Solve by Substitution 2x+y=20 , y=3x

$2 x + y = 20$ , $y = 3 x$

Step 1

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

Replace all occurrences of $y$ in $2 x + y = 20$ with $3 x$ .

$2 x + 3 x = 20$

$y = 3 x$

Simplify the left side.

Simplify $2 x + 3 x$ .

Remove parentheses.

$2 x + 3 x = 20$

$y = 3 x$

Add $2 x$ and $3 x$ .

$5 x = 20$

$y = 3 x$

$5 x = 20$

$y = 3 x$

$5 x = 20$

$y = 3 x$

$5 x = 20$

$y = 3 x$

Step 2

Divide each term in $5 x = 20$ by $5$ and simplify.

Divide each term in $5 x = 20$ by $5$ .

$\frac{5 x}{5} = \frac{20}{5}$

$y = 3 x$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{20}{5}$

$y = 3 x$

Divide $x$ by $1$ .

$x = \frac{20}{5}$

$y = 3 x$

$x = \frac{20}{5}$

$y = 3 x$

$x = \frac{20}{5}$

$y = 3 x$

Simplify the right side.

Divide $20$ by $5$ .

$x = 4$

$y = 3 x$

$x = 4$

$y = 3 x$

$x = 4$

$y = 3 x$

Step 3

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

Replace all occurrences of $x$ in $y = 3 x$ with $4$ .

$y = 3 (4)$

$x = 4$

Simplify the right side.

Multiply $3$ by $4$ .

$y = 12$

$x = 4$

$y = 12$

$x = 4$

$y = 12$

$x = 4$

Step 4

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

$(4, 12)$

Step 5

The result can be shown in multiple forms.

Point Form:

$(4, 12)$

Equation Form:

$x = 4, y = 12$

Step 6

Solve by Substitution 2x+y=20 , y=3x

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