Solve by Substitution 3x-y=4 , 2x+y=16

$3 x - y = 4$ , $2 x + y = 16$

Step 1

Subtract $2 x$ from both sides of the equation.

$y = 16 - 2 x$

$3 x - y = 4$

Step 2

Replace all occurrences of $y$ with $16 - 2 x$ in each equation.

Replace all occurrences of $y$ in $3 x - y = 4$ with $16 - 2 x$ .

$3 x - (16 - 2 x) = 4$

$y = 16 - 2 x$

Simplify the left side.

Simplify $3 x - (16 - 2 x)$ .

Simplify each term.

Apply the distributive property.

$3 x - 1 \cdot 16 - (- 2 x) = 4$

$y = 16 - 2 x$

Multiply $- 1$ by $16$ .

$3 x - 16 - (- 2 x) = 4$

$y = 16 - 2 x$

Multiply $- 2$ by $- 1$ .

$3 x - 16 + 2 x = 4$

$y = 16 - 2 x$

$3 x - 16 + 2 x = 4$

$y = 16 - 2 x$

Add $3 x$ and $2 x$ .

$5 x - 16 = 4$

$y = 16 - 2 x$

$5 x - 16 = 4$

$y = 16 - 2 x$

$5 x - 16 = 4$

$y = 16 - 2 x$

$5 x - 16 = 4$

$y = 16 - 2 x$

Step 3

Solve for $x$ in $5 x - 16 = 4$ .

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

Add $16$ to both sides of the equation.

$5 x = 4 + 16$

$y = 16 - 2 x$

Add $4$ and $16$ .

$5 x = 20$

$y = 16 - 2 x$

$5 x = 20$

$y = 16 - 2 x$

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

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

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

$y = 16 - 2 x$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

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

$y = 16 - 2 x$

Divide $x$ by $1$ .

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

$y = 16 - 2 x$

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

$y = 16 - 2 x$

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

$y = 16 - 2 x$

Simplify the right side.

Divide $20$ by $5$ .

$x = 4$

$y = 16 - 2 x$

$x = 4$

$y = 16 - 2 x$

$x = 4$

$y = 16 - 2 x$

$x = 4$

$y = 16 - 2 x$

Step 4

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

Replace all occurrences of $x$ in $y = 16 - 2 x$ with $4$ .

$y = 16 - 2 \cdot 4$

$x = 4$

Simplify the right side.

Simplify $16 - 2 \cdot 4$ .

Multiply $- 2$ by $4$ .

$y = 16 - 8$

$x = 4$

Subtract $8$ from $16$ .

$y = 8$

$x = 4$

$y = 8$

$x = 4$

$y = 8$

$x = 4$

$y = 8$

$x = 4$

Step 5

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

$(4, 8)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(4, 8)$

Equation Form:

$x = 4, y = 8$

Step 7

Solve by Substitution 3x-y=4 , 2x+y=16

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