Solve by Substitution 2x+2y=10 y=5+x

$2 x + 2 y = 10$ $y = 5 + x$

Step 1

Reorder $5$ and $x$ .

$y = x + 5$

$2 x + 2 y = 10$

Step 2

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

Replace all occurrences of $y$ in $2 x + 2 y = 10$ with $x + 5$ .

$2 x + 2 (x + 5) = 10$

$y = x + 5$

Simplify the left side.

Simplify $2 x + 2 (x + 5)$ .

Simplify each term.

Apply the distributive property.

$2 x + 2 x + 2 \cdot 5 = 10$

$y = x + 5$

Multiply $2$ by $5$ .

$2 x + 2 x + 10 = 10$

$y = x + 5$

$2 x + 2 x + 10 = 10$

$y = x + 5$

Add $2 x$ and $2 x$ .

$4 x + 10 = 10$

$y = x + 5$

$4 x + 10 = 10$

$y = x + 5$

$4 x + 10 = 10$

$y = x + 5$

$4 x + 10 = 10$

$y = x + 5$

Step 3

Solve for $x$ in $4 x + 10 = 10$ .

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

Subtract $10$ from both sides of the equation.

$4 x = 10 - 10$

$y = x + 5$

Subtract $10$ from $10$ .

$4 x = 0$

$y = x + 5$

$4 x = 0$

$y = x + 5$

Divide each term in $4 x = 0$ by $4$ and simplify.

Divide each term in $4 x = 0$ by $4$ .

$\frac{4 x}{4} = \frac{0}{4}$

$y = x + 5$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{0}{4}$

$y = x + 5$

Divide $x$ by $1$ .

$x = \frac{0}{4}$

$y = x + 5$

$x = \frac{0}{4}$

$y = x + 5$

$x = \frac{0}{4}$

$y = x + 5$

Simplify the right side.

Divide $0$ by $4$ .

$x = 0$

$y = x + 5$

$x = 0$

$y = x + 5$

$x = 0$

$y = x + 5$

$x = 0$

$y = x + 5$

Step 4

Replace all occurrences of $x$ with $0$ in each equation.

Replace all occurrences of $x$ in $y = x + 5$ with $0$ .

$y = (0) + 5$

$x = 0$

Simplify the right side.

Add $0$ and $5$ .

$y = 5$

$x = 0$

$y = 5$

$x = 0$

$y = 5$

$x = 0$

Step 5

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

$(0, 5)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(0, 5)$

Equation Form:

$x = 0, y = 5$

Step 7

Solve by Substitution 2x+2y=10 y=5+x

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