Linear Algebra Examples

$4 x + 4 y = 1$ , $6 x - y = 1$

Step 1

Solve for $x$ in $4 x + 4 y = 1$ .

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Step 1.1

Subtract $4 y$ from both sides of the equation.

$4 x = 1 - 4 y$

$6 x - y = 1$

Step 1.2

Divide each term in $4 x = 1 - 4 y$ by $4$ and simplify.

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Step 1.2.1

Divide each term in $4 x = 1 - 4 y$ by $4$ .

$\frac{4 x}{4} = \frac{1}{4} + \frac{- 4 y}{4}$

$6 x - y = 1$

Step 1.2.2

Simplify the left side.

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Step 1.2.2.1

Cancel the common factor of $4$ .

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Step 1.2.2.1.1

Cancel the common factor.

$\frac{4 x}{4} = \frac{1}{4} + \frac{- 4 y}{4}$

$6 x - y = 1$

Step 1.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{1}{4} + \frac{- 4 y}{4}$

$6 x - y = 1$

$x = \frac{1}{4} + \frac{- 4 y}{4}$

$6 x - y = 1$

$x = \frac{1}{4} + \frac{- 4 y}{4}$

$6 x - y = 1$

Step 1.2.3

Simplify the right side.

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Step 1.2.3.1

Cancel the common factor of $- 4$ and $4$ .

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Step 1.2.3.1.1

Factor $4$ out of $- 4 y$ .

$x = \frac{1}{4} + \frac{4 (- y)}{4}$

$6 x - y = 1$

Step 1.2.3.1.2

Cancel the common factors.

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Step 1.2.3.1.2.1

Factor $4$ out of $4$ .

$x = \frac{1}{4} + \frac{4 (- y)}{4 (1)}$

$6 x - y = 1$

Step 1.2.3.1.2.2

Cancel the common factor.

$x = \frac{1}{4} + \frac{4 (- y)}{4 \cdot 1}$

$6 x - y = 1$

Step 1.2.3.1.2.3

Rewrite the expression.

$x = \frac{1}{4} + \frac{- y}{1}$

$6 x - y = 1$

Step 1.2.3.1.2.4

Divide $- y$ by $1$ .

$x = \frac{1}{4} - y$

$6 x - y = 1$

$x = \frac{1}{4} - y$

$6 x - y = 1$

$x = \frac{1}{4} - y$

$6 x - y = 1$

$x = \frac{1}{4} - y$

$6 x - y = 1$

$x = \frac{1}{4} - y$

$6 x - y = 1$

$x = \frac{1}{4} - y$

$6 x - y = 1$

Step 2

Replace all occurrences of $x$ with $\frac{1}{4} - y$ in each equation.

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Step 2.1

Replace all occurrences of $x$ in $6 x - y = 1$ with $\frac{1}{4} - y$ .

$6 (\frac{1}{4} - y) - y = 1$

$x = \frac{1}{4} - y$

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify $6 (\frac{1}{4} - y) - y$ .

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Step 2.2.1.1

Simplify each term.

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Step 2.2.1.1.1

Apply the distributive property.

$6 (\frac{1}{4}) + 6 (- y) - y = 1$

$x = \frac{1}{4} - y$

Step 2.2.1.1.2

Cancel the common factor of $2$ .

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Step 2.2.1.1.2.1

Factor $2$ out of $6$ .

$2 (3) (\frac{1}{4}) + 6 (- y) - y = 1$

$x = \frac{1}{4} - y$

Step 2.2.1.1.2.2

Factor $2$ out of $4$ .

$2 \cdot (3 (\frac{1}{2 \cdot 2})) + 6 (- y) - y = 1$

$x = \frac{1}{4} - y$

Step 2.2.1.1.2.3

Cancel the common factor.

$2 \cdot (3 (\frac{1}{2 \cdot 2})) + 6 (- y) - y = 1$

$x = \frac{1}{4} - y$

Step 2.2.1.1.2.4

Rewrite the expression.

$3 (\frac{1}{2}) + 6 (- y) - y = 1$

$x = \frac{1}{4} - y$

$3 (\frac{1}{2}) + 6 (- y) - y = 1$

$x = \frac{1}{4} - y$

Step 2.2.1.1.3

Combine $3$ and $\frac{1}{2}$ .

$\frac{3}{2} + 6 (- y) - y = 1$

$x = \frac{1}{4} - y$

Step 2.2.1.1.4

Multiply $- 1$ by $6$ .

$\frac{3}{2} - 6 y - y = 1$

$x = \frac{1}{4} - y$

$\frac{3}{2} - 6 y - y = 1$

$x = \frac{1}{4} - y$

Step 2.2.1.2

Subtract $y$ from $- 6 y$ .

$\frac{3}{2} - 7 y = 1$

$x = \frac{1}{4} - y$

$\frac{3}{2} - 7 y = 1$

$x = \frac{1}{4} - y$

$\frac{3}{2} - 7 y = 1$

$x = \frac{1}{4} - y$

$\frac{3}{2} - 7 y = 1$

$x = \frac{1}{4} - y$

Step 3

Solve for $y$ in $\frac{3}{2} - 7 y = 1$ .

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Step 3.1

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

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Step 3.1.1

Subtract $\frac{3}{2}$ from both sides of the equation.

$- 7 y = 1 - \frac{3}{2}$

$x = \frac{1}{4} - y$

Step 3.1.2

Write $1$ as a fraction with a common denominator.

$- 7 y = \frac{2}{2} - \frac{3}{2}$

$x = \frac{1}{4} - y$

Step 3.1.3

Combine the numerators over the common denominator.

