Linear Algebra Examples

$3 x - 7 y = 1$ , $- 6 x + 14 y = - 2$

Step 1

Solve for $x$ in $3 x - 7 y = 1$ .

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Step 1.1

Add $7 y$ to both sides of the equation.

$3 x = 1 + 7 y$

$- 6 x + 14 y = - 2$

Step 1.2

Divide each term in $3 x = 1 + 7 y$ by $3$ and simplify.

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Step 1.2.1

Divide each term in $3 x = 1 + 7 y$ by $3$ .

$\frac{3 x}{3} = \frac{1}{3} + \frac{7 y}{3}$

$- 6 x + 14 y = - 2$

Step 1.2.2

Simplify the left side.

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Step 1.2.2.1

Cancel the common factor of $3$ .

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Step 1.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{1}{3} + \frac{7 y}{3}$

$- 6 x + 14 y = - 2$

Step 1.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{1}{3} + \frac{7 y}{3}$

$- 6 x + 14 y = - 2$

$x = \frac{1}{3} + \frac{7 y}{3}$

$- 6 x + 14 y = - 2$

$x = \frac{1}{3} + \frac{7 y}{3}$

$- 6 x + 14 y = - 2$

$x = \frac{1}{3} + \frac{7 y}{3}$

$- 6 x + 14 y = - 2$

$x = \frac{1}{3} + \frac{7 y}{3}$

$- 6 x + 14 y = - 2$

Step 2

Replace all occurrences of $x$ with $\frac{1}{3} + \frac{7 y}{3}$ in each equation.

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Step 2.1

Replace all occurrences of $x$ in $- 6 x + 14 y = - 2$ with $\frac{1}{3} + \frac{7 y}{3}$ .

$- 6 (\frac{1}{3} + \frac{7 y}{3}) + 14 y = - 2$

$x = \frac{1}{3} + \frac{7 y}{3}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify $- 6 (\frac{1}{3} + \frac{7 y}{3}) + 14 y$ .

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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.

$- 6 (\frac{1}{3}) - 6 \frac{7 y}{3} + 14 y = - 2$

$x = \frac{1}{3} + \frac{7 y}{3}$

Step 2.2.1.1.2

Cancel the common factor of $3$ .

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Step 2.2.1.1.2.1

Factor $3$ out of $- 6$ .

$3 (- 2) (\frac{1}{3}) - 6 \frac{7 y}{3} + 14 y = - 2$

$x = \frac{1}{3} + \frac{7 y}{3}$

Step 2.2.1.1.2.2

Cancel the common factor.

$3 \cdot (- 2 (\frac{1}{3})) - 6 \frac{7 y}{3} + 14 y = - 2$

$x = \frac{1}{3} + \frac{7 y}{3}$

Step 2.2.1.1.2.3

Rewrite the expression.

$- 2 - 6 \frac{7 y}{3} + 14 y = - 2$

$x = \frac{1}{3} + \frac{7 y}{3}$

$- 2 - 6 \frac{7 y}{3} + 14 y = - 2$

$x = \frac{1}{3} + \frac{7 y}{3}$

Step 2.2.1.1.3

Cancel the common factor of $3$ .

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Step 2.2.1.1.3.1

Factor $3$ out of $- 6$ .

$- 2 + 3 (- 2) (\frac{7 y}{3}) + 14 y = - 2$

$x = \frac{1}{3} + \frac{7 y}{3}$

Step 2.2.1.1.3.2

Cancel the common factor.

$- 2 + 3 \cdot (- 2 \frac{7 y}{3}) + 14 y = - 2$

$x = \frac{1}{3} + \frac{7 y}{3}$

Step 2.2.1.1.3.3

Rewrite the expression.

$- 2 - 2 (7 y) + 14 y = - 2$

$x = \frac{1}{3} + \frac{7 y}{3}$

$- 2 - 2 (7 y) + 14 y = - 2$

$x = \frac{1}{3} + \frac{7 y}{3}$

Step 2.2.1.1.4

Multiply $7$ by $- 2$ .

$- 2 - 14 y + 14 y = - 2$

$x = \frac{1}{3} + \frac{7 y}{3}$

$- 2 - 14 y + 14 y = - 2$

$x = \frac{1}{3} + \frac{7 y}{3}$

Step 2.2.1.2

Combine the opposite terms in $- 2 - 14 y + 14 y$ .

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Step 2.2.1.2.1

Add $- 14 y$ and $14 y$ .

$- 2 + 0 = - 2$

$x = \frac{1}{3} + \frac{7 y}{3}$

Step 2.2.1.2.2

Add $- 2$ and $0$ .

$- 2 = - 2$

$x = \frac{1}{3} + \frac{7 y}{3}$

$- 2 = - 2$

$x = \frac{1}{3} + \frac{7 y}{3}$

$- 2 = - 2$

$x = \frac{1}{3} + \frac{7 y}{3}$

$- 2 = - 2$

$x = \frac{1}{3} + \frac{7 y}{3}$

$- 2 = - 2$

$x = \frac{1}{3} + \frac{7 y}{3}$

Step 3

Remove any equations from the system that are always true.

$x = \frac{1}{3} + \frac{7 y}{3}$

Step 4