Solve by Substitution 2x+y=15 , y=3x

$2 x + y = 15$ , $y = 3 x$

Step 1

Replace all occurrences of $y$ with $3 x$ in each equation.

Replace all occurrences of $y$ in $2 x + y = 15$ with $3 x$ .

$2 x + 3 x = 15$

$y = 3 x$

Simplify the left side.

Simplify $2 x + 3 x$ .

Remove parentheses.

$2 x + 3 x = 15$

$y = 3 x$

Add $2 x$ and $3 x$ .

$5 x = 15$

$y = 3 x$

$5 x = 15$

$y = 3 x$

$5 x = 15$

$y = 3 x$

$5 x = 15$

$y = 3 x$

Step 2

Divide each term in $5 x = 15$ by $5$ and simplify.

Divide each term in $5 x = 15$ by $5$ .

$\frac{5 x}{5} = \frac{15}{5}$

$y = 3 x$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{15}{5}$

$y = 3 x$

Divide $x$ by $1$ .

$x = \frac{15}{5}$

$y = 3 x$

$x = \frac{15}{5}$

$y = 3 x$

$x = \frac{15}{5}$

$y = 3 x$

Simplify the right side.

Divide $15$ by $5$ .

$x = 3$

$y = 3 x$

$x = 3$

$y = 3 x$

$x = 3$

$y = 3 x$

Step 3

Replace all occurrences of $x$ with $3$ in each equation.

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

$y = 3 (3)$

$x = 3$

Simplify the right side.

Multiply $3$ by $3$ .

$y = 9$

$x = 3$

$y = 9$

$x = 3$

$y = 9$

$x = 3$

Step 4

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

$(3, 9)$

Step 5

The result can be shown in multiple forms.

Point Form:

$(3, 9)$

Equation Form:

$x = 3, y = 9$

Step 6

Solve by Substitution 2x+y=15 , y=3x

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