Linear Algebra Examples

$x + b y = 5$ , $x + 5 y = b$

Step 1

Subtract $b$ from both sides of the equation.

$x + b y = 5, x + 5 y - b = 0$

Step 2

Write the system of equations in matrix form.

$[\begin{array}{c} y & 1 & 0 & 5 \\ - 1 & 1 & 5 & 0 \end{array}]$

Step 3

Find the reduced row echelon form.

Tap for more steps...

Step 3.1

Multiply each element of $R_{1}$ by $\frac{1}{y}$ to make the entry at $1, 1$ a $1$ .

Tap for more steps...

Step 3.1.1

Multiply each element of $R_{1}$ by $\frac{1}{y}$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} \frac{y}{y} & \frac{1}{y} & \frac{0}{y} & \frac{5}{y} \\ - 1 & 1 & 5 & 0 \end{array}]$

Step 3.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & \frac{1}{y} & 0 & \frac{5}{y} \\ - 1 & 1 & 5 & 0 \end{array}]$

Step 3.2

Perform the row operation $R_{2} = R_{2} + R_{1}$ to make the entry at $2, 1$ a $0$ .

Tap for more steps...

Step 3.2.1

Perform the row operation $R_{2} = R_{2} + R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & \frac{1}{y} & 0 & \frac{5}{y} \\ - 1 + 1 \cdot 1 & 1 + \frac{1}{y} & 5 + 0 & 0 + \frac{5}{y} \end{array}]$

Step 3.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{1}{y} & 0 & \frac{5}{y} \\ 0 & 1 + \frac{1}{y} & 5 & \frac{5}{y} \end{array}]$

Step 3.3

Multiply each element of $R_{2}$ by $\frac{1}{1 + \frac{1}{y}}$ to make the entry at $2, 2$ a $1$ .

Tap for more steps...

Step 3.3.1

Multiply each element of $R_{2}$ by $\frac{1}{1 + \frac{1}{y}}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & \frac{1}{y} & 0 & \frac{5}{y} \\ \frac{0}{1 + \frac{1}{y}} & \frac{1 + \frac{1}{y}}{1 + \frac{1}{y}} & \frac{5}{1 + \frac{1}{y}} & \frac{\frac{5}{y}}{1 + \frac{1}{y}} \end{array}]$

Step 3.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & \frac{1}{y} & 0 & \frac{5}{y} \\ 0 & 1 & \frac{5 y}{y + 1} & \frac{5}{y + 1} \end{array}]$

Step 3.4

Perform the row operation $R_{1} = R_{1} - \frac{1}{y} R_{2}$ to make the entry at $1, 2$ a $0$ .

Tap for more steps...

Step 3.4.1

Perform the row operation $R_{1} = R_{1} - \frac{1}{y} R_{2}$ to make the entry at $1, 2$ a $0$ .

$[\begin{array}{c} 1 - \frac{1}{y} \cdot 0 & \frac{1}{y} - \frac{1}{y} \cdot 1 & 0 - \frac{1}{y} \cdot \frac{5 y}{y + 1} & \frac{5}{y} - \frac{1}{y} \cdot \frac{5}{y + 1} \\ 0 & 1 & \frac{5 y}{y + 1} & \frac{5}{y + 1} \end{array}]$

Step 3.4.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & - \frac{5}{y + 1} & \frac{5}{y} - \frac{5}{y (y + 1)} \\ 0 & 1 & \frac{5 y}{y + 1} & \frac{5}{y + 1} \end{array}]$

Step 4

Use the result matrix to declare the final solutions to the system of equations.

$b - \frac{5 y}{y + 1} = \frac{5}{y} - \frac{5}{y (y + 1)}$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

Step 5

Solve the equation for $b$ .

Tap for more steps...

Step 5.1

Move all terms containing variables to the left side of the equation.

Tap for more steps...

Step 5.1.1

Subtract $\frac{5}{y}$ from both sides of the equation.

$b - \frac{5 y}{y + 1} - \frac{5}{y} = - \frac{5}{y (y + 1)}$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

Step 5.1.2

Add $\frac{5}{y (y + 1)}$ to both sides of the equation.

$b - \frac{5 y}{y + 1} - \frac{5}{y} + \frac{5}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

Step 5.1.3

To write $b$ as a fraction with a common denominator, multiply by $\frac{y + 1}{y + 1}$ .

