Linear Algebra Examples

$x + 5 y = 4$ , $6 x - 5 y = 3$

Step 1

Subtract $5 y$ from both sides of the equation.

$x = 4 - 5 y$

$6 x - 5 y = 3$

Step 2

Replace all occurrences of $x$ with $4 - 5 y$ in each equation.

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Step 2.1

Replace all occurrences of $x$ in $6 x - 5 y = 3$ with $4 - 5 y$ .

$6 (4 - 5 y) - 5 y = 3$

$x = 4 - 5 y$

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify $6 (4 - 5 y) - 5 y$ .

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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.

$6 \cdot 4 + 6 (- 5 y) - 5 y = 3$

$x = 4 - 5 y$

Step 2.2.1.1.2

Multiply $6$ by $4$ .

$24 + 6 (- 5 y) - 5 y = 3$

$x = 4 - 5 y$

Step 2.2.1.1.3

Multiply $- 5$ by $6$ .

$24 - 30 y - 5 y = 3$

$x = 4 - 5 y$

$24 - 30 y - 5 y = 3$

$x = 4 - 5 y$

Step 2.2.1.2

Subtract $5 y$ from $- 30 y$ .

$24 - 35 y = 3$

$x = 4 - 5 y$

$24 - 35 y = 3$

$x = 4 - 5 y$

$24 - 35 y = 3$

$x = 4 - 5 y$

$24 - 35 y = 3$

$x = 4 - 5 y$

Step 3

Solve for $y$ in $24 - 35 y = 3$ .

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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 $24$ from both sides of the equation.

$- 35 y = 3 - 24$

$x = 4 - 5 y$

Step 3.1.2

Subtract $24$ from $3$ .

$- 35 y = - 21$

$x = 4 - 5 y$

$- 35 y = - 21$

$x = 4 - 5 y$

Step 3.2

Divide each term in $- 35 y = - 21$ by $- 35$ and simplify.

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Step 3.2.1

Divide each term in $- 35 y = - 21$ by $- 35$ .

$\frac{- 35 y}{- 35} = \frac{- 21}{- 35}$

$x = 4 - 5 y$

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Cancel the common factor of $- 35$ .

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Step 3.2.2.1.1

Cancel the common factor.

$\frac{- 35 y}{- 35} = \frac{- 21}{- 35}$

$x = 4 - 5 y$

Step 3.2.2.1.2

Divide $y$ by $1$ .

$y = \frac{- 21}{- 35}$

$x = 4 - 5 y$

$y = \frac{- 21}{- 35}$

$x = 4 - 5 y$

$y = \frac{- 21}{- 35}$

$x = 4 - 5 y$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

Cancel the common factor of $- 21$ and $- 35$ .

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Step 3.2.3.1.1

Factor $- 7$ out of $- 21$ .

$y = \frac{- 7 \cdot 3}{- 35}$

$x = 4 - 5 y$

Step 3.2.3.1.2

Cancel the common factors.

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Step 3.2.3.1.2.1

Factor $- 7$ out of $- 35$ .

$y = \frac{- 7 \cdot 3}{- 7 \cdot 5}$

$x = 4 - 5 y$

Step 3.2.3.1.2.2

Cancel the common factor.

$y = \frac{- 7 \cdot 3}{- 7 \cdot 5}$

$x = 4 - 5 y$

Step 3.2.3.1.2.3

Rewrite the expression.

$y = \frac{3}{5}$

$x = 4 - 5 y$

$y = \frac{3}{5}$

$x = 4 - 5 y$

$y = \frac{3}{5}$

$x = 4 - 5 y$

$y = \frac{3}{5}$

$x = 4 - 5 y$

$y = \frac{3}{5}$

$x = 4 - 5 y$

$y = \frac{3}{5}$

$x = 4 - 5 y$

Step 4

Replace all occurrences of $y$ with $\frac{3}{5}$ in each equation.

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Step 4.1

Replace all occurrences of $y$ in $x = 4 - 5 y$ with $\frac{3}{5}$ .

$x = 4 - 5 (\frac{3}{5})$

$y = \frac{3}{5}$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify $4 - 5 (\frac{3}{5})$ .

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Step 4.2.1.1

Simplify each term.

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Step 4.2.1.1.1

Cancel the common factor of $5$ .

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Step 4.2.1.1.1.1

Factor $5$ out of $- 5$ .

$x = 4 + 5 (- 1) (\frac{3}{5})$

$y = \frac{3}{5}$

Step 4.2.1.1.1.2

Cancel the common factor.

$x = 4 + 5 \cdot (- 1 (\frac{3}{5}))$

$y = \frac{3}{5}$

Step 4.2.1.1.1.3

Rewrite the expression.

$x = 4 - 1 \cdot 3$

$y = \frac{3}{5}$

$x = 4 - 1 \cdot 3$

$y = \frac{3}{5}$

Step 4.2.1.1.2

Multiply $- 1$ by $3$ .

$x = 4 - 3$

$y = \frac{3}{5}$

$x = 4 - 3$

$y = \frac{3}{5}$

Step 4.2.1.2

Subtract $3$ from $4$ .

$x = 1$

$y = \frac{3}{5}$

$x = 1$

$y = \frac{3}{5}$

$x = 1$

$y = \frac{3}{5}$

$x = 1$

$y = \frac{3}{5}$

Step 5

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

$(1, \frac{3}{5})$

Step 6

The result can be shown in multiple forms.

Point Form:

$(1, \frac{3}{5})$

Equation Form:

$x = 1, y = \frac{3}{5}$

Step 7