Linear Algebra Examples

$\frac{2 x + 3}{3 y - 2} = 1$ , $x (2 y - 5) - 2 y (x + 3) = 2 x + 1$

Step 1

Simplify.

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Step 1.1

Simplify each term.

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Step 1.1.1

Apply the distributive property.

$\frac{2 x + 3}{3 y - 2} = 1, x (2 y) + x \cdot - 5 - 2 y (x + 3) = 2 x + 1$

Step 1.1.2

Rewrite using the commutative property of multiplication.

$\frac{2 x + 3}{3 y - 2} = 1, 2 x y + x \cdot - 5 - 2 y (x + 3) = 2 x + 1$

Step 1.1.3

Move $- 5$ to the left of $x$ .

$\frac{2 x + 3}{3 y - 2} = 1, 2 x y - 5 \cdot x - 2 y (x + 3) = 2 x + 1$

Step 1.1.4

Apply the distributive property.

$\frac{2 x + 3}{3 y - 2} = 1, 2 x y - 5 x - 2 y x - 2 y \cdot 3 = 2 x + 1$

Step 1.1.5

Multiply $3$ by $- 2$ .

$\frac{2 x + 3}{3 y - 2} = 1, 2 x y - 5 x - 2 y x - 6 y = 2 x + 1$

Step 1.2

Combine the opposite terms in $2 x y - 5 x - 2 y x - 6 y$ .

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Step 1.2.1

Reorder the factors in the terms $2 x y$ and $- 2 y x$ .

$\frac{2 x + 3}{3 y - 2} = 1, 2 x y - 5 x - 2 x y - 6 y = 2 x + 1$

Step 1.2.2

Subtract $2 x y$ from $2 x y$ .

$\frac{2 x + 3}{3 y - 2} = 1, - 5 x + 0 - 6 y = 2 x + 1$

Step 1.2.3

Add $- 5 x$ and $0$ .

$\frac{2 x + 3}{3 y - 2} = 1, - 5 x - 6 y = 2 x + 1$

Step 2

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

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Step 2.1

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

$\frac{2 x + 3}{3 y - 2} = 1, - 5 x - 6 y - 2 x = 1$

Step 2.2

Subtract $2 x$ from $- 5 x$ .

$\frac{2 x + 3}{3 y - 2} = 1, - 7 x - 6 y = 1$

Step 3

Write the system of equations in matrix form.

$[\begin{array}{c} \frac{1}{3 y - 2} & 0 & 1 \\ - 7 & - 6 & 1 \end{array}]$

Step 4

Find the reduced row echelon form.

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Step 4.1

Multiply each element of $R_{1}$ by $3 y - 2$ to make the entry at $1, 1$ a $1$ .

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Step 4.1.1

Multiply each element of $R_{1}$ by $3 y - 2$ to make the entry at $1, 1$ a $1$ .

$[\begin{array}{c} (3 y - 2) \frac{1}{3 y - 2} & (3 y - 2) \cdot 0 & (3 y - 2) \cdot 1 \\ - 7 & - 6 & 1 \end{array}]$

Step 4.1.2

Simplify $R_{1}$ .

$[\begin{array}{c} 1 & 0 & 3 y - 2 \\ - 7 & - 6 & 1 \end{array}]$

Step 4.2

Perform the row operation $R_{2} = R_{2} + 7 R_{1}$ to make the entry at $2, 1$ a $0$ .

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Step 4.2.1

Perform the row operation $R_{2} = R_{2} + 7 R_{1}$ to make the entry at $2, 1$ a $0$ .

$[\begin{array}{c} 1 & 0 & 3 y - 2 \\ - 7 + 7 \cdot 1 & - 6 + 7 \cdot 0 & 1 + 7 (3 y - 2) \end{array}]$

Step 4.2.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 0 & 3 y - 2 \\ 0 & - 6 & 21 y - 13 \end{array}]$

Step 4.3

Multiply each element of $R_{2}$ by $- \frac{1}{6}$ to make the entry at $2, 2$ a $1$ .

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Step 4.3.1

Multiply each element of $R_{2}$ by $- \frac{1}{6}$ to make the entry at $2, 2$ a $1$ .

$[\begin{array}{c} 1 & 0 & 3 y - 2 \\ - \frac{1}{6} \cdot 0 & - \frac{1}{6} \cdot - 6 & - \frac{1}{6} (21 y - 13) \end{array}]$

Step 4.3.2

Simplify $R_{2}$ .

$[\begin{array}{c} 1 & 0 & 3 y - 2 \\ 0 & 1 & - \frac{7 y}{2} + \frac{13}{6} \end{array}]$

Step 5

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

$x = 3 y - 2$

$y = - \frac{7 y}{2} + \frac{13}{6}$

Step 6

Solve the equation for $y$ .

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Step 6.1

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

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Step 6.1.1

Add $\frac{7 y}{2}$ to both sides of the equation.