$b \cdot \frac{y + 1}{y + 1} - \frac{5 y}{y + 1} - \frac{5}{y} + \frac{5}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

Step 5.1.4

Combine $b$ and $\frac{y + 1}{y + 1}$ .

$\frac{b (y + 1)}{y + 1} - \frac{5 y}{y + 1} - \frac{5}{y} + \frac{5}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

Step 5.1.5

Combine the numerators over the common denominator.

$\frac{b (y + 1) - 5 y}{y + 1} - \frac{5}{y} + \frac{5}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

Step 5.1.6

Simplify the numerator.

Tap for more steps...

Step 5.1.6.1

Apply the distributive property.

$\frac{b y + b \cdot 1 - 5 y}{y + 1} - \frac{5}{y} + \frac{5}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

Step 5.1.6.2

Multiply $b$ by $1$ .

$\frac{b y + b - 5 y}{y + 1} - \frac{5}{y} + \frac{5}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

$\frac{b y + b - 5 y}{y + 1} - \frac{5}{y} + \frac{5}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

Step 5.1.7

To write $\frac{b y + b - 5 y}{y + 1}$ as a fraction with a common denominator, multiply by $\frac{y}{y}$ .

$\frac{b y + b - 5 y}{y + 1} \cdot \frac{y}{y} - \frac{5}{y} + \frac{5}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

Step 5.1.8

To write $- \frac{5}{y}$ as a fraction with a common denominator, multiply by $\frac{y + 1}{y + 1}$ .

$\frac{b y + b - 5 y}{y + 1} \cdot \frac{y}{y} - \frac{5}{y} \cdot \frac{y + 1}{y + 1} + \frac{5}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

Step 5.1.9

Write each expression with a common denominator of $(y + 1) y$ , by multiplying each by an appropriate factor of $1$ .

Tap for more steps...

Step 5.1.9.1

Multiply $\frac{b y + b - 5 y}{y + 1}$ by $\frac{y}{y}$ .

$\frac{(b y + b - 5 y) y}{(y + 1) y} - \frac{5}{y} \cdot \frac{y + 1}{y + 1} + \frac{5}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

Step 5.1.9.2

Multiply $\frac{5}{y}$ by $\frac{y + 1}{y + 1}$ .

$\frac{(b y + b - 5 y) y}{(y + 1) y} - \frac{5 (y + 1)}{y (y + 1)} + \frac{5}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

Step 5.1.9.3

Reorder the factors of $(y + 1) y$ .

$\frac{(b y + b - 5 y) y}{y (y + 1)} - \frac{5 (y + 1)}{y (y + 1)} + \frac{5}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

$\frac{(b y + b - 5 y) y}{y (y + 1)} - \frac{5 (y + 1)}{y (y + 1)} + \frac{5}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

Step 5.1.10

Combine the numerators over the common denominator.

$\frac{(b y + b - 5 y) y - 5 (y + 1)}{y (y + 1)} + \frac{5}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

Step 5.1.11

Simplify the numerator.

Tap for more steps...

Step 5.1.11.1

Apply the distributive property.

$\frac{b y \cdot y + b y - 5 y \cdot y - 5 (y + 1)}{y (y + 1)} + \frac{5}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

Step 5.1.11.2

Simplify.

Tap for more steps...

Step 5.1.11.2.1

Multiply $y$ by $y$ by adding the exponents.

Tap for more steps...

Step 5.1.11.2.1.1

Move $y$ .

$\frac{b (y \cdot y) + b y - 5 y \cdot y - 5 (y + 1)}{y (y + 1)} + \frac{5}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

Step 5.1.11.2.1.2

Multiply $y$ by $y$ .

$\frac{b y^{2} + b y - 5 y \cdot y - 5 (y + 1)}{y (y + 1)} + \frac{5}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

$\frac{b y^{2} + b y - 5 y \cdot y - 5 (y + 1)}{y (y + 1)} + \frac{5}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

Step 5.1.11.2.2

Multiply $y$ by $y$ by adding the exponents.

Tap for more steps...

Step 5.1.11.2.2.1

Move $y$ .

$\frac{b y^{2} + b y - 5 (y \cdot y) - 5 (y + 1)}{y (y + 1)} + \frac{5}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

Step 5.1.11.2.2.2

Multiply $y$ by $y$ .