$- 7 y = \frac{2 - 3}{2}$

$x = \frac{1}{4} - y$

Step 3.1.4

Subtract $3$ from $2$ .

$- 7 y = \frac{- 1}{2}$

$x = \frac{1}{4} - y$

Step 3.1.5

Move the negative in front of the fraction.

$- 7 y = - \frac{1}{2}$

$x = \frac{1}{4} - y$

$- 7 y = - \frac{1}{2}$

$x = \frac{1}{4} - y$

Step 3.2

Divide each term in $- 7 y = - \frac{1}{2}$ by $- 7$ and simplify.

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Step 3.2.1

Divide each term in $- 7 y = - \frac{1}{2}$ by $- 7$ .

$\frac{- 7 y}{- 7} = \frac{- \frac{1}{2}}{- 7}$

$x = \frac{1}{4} - y$

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Cancel the common factor of $- 7$ .

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Step 3.2.2.1.1

Cancel the common factor.

$\frac{- 7 y}{- 7} = \frac{- \frac{1}{2}}{- 7}$

$x = \frac{1}{4} - y$

Step 3.2.2.1.2

Divide $y$ by $1$ .

$y = \frac{- \frac{1}{2}}{- 7}$

$x = \frac{1}{4} - y$

$y = \frac{- \frac{1}{2}}{- 7}$

$x = \frac{1}{4} - y$

$y = \frac{- \frac{1}{2}}{- 7}$

$x = \frac{1}{4} - y$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

Multiply the numerator by the reciprocal of the denominator.

$y = - \frac{1}{2} \cdot \frac{1}{- 7}$

$x = \frac{1}{4} - y$

Step 3.2.3.2

Move the negative in front of the fraction.

$y = - \frac{1}{2} \cdot (- \frac{1}{7})$

$x = \frac{1}{4} - y$

Step 3.2.3.3

Multiply $- \frac{1}{2} (- \frac{1}{7})$ .

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Step 3.2.3.3.1

Multiply $- 1$ by $- 1$ .

$y = 1 (\frac{1}{2}) \cdot \frac{1}{7}$

$x = \frac{1}{4} - y$

Step 3.2.3.3.2

Multiply $\frac{1}{2}$ by $1$ .

$y = \frac{1}{2} \cdot \frac{1}{7}$

$x = \frac{1}{4} - y$

Step 3.2.3.3.3

Multiply $\frac{1}{2}$ by $\frac{1}{7}$ .

$y = \frac{1}{2 \cdot 7}$

$x = \frac{1}{4} - y$

Step 3.2.3.3.4

Multiply $2$ by $7$ .

$y = \frac{1}{14}$

$x = \frac{1}{4} - y$

$y = \frac{1}{14}$

$x = \frac{1}{4} - y$

$y = \frac{1}{14}$

$x = \frac{1}{4} - y$

$y = \frac{1}{14}$

$x = \frac{1}{4} - y$

$y = \frac{1}{14}$

$x = \frac{1}{4} - y$

Step 4

Replace all occurrences of $y$ with $\frac{1}{14}$ in each equation.

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Step 4.1

Replace all occurrences of $y$ in $x = \frac{1}{4} - y$ with $\frac{1}{14}$ .

$x = \frac{1}{4} - (\frac{1}{14})$

$y = \frac{1}{14}$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify $\frac{1}{4} - (\frac{1}{14})$ .

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Step 4.2.1.1

To write $\frac{1}{4}$ as a fraction with a common denominator, multiply by $\frac{7}{7}$ .

$x = \frac{1}{4} \cdot \frac{7}{7} - \frac{1}{14}$

$y = \frac{1}{14}$

Step 4.2.1.2

To write $- \frac{1}{14}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$x = \frac{1}{4} \cdot \frac{7}{7} - \frac{1}{14} \cdot \frac{2}{2}$

$y = \frac{1}{14}$

Step 4.2.1.3

Write each expression with a common denominator of $28$ , by multiplying each by an appropriate factor of $1$ .

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Step 4.2.1.3.1

Multiply $\frac{1}{4}$ by $\frac{7}{7}$ .

$x = \frac{7}{4 \cdot 7} - \frac{1}{14} \cdot \frac{2}{2}$

$y = \frac{1}{14}$

Step 4.2.1.3.2

Multiply $4$ by $7$ .

$x = \frac{7}{28} - \frac{1}{14} \cdot \frac{2}{2}$

$y = \frac{1}{14}$

Step 4.2.1.3.3

Multiply $\frac{1}{14}$ by $\frac{2}{2}$ .

$x = \frac{7}{28} - \frac{2}{14 \cdot 2}$

$y = \frac{1}{14}$

Step 4.2.1.3.4

Multiply $14$ by $2$ .

$x = \frac{7}{28} - \frac{2}{28}$

$y = \frac{1}{14}$

$x = \frac{7}{28} - \frac{2}{28}$

$y = \frac{1}{14}$

Step 4.2.1.4

Combine the numerators over the common denominator.

$x = \frac{7 - 2}{28}$

$y = \frac{1}{14}$

Step 4.2.1.5

Subtract $2$ from $7$ .

$x = \frac{5}{28}$

$y = \frac{1}{14}$

$x = \frac{5}{28}$

$y = \frac{1}{14}$

$x = \frac{5}{28}$

$y = \frac{1}{14}$

$x = \frac{5}{28}$

$y = \frac{1}{14}$

Step 5

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

$(\frac{5}{28}, \frac{1}{14})$

Step 6

The result can be shown in multiple forms.

Point Form:

$(\frac{5}{28}, \frac{1}{14})$

Equation Form:

$x = \frac{5}{28}, y = \frac{1}{14}$

Step 7

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