Solve the System of @WORD 3x+6y=6 , 5x+y=10

$3 x + 6 y = 6$ , $5 x + y = 10$

Step 1

Subtract $5 x$ from both sides of the equation.

$y = 10 - 5 x$

$3 x + 6 y = 6$

Step 2

Replace all occurrences of $y$ with $10 - 5 x$ in each equation.

Replace all occurrences of $y$ in $3 x + 6 y = 6$ with $10 - 5 x$ .

$3 x + 6 (10 - 5 x) = 6$

$y = 10 - 5 x$

Simplify the left side.

Simplify $3 x + 6 (10 - 5 x)$ .

Simplify each term.

Apply the distributive property.

$3 x + 6 \cdot 10 + 6 (- 5 x) = 6$

$y = 10 - 5 x$

Multiply $6$ by $10$ .

$3 x + 60 + 6 (- 5 x) = 6$

$y = 10 - 5 x$

Multiply $- 5$ by $6$ .

$3 x + 60 - 30 x = 6$

$y = 10 - 5 x$

$3 x + 60 - 30 x = 6$

$y = 10 - 5 x$

Subtract $30 x$ from $3 x$ .

$- 27 x + 60 = 6$

$y = 10 - 5 x$

$- 27 x + 60 = 6$

$y = 10 - 5 x$

$- 27 x + 60 = 6$

$y = 10 - 5 x$

$- 27 x + 60 = 6$

$y = 10 - 5 x$

Step 3

Solve for $x$ in $- 27 x + 60 = 6$ .

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

Subtract $60$ from both sides of the equation.

$- 27 x = 6 - 60$

$y = 10 - 5 x$

Subtract $60$ from $6$ .

$- 27 x = - 54$

$y = 10 - 5 x$

$- 27 x = - 54$

$y = 10 - 5 x$

Divide each term in $- 27 x = - 54$ by $- 27$ and simplify.

Divide each term in $- 27 x = - 54$ by $- 27$ .

$\frac{- 27 x}{- 27} = \frac{- 54}{- 27}$

$y = 10 - 5 x$

Simplify the left side.

Cancel the common factor of $- 27$ .

Cancel the common factor.

$\frac{- 27 x}{- 27} = \frac{- 54}{- 27}$

$y = 10 - 5 x$

Divide $x$ by $1$ .

$x = \frac{- 54}{- 27}$

$y = 10 - 5 x$

$x = \frac{- 54}{- 27}$

$y = 10 - 5 x$

$x = \frac{- 54}{- 27}$

$y = 10 - 5 x$

Simplify the right side.

Divide $- 54$ by $- 27$ .

$x = 2$

$y = 10 - 5 x$

$x = 2$

$y = 10 - 5 x$

$x = 2$

$y = 10 - 5 x$

$x = 2$

$y = 10 - 5 x$

Step 4

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

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

$y = 10 - 5 \cdot 2$

$x = 2$

Simplify the right side.

Simplify $10 - 5 \cdot 2$ .

Multiply $- 5$ by $2$ .

$y = 10 - 10$

$x = 2$

Subtract $10$ from $10$ .

$y = 0$

$x = 2$

$y = 0$

$x = 2$

$y = 0$

$x = 2$

$y = 0$

$x = 2$

Step 5

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

$(2, 0)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(2, 0)$

Equation Form:

$x = 2, y = 0$

Step 7

Solve the System of @WORD 3x+6y=6 , 5x+y=10

Related Questions

Support LCMGCF.com by disabling your adblocker.