Solve the System of Equations y-2x=0 y+x=3

$y - 2 x = 0$ $y + x = 3$

Add $2 x$ to both sides of the equation.

$y = 2 x$

$y + x = 3$

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

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

$(2 x) + x = 3$

$y = 2 x$

Simplify the left side.

Add $2 x$ and $x$ .

$3 x = 3$

$y = 2 x$

$3 x = 3$

$y = 2 x$

$3 x = 3$

$y = 2 x$

Divide each term in $3 x = 3$ by $3$ and simplify.

Divide each term in $3 x = 3$ by $3$ .

$\frac{3 x}{3} = \frac{3}{3}$

$y = 2 x$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{3}{3}$

$y = 2 x$

Divide $x$ by $1$ .

$x = \frac{3}{3}$

$y = 2 x$

$x = \frac{3}{3}$

$y = 2 x$

$x = \frac{3}{3}$

$y = 2 x$

Simplify the right side.

Divide $3$ by $3$ .

$x = 1$

$y = 2 x$

$x = 1$

$y = 2 x$

$x = 1$

$y = 2 x$

Replace all occurrences of $x$ with $1$ in each equation.

Replace all occurrences of $x$ in $y = 2 x$ with $1$ .

$y = 2 (1)$

$x = 1$

Simplify the right side.

Multiply $2$ by $1$ .

$y = 2$

$x = 1$

$y = 2$

$x = 1$

$y = 2$

$x = 1$

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

$(1, 2)$

The result can be shown in multiple forms.

Point Form:

$(1, 2)$

Equation Form:

$x = 1, y = 2$

Solve the System of Equations y-2x=0 y+x=3

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