Solve by Substitution x+3y=9 y=x-5

$x + 3 y = 9$ $y = x - 5$

Step 1

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

Replace all occurrences of $y$ in $x + 3 y = 9$ with $x - 5$ .

$x + 3 (x - 5) = 9$

$y = x - 5$

Simplify the left side.

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

Simplify each term.

Apply the distributive property.

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

$y = x - 5$

Multiply $3$ by $- 5$ .

$x + 3 x - 15 = 9$

$y = x - 5$

$x + 3 x - 15 = 9$

$y = x - 5$

Add $x$ and $3 x$ .

$4 x - 15 = 9$

$y = x - 5$

$4 x - 15 = 9$

$y = x - 5$

$4 x - 15 = 9$

$y = x - 5$

$4 x - 15 = 9$

$y = x - 5$

Step 2

Solve for $x$ in $4 x - 15 = 9$ .

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

Add $15$ to both sides of the equation.

$4 x = 9 + 15$

$y = x - 5$

Add $9$ and $15$ .

$4 x = 24$

$y = x - 5$

$4 x = 24$

$y = x - 5$

Divide each term in $4 x = 24$ by $4$ and simplify.

Divide each term in $4 x = 24$ by $4$ .

$\frac{4 x}{4} = \frac{24}{4}$

$y = x - 5$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{24}{4}$

$y = x - 5$

Divide $x$ by $1$ .

$x = \frac{24}{4}$

$y = x - 5$

$x = \frac{24}{4}$

$y = x - 5$

$x = \frac{24}{4}$

$y = x - 5$

Simplify the right side.

Divide $24$ by $4$ .

$x = 6$

$y = x - 5$

$x = 6$

$y = x - 5$

$x = 6$

$y = x - 5$

$x = 6$

$y = x - 5$

Step 3

Replace all occurrences of $x$ with $6$ in each equation.

Replace all occurrences of $x$ in $y = x - 5$ with $6$ .

$y = (6) - 5$

$x = 6$

Simplify the right side.

Subtract $5$ from $6$ .

$y = 1$

$x = 6$

$y = 1$

$x = 6$

$y = 1$

$x = 6$

Step 4

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

$(6, 1)$

Step 5

The result can be shown in multiple forms.

Point Form:

$(6, 1)$

Equation Form:

$x = 6, y = 1$

Step 6

Solve by Substitution x+3y=9 y=x-5

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