4 \cdot 1}{4 \cdot 2}$

$y = 4$

Cancel the common factor.

$x = \frac{1}{2} + \frac{4 \cdot 1}{4 \cdot 2}$

$y = 4$

Rewrite the expression.

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

$y = 4$

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

$y = 4$

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

$y = 4$

Combine the numerators over the common denominator.

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

$y = 4$

Simplify the expression.

Add $1$ and $1$ .

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

$y = 4$

Divide $2$ by $2$ .

$x = 1$

$y = 4$

$x = 1$

$y = 4$

$x = 1$

$y = 4$

$x = 1$

$y = 4$

$x = 1$

$y = 4$

Step 5

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

$(1, 4)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(1, 4)$

Equation Form:

$x = 1, y = 4$

Step 7

Solve by Substitution 8x-y=4 5x+3y=17

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