$\frac{b y^{2} + b y - 5 y^{2} - 5 (y + 1)}{y (y + 1)} + \frac{5}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

$\frac{b y^{2} + b y - 5 y^{2} - 5 (y + 1)}{y (y + 1)} + \frac{5}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

$\frac{b y^{2} + b y - 5 y^{2} - 5 (y + 1)}{y (y + 1)} + \frac{5}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

Step 5.1.11.3

Apply the distributive property.

$\frac{b y^{2} + b y - 5 y^{2} - 5 y - 5 \cdot 1}{y (y + 1)} + \frac{5}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

Step 5.1.11.4

Multiply $- 5$ by $1$ .

$\frac{b y^{2} + b y - 5 y^{2} - 5 y - 5}{y (y + 1)} + \frac{5}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

$\frac{b y^{2} + b y - 5 y^{2} - 5 y - 5}{y (y + 1)} + \frac{5}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

Step 5.1.12

Combine the numerators over the common denominator.

$\frac{b y^{2} + b y - 5 y^{2} - 5 y - 5 + 5}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

Step 5.1.13

Combine the opposite terms in $b y^{2} + b y - 5 y^{2} - 5 y - 5 + 5$ .

Tap for more steps...

Step 5.1.13.1

Add $- 5$ and $5$ .

$\frac{b y^{2} + b y - 5 y^{2} - 5 y + 0}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

Step 5.1.13.2

Add $b y^{2} + b y - 5 y^{2} - 5 y$ and $0$ .

$\frac{b y^{2} + b y - 5 y^{2} - 5 y}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

$\frac{b y^{2} + b y - 5 y^{2} - 5 y}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

Step 5.1.14

Simplify the numerator.

Tap for more steps...

Step 5.1.14.1

Factor $y$ out of $b y^{2} + b y - 5 y^{2} - 5 y$ .

Tap for more steps...

Step 5.1.14.1.1

Factor $y$ out of $b y^{2}$ .

$\frac{y (b y) + b y - 5 y^{2} - 5 y}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

Step 5.1.14.1.2

Factor $y$ out of $b y$ .

$\frac{y (b y) + y b - 5 y^{2} - 5 y}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

Step 5.1.14.1.3

Factor $y$ out of $- 5 y^{2}$ .

$\frac{y (b y) + y b + y (- 5 y) - 5 y}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

Step 5.1.14.1.4

Factor $y$ out of $- 5 y$ .

$\frac{y (b y) + y b + y (- 5 y) + y \cdot - 5}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

Step 5.1.14.1.5

Factor $y$ out of $y (b y) + y b$ .

$\frac{y (b y + b) + y (- 5 y) + y \cdot - 5}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

Step 5.1.14.1.6

Factor $y$ out of $y (b y + b) + y (- 5 y)$ .

$\frac{y (b y + b - 5 y) + y \cdot - 5}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

Step 5.1.14.1.7

Factor $y$ out of $y (b y + b - 5 y) + y \cdot - 5$ .

$\frac{y (b y + b - 5 y - 5)}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

$\frac{y (b y + b - 5 y - 5)}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

Step 5.1.14.2

Factor out the greatest common factor from each group.

Tap for more steps...

Step 5.1.14.2.1

Group the first two terms and the last two terms.

$\frac{y ((b y + b) - 5 y - 5)}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

Step 5.1.14.2.2

Factor out the greatest common factor (GCF) from each group.

$\frac{y (b (y + 1) - 5 (y + 1))}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

$\frac{y (b (y + 1) - 5 (y + 1))}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

Step 5.1.14.3

Factor the polynomial by factoring out the greatest common factor, $y + 1$ .

$\frac{y (y + 1) (b - 5)}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

$\frac{y (y + 1) (b - 5)}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

Step 5.1.15

Cancel the common factor of $y$ .

Tap for more steps...

Step 5.1.15.1

Cancel the common factor.

$\frac{y (y + 1) (b - 5)}{y (y + 1)} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

Step 5.1.15.2

Rewrite the expression.

$\frac{(y + 1) (b - 5)}{y + 1} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

$\frac{(y + 1) (b - 5)}{y + 1} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

Step 5.1.16

Cancel the common factor of $y + 1$ .

Tap for more steps...

Step 5.1.16.1

Cancel the common factor.

$\frac{(y + 1) (b - 5)}{y + 1} = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

Step 5.1.16.2

Divide $b - 5$ by $1$ .

$b - 5 = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

$b - 5 = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

$b - 5 = 0$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

Step 5.2

Add $5$ to both sides of the equation.

