Solve by Substitution x=17-4y and y=x-2

$x = 17 - 4 y$ and $y = x - 2$

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

Replace all occurrences of $y$ in $x = 17 - 4 y$ with $x - 2$ .

$x = 17 - 4 (x - 2)$

$y = x - 2$

Simplify the right side.

Simplify $17 - 4 (x - 2)$ .

Simplify each term.

Apply the distributive property.

$x = 17 - 4 x - 4 \cdot - 2$

$y = x - 2$

Multiply $- 4$ by $- 2$ .

$x = 17 - 4 x + 8$

$y = x - 2$

$x = 17 - 4 x + 8$

$y = x - 2$

Add $17$ and $8$ .

$x = - 4 x + 25$

$y = x - 2$

$x = - 4 x + 25$

$y = x - 2$

$x = - 4 x + 25$

$y = x - 2$

$x = - 4 x + 25$

$y = x - 2$

Solve for $x$ in $x = - 4 x + 25$ .

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

Add $4 x$ to both sides of the equation.

$x + 4 x = 25$

$y = x - 2$

Add $x$ and $4 x$ .

$5 x = 25$

$y = x - 2$

$5 x = 25$

$y = x - 2$

Divide each term in $5 x = 25$ by $5$ and simplify.

Divide each term in $5 x = 25$ by $5$ .

$\frac{5 x}{5} = \frac{25}{5}$

$y = x - 2$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{25}{5}$

$y = x - 2$

Divide $x$ by $1$ .

$x = \frac{25}{5}$

$y = x - 2$

$x = \frac{25}{5}$

$y = x - 2$

$x = \frac{25}{5}$

$y = x - 2$

Simplify the right side.

Divide $25$ by $5$ .

$x = 5$

$y = x - 2$

$x = 5$

$y = x - 2$

$x = 5$

$y = x - 2$

$x = 5$

$y = x - 2$

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

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

$y = (5) - 2$

$x = 5$

Simplify the right side.

Subtract $2$ from $5$ .

$y = 3$

$x = 5$

$y = 3$

$x = 5$

$y = 3$

$x = 5$

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

$(5, 3)$

The result can be shown in multiple forms.

Point Form:

$(5, 3)$

Equation Form:

$x = 5, y = 3$

Solve by Substitution x=17-4y and y=x-2

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