Solve the System of @WORD y=x+3 , y=4-x

$y = x + 3$ , $y = 4 - x$

Step 1

Eliminate the equal sides of each equation and combine.

$x + 3 = 4 - x$

Step 2

Solve $x + 3 = 4 - x$ for $x$ .

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

Add $x$ to both sides of the equation.

$x + 3 + x = 4$

Add $x$ and $x$ .

$2 x + 3 = 4$

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

Subtract $3$ from both sides of the equation.

$2 x = 4 - 3$

Subtract $3$ from $4$ .

$2 x = 1$

Divide each term in $2 x = 1$ by $2$ and simplify.

Divide each term in $2 x = 1$ by $2$ .

$\frac{2 x}{2} = \frac{1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{1}{2}$

Divide $x$ by $1$ .

$x = \frac{1}{2}$

Step 3

Evaluate $y$ when $x = \frac{1}{2}$ .

Substitute $\frac{1}{2}$ for $x$ .

$y = 4 - (\frac{1}{2})$

Substitute $\frac{1}{2}$ for $x$ in $y = 4 - (\frac{1}{2})$ and solve for $y$ .

Multiply $- 1$ by $\frac{1}{2}$ .

$y = 4 - \frac{1}{2}$

Simplify $4 - \frac{1}{2}$ .

To write $4$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$y = 4 \cdot \frac{2}{2} - \frac{1}{2}$

Combine $4$ and $\frac{2}{2}$ .

$y = \frac{4 \cdot 2}{2} - \frac{1}{2}$

Combine the numerators over the common denominator.

$y = \frac{4 \cdot 2 - 1}{2}$

Simplify the numerator.

Multiply $4$ by $2$ .

$y = \frac{8 - 1}{2}$

Subtract $1$ from $8$ .

$y = \frac{7}{2}$

Step 4

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

$(\frac{1}{2}, \frac{7}{2})$

Step 5

The result can be shown in multiple forms.

Point Form:

$(\frac{1}{2}, \frac{7}{2})$

Equation Form:

$x = \frac{1}{2}, y = \frac{7}{2}$

Step 6

Solve the System of @WORD y=x+3 , y=4-x

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