$b = 5$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

$b = 5$

$x + \frac{5 y \cdot y}{y + 1} = \frac{5}{y + 1}$

Step 6

Solve the equation for $x$ .

Tap for more steps...

Step 6.1

Move all terms containing variables to the left side of the equation.

Tap for more steps...

Step 6.1.1

Subtract $\frac{5}{y + 1}$ from both sides of the equation.

$x + \frac{5 y \cdot y}{y + 1} - \frac{5}{y + 1} = 0$

$b = 5$

Step 6.1.2

Multiply $y$ by $y$ by adding the exponents.

Tap for more steps...

Step 6.1.2.1

Move $y$ .

$x + \frac{5 (y \cdot y)}{y + 1} - \frac{5}{y + 1} = 0$

$b = 5$

Step 6.1.2.2

Multiply $y$ by $y$ .

$x + \frac{5 y^{2}}{y + 1} - \frac{5}{y + 1} = 0$

$b = 5$

$x + \frac{5 y^{2}}{y + 1} - \frac{5}{y + 1} = 0$

$b = 5$

Step 6.1.3

To write $x$ as a fraction with a common denominator, multiply by $\frac{y + 1}{y + 1}$ .

$x \cdot \frac{y + 1}{y + 1} + \frac{5 y^{2}}{y + 1} - \frac{5}{y + 1} = 0$

$b = 5$

Step 6.1.4

Combine $x$ and $\frac{y + 1}{y + 1}$ .

$\frac{x (y + 1)}{y + 1} + \frac{5 y^{2}}{y + 1} - \frac{5}{y + 1} = 0$

$b = 5$

Step 6.1.5

Combine the numerators over the common denominator.

$\frac{x (y + 1) + 5 y^{2}}{y + 1} - \frac{5}{y + 1} = 0$

$b = 5$

Step 6.1.6

Simplify the numerator.

Tap for more steps...

Step 6.1.6.1

Apply the distributive property.

$\frac{x y + x \cdot 1 + 5 y^{2}}{y + 1} - \frac{5}{y + 1} = 0$

$b = 5$

Step 6.1.6.2

Multiply $x$ by $1$ .

$\frac{x y + x + 5 y^{2}}{y + 1} - \frac{5}{y + 1} = 0$

$b = 5$

$\frac{x y + x + 5 y^{2}}{y + 1} - \frac{5}{y + 1} = 0$

$b = 5$

Step 6.1.7

Combine the numerators over the common denominator.

$\frac{x y + x + 5 y^{2} - 5}{y + 1} = 0$

$b = 5$

$\frac{x y + x + 5 y^{2} - 5}{y + 1} = 0$

$b = 5$

Step 6.2

Set the numerator equal to zero.

$x y + x + 5 y^{2} - 5 = 0$

$b = 5$

Step 6.3

Solve the equation for $x$ .

Tap for more steps...

Step 6.3.1

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

Tap for more steps...

Step 6.3.1.1

Subtract $5 y^{2}$ from both sides of the equation.

$x y + x - 5 = - 5 y^{2}$

$b = 5$

Step 6.3.1.2

Add $5$ to both sides of the equation.

$x y + x = - 5 y^{2} + 5$

$b = 5$

$x y + x = - 5 y^{2} + 5$

$b = 5$

Step 6.3.2

Factor $x$ out of $x y + x$ .

Tap for more steps...

Step 6.3.2.1

Factor $x$ out of $x y$ .

$x (y) + x = - 5 y^{2} + 5$

$b = 5$

Step 6.3.2.2

Raise $x$ to the power of $1$ .

$x (y) + x = - 5 y^{2} + 5$

$b = 5$

Step 6.3.2.3

Factor $x$ out of $x^{1}$ .

$x (y) + x \cdot 1 = - 5 y^{2} + 5$

$b = 5$

Step 6.3.2.4

Factor $x$ out of $x (y) + x \cdot 1$ .

$x (y + 1) = - 5 y^{2} + 5$

$b = 5$

$x (y + 1) = - 5 y^{2} + 5$

$b = 5$

Step 6.3.3

Divide each term in $x (y + 1) = - 5 y^{2} + 5$ by $y + 1$ and simplify.

Tap for more steps...

Step 6.3.3.1

Divide each term in $x (y + 1) = - 5 y^{2} + 5$ by $y + 1$ .

$\frac{x (y + 1)}{y + 1} = \frac{- 5 y^{2}}{y + 1} + \frac{5}{y + 1}$

$b = 5$

Step 6.3.3.2

Simplify the left side.

