Solve the System of Equations x-y=1 3x=3y+3

$x - y = 1$ $3 x = 3 y + 3$

Step 1

Add $y$ to both sides of the equation.

$x = 1 + y$

$3 x = 3 y + 3$

Step 2

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

Replace all occurrences of $x$ in $3 x = 3 y + 3$ with $1 + y$ .

$3 (1 + y) = 3 y + 3$

$x = 1 + y$

Simplify the left side.

Simplify $3 (1 + y)$ .

Apply the distributive property.

$3 \cdot 1 + 3 y = 3 y + 3$

$x = 1 + y$

Multiply $3$ by $1$ .

$3 + 3 y = 3 y + 3$

$x = 1 + y$

$3 + 3 y = 3 y + 3$

$x = 1 + y$

$3 + 3 y = 3 y + 3$

$x = 1 + y$

$3 + 3 y = 3 y + 3$

$x = 1 + y$

Step 3

Solve for $y$ in $3 + 3 y = 3 y + 3$ .

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

Subtract $3 y$ from both sides of the equation.

$3 + 3 y - 3 y = 3$

$x = 1 + y$

Combine the opposite terms in $3 + 3 y - 3 y$ .

Subtract $3 y$ from $3 y$ .

$3 + 0 = 3$

$x = 1 + y$

Add $3$ and $0$ .

$3 = 3$

$x = 1 + y$

$3 = 3$

$x = 1 + y$

$3 = 3$

$x = 1 + y$

Since $3 = 3$ , the equation will always be true.

Always true

$x = 1 + y$

Always true

$x = 1 + y$

Step 4

Remove any equations from the system that are always true.

$x = 1 + y$

Step 5

Solve the System of Equations x-y=1 3x=3y+3

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