Solve the System of Equations 5x-y=6 -2x+y=8

$5 x - y = 6$ $- 2 x + y = 8$

Step 1

Add $2 x$ to both sides of the equation.

$y = 8 + 2 x$

$5 x - y = 6$

Step 2

Replace all occurrences of $y$ with $8 + 2 x$ in each equation.

Replace all occurrences of $y$ in $5 x - y = 6$ with $8 + 2 x$ .

$5 x - (8 + 2 x) = 6$

$y = 8 + 2 x$

Simplify the left side.

Simplify $5 x - (8 + 2 x)$ .

Simplify each term.

Apply the distributive property.

$5 x - 1 \cdot 8 - (2 x) = 6$

$y = 8 + 2 x$

Multiply $- 1$ by $8$ .

$5 x - 8 - (2 x) = 6$

$y = 8 + 2 x$

Multiply $2$ by $- 1$ .

$5 x - 8 - 2 x = 6$

$y = 8 + 2 x$

$5 x - 8 - 2 x = 6$

$y = 8 + 2 x$

Subtract $2 x$ from $5 x$ .

$3 x - 8 = 6$

$y = 8 + 2 x$

$3 x - 8 = 6$

$y = 8 + 2 x$

$3 x - 8 = 6$

$y = 8 + 2 x$

$3 x - 8 = 6$

$y = 8 + 2 x$

Step 3

Solve for $x$ in $3 x - 8 = 6$ .

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

Add $8$ to both sides of the equation.

$3 x = 6 + 8$

$y = 8 + 2 x$

Add $6$ and $8$ .

$3 x = 14$

$y = 8 + 2 x$

$3 x = 14$

$y = 8 + 2 x$

Divide each term in $3 x = 14$ by $3$ and simplify.

Divide each term in $3 x = 14$ by $3$ .

$\frac{3 x}{3} = \frac{14}{3}$

$y = 8 + 2 x$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{14}{3}$

$y = 8 + 2 x$

Divide $x$ by $1$ .

$x = \frac{14}{3}$

$y = 8 + 2 x$

$x = \frac{14}{3}$

$y = 8 + 2 x$

$x = \frac{14}{3}$

$y = 8 + 2 x$

$x = \frac{14}{3}$

$y = 8 + 2 x$

$x = \frac{14}{3}$

$y = 8 + 2 x$

Step 4

Replace all occurrences of $x$ with $\frac{14}{3}$ in each equation.

Replace all occurrences of $x$ in $y = 8 + 2 x$ with $\frac{14}{3}$ .

$y = 8 + 2 (\frac{14}{3})$

$x = \frac{14}{3}$

Simplify the right side.

Simplify $8 + 2 (\frac{14}{3})$ .

Multiply $2 (\frac{14}{3})$ .

Combine $2$ and $\frac{14}{3}$ .

$y = 8 + \frac{2 \cdot 14}{3}$

$x = \frac{14}{3}$

Multiply $2$ by $14$ .

$y = 8 + \frac{28}{3}$

$x = \frac{14}{3}$

$y = 8 + \frac{28}{3}$

$x = \frac{14}{3}$

To write $8$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$y = 8 \cdot \frac{3}{3} + \frac{28}{3}$

$x = \frac{14}{3}$

Combine $8$ and $\frac{3}{3}$ .

$y = \frac{8 \cdot 3}{3} + \frac{28}{3}$

$x = \frac{14}{3}$

Combine the numerators over the common denominator.

$y = \frac{8 \cdot 3 + 28}{3}$

$x = \frac{14}{3}$

Simplify the numerator.

Multiply $8$ by $3$ .

$y = \frac{24 + 28}{3}$

$x = \frac{14}{3}$

Add $24$ and $28$ .

$y = \frac{52}{3}$

$x = \frac{14}{3}$

$y = \frac{52}{3}$

$x = \frac{14}{3}$

$y = \frac{52}{3}$

$x = \frac{14}{3}$

$y = \frac{52}{3}$

$x = \frac{14}{3}$

$y = \frac{52}{3}$

$x = \frac{14}{3}$

Step 5

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

$(\frac{14}{3}, \frac{52}{3})$

Step 6

The result can be shown in multiple forms.

Point Form:

$(\frac{14}{3}, \frac{52}{3})$

Equation Form:

$x = \frac{14}{3}, y = \frac{52}{3}$

Step 7

Solve the System of Equations 5x-y=6 -2x+y=8

Related Questions

Support LCMGCF.com by disabling your adblocker.