$y + \frac{7 y}{2} = \frac{13}{6}$

$x = 3 y - 2$

Step 6.1.2

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

$y \cdot \frac{2}{2} + \frac{7 y}{2} = \frac{13}{6}$

$x = 3 y - 2$

Step 6.1.3

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

$\frac{y \cdot 2}{2} + \frac{7 y}{2} = \frac{13}{6}$

$x = 3 y - 2$

Step 6.1.4

Combine the numerators over the common denominator.

$\frac{y \cdot 2 + 7 y}{2} = \frac{13}{6}$

$x = 3 y - 2$

Step 6.1.5

Simplify the numerator.

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Step 6.1.5.1

Move $2$ to the left of $y$ .

$\frac{2 \cdot y + 7 y}{2} = \frac{13}{6}$

$x = 3 y - 2$

Step 6.1.5.2

Add $2 y$ and $7 y$ .

$\frac{9 y}{2} = \frac{13}{6}$

$x = 3 y - 2$

$\frac{9 y}{2} = \frac{13}{6}$

$x = 3 y - 2$

$\frac{9 y}{2} = \frac{13}{6}$

$x = 3 y - 2$

Step 6.2

Multiply both sides of the equation by $\frac{2}{9}$ .

$\frac{2}{9} \cdot \frac{9 y}{2} = \frac{2}{9} \cdot \frac{13}{6}$

$x = 3 y - 2$

Step 6.3

Simplify both sides of the equation.

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Step 6.3.1

Simplify the left side.

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Step 6.3.1.1

Simplify $\frac{2}{9} \cdot \frac{9 y}{2}$ .

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Step 6.3.1.1.1

Cancel the common factor of $2$ .

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Step 6.3.1.1.1.1

Cancel the common factor.

$\frac{2}{9} \cdot \frac{9 y}{2} = \frac{2}{9} \cdot \frac{13}{6}$

$x = 3 y - 2$

Step 6.3.1.1.1.2

Rewrite the expression.

$\frac{1}{9} \cdot (9 y) = \frac{2}{9} \cdot \frac{13}{6}$

$x = 3 y - 2$

$\frac{1}{9} \cdot (9 y) = \frac{2}{9} \cdot \frac{13}{6}$

$x = 3 y - 2$

Step 6.3.1.1.2

Cancel the common factor of $9$ .

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Step 6.3.1.1.2.1

Factor $9$ out of $9 y$ .

$\frac{1}{9} \cdot (9 (y)) = \frac{2}{9} \cdot \frac{13}{6}$

$x = 3 y - 2$

Step 6.3.1.1.2.2

Cancel the common factor.

$\frac{1}{9} \cdot (9 y) = \frac{2}{9} \cdot \frac{13}{6}$

$x = 3 y - 2$

Step 6.3.1.1.2.3

Rewrite the expression.

$y = \frac{2}{9} \cdot \frac{13}{6}$

$x = 3 y - 2$

$y = \frac{2}{9} \cdot \frac{13}{6}$

$x = 3 y - 2$

$y = \frac{2}{9} \cdot \frac{13}{6}$

$x = 3 y - 2$

$y = \frac{2}{9} \cdot \frac{13}{6}$

$x = 3 y - 2$

Step 6.3.2

Simplify the right side.

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Step 6.3.2.1

Simplify $\frac{2}{9} \cdot \frac{13}{6}$ .

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Step 6.3.2.1.1

Cancel the common factor of $2$ .

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Step 6.3.2.1.1.1

Factor $2$ out of $6$ .

$y = \frac{2}{9} \cdot \frac{13}{2 (3)}$

$x = 3 y - 2$

Step 6.3.2.1.1.2

Cancel the common factor.

$y = \frac{2}{9} \cdot \frac{13}{2 \cdot 3}$

$x = 3 y - 2$

Step 6.3.2.1.1.3

Rewrite the expression.

$y = \frac{1}{9} \cdot \frac{13}{3}$

$x = 3 y - 2$

$y = \frac{1}{9} \cdot \frac{13}{3}$

$x = 3 y - 2$

Step 6.3.2.1.2

Multiply $\frac{1}{9}$ by $\frac{13}{3}$ .

$y = \frac{13}{9 \cdot 3}$

$x = 3 y - 2$

Step 6.3.2.1.3

Multiply $9$ by $3$ .

$y = \frac{13}{27}$

$x = 3 y - 2$

$y = \frac{13}{27}$

$x = 3 y - 2$

$y = \frac{13}{27}$

$x = 3 y - 2$

$y = \frac{13}{27}$

$x = 3 y - 2$

$y = \frac{13}{27}$

$x = 3 y - 2$

Step 7

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

$(3 y - 2, \frac{13}{27})$

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.

$X = [\begin{array}{c} x \\ y \end{array}] = [\begin{array}{c} 3 y - 2 \\ \frac{13}{27} \end{array}]$