Tap for more steps...

Step 6.3.3.2.1

Cancel the common factor of $y + 1$ .

Tap for more steps...

Step 6.3.3.2.1.1

Cancel the common factor.

$\frac{x (y + 1)}{y + 1} = \frac{- 5 y^{2}}{y + 1} + \frac{5}{y + 1}$

$b = 5$

Step 6.3.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 5 y^{2}}{y + 1} + \frac{5}{y + 1}$

$b = 5$

$x = \frac{- 5 y^{2}}{y + 1} + \frac{5}{y + 1}$

$b = 5$

$x = \frac{- 5 y^{2}}{y + 1} + \frac{5}{y + 1}$

$b = 5$

Step 6.3.3.3

Simplify the right side.

Tap for more steps...

Step 6.3.3.3.1

Combine the numerators over the common denominator.

$x = \frac{- 5 y^{2} + 5}{y + 1}$

$b = 5$

Step 6.3.3.3.2

Simplify the numerator.

Tap for more steps...

Step 6.3.3.3.2.1

Factor $5$ out of $- 5 y^{2} + 5$ .

Tap for more steps...

Step 6.3.3.3.2.1.1

Factor $5$ out of $- 5 y^{2}$ .

$x = \frac{5 (- y^{2}) + 5}{y + 1}$

$b = 5$

Step 6.3.3.3.2.1.2

Factor $5$ out of $5$ .

$x = \frac{5 (- y^{2}) + 5 (1)}{y + 1}$

$b = 5$

Step 6.3.3.3.2.1.3

Factor $5$ out of $5 (- y^{2}) + 5 (1)$ .

$x = \frac{5 (- y^{2} + 1)}{y + 1}$

$b = 5$

$x = \frac{5 (- y^{2} + 1)}{y + 1}$

$b = 5$

Step 6.3.3.3.2.2

Rewrite $1$ as $1^{2}$ .

$x = \frac{5 (- y^{2} + 1^{2})}{y + 1}$

$b = 5$

Step 6.3.3.3.2.3

Reorder $- y^{2}$ and $1^{2}$ .

$x = \frac{5 (1^{2} - y^{2})}{y + 1}$

$b = 5$

Step 6.3.3.3.2.4

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = y$ .

$x = \frac{5 (1 + y) (1 - y)}{y + 1}$

$b = 5$

$x = \frac{5 (1 + y) (1 - y)}{y + 1}$

$b = 5$

Step 6.3.3.3.3

Simplify terms.

Tap for more steps...

Step 6.3.3.3.3.1

Cancel the common factor of $1 + y$ and $y + 1$ .

Tap for more steps...

Step 6.3.3.3.3.1.1

Reorder terms.

$x = \frac{5 (y + 1) (1 - y)}{y + 1}$

$b = 5$

Step 6.3.3.3.3.1.2

Cancel the common factor.

$x = \frac{5 (y + 1) (1 - y)}{y + 1}$

$b = 5$

Step 6.3.3.3.3.1.3

Divide $5 (1 - y)$ by $1$ .

$x = 5 (1 - y)$

$b = 5$

$x = 5 (1 - y)$

$b = 5$

Step 6.3.3.3.3.2

Apply the distributive property.

$x = 5 \cdot 1 + 5 (- y)$

$b = 5$

Step 6.3.3.3.3.3

Multiply.

Tap for more steps...

Step 6.3.3.3.3.3.1

Multiply $5$ by $1$ .

$x = 5 + 5 (- y)$

$b = 5$

Step 6.3.3.3.3.3.2

Multiply $- 1$ by $5$ .

$x = 5 - 5 y$

$b = 5$

$x = 5 - 5 y$

$b = 5$

$x = 5 - 5 y$

$b = 5$

$x = 5 - 5 y$

$b = 5$

$x = 5 - 5 y$

$b = 5$

$x = 5 - 5 y$

$b = 5$

$x = 5 - 5 y$

$b = 5$

Step 7

The solution is the set of ordered pairs that makes the system true.

$(5, 5 - 5 y, y)$

Step 8

Decompose a solution vector by re-arranging each equation represented in the row-reduced form of the augmented matrix by solving for the dependent variable in each row yields the vector equality.

$X = [\begin{array}{c} b \\ x \\ y \end{array}] = [\begin{array}{c} 5 \\ 5 - 5 y \\ y \end{array}]$