Algebra Examples

$8 x - \frac{1}{3} y = 0$ $12 x + 3 = y$

Step 1

Rewrite the equation as $y = 12 x + 3$ .

$y = 12 x + 3$

$8 x - \frac{1}{3} y = 0$

Step 2

Replace all occurrences of $y$ with $12 x + 3$ in each equation.

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Step 2.1

Replace all occurrences of $y$ in $8 x - \frac{1}{3} y = 0$ with $12 x + 3$ .

$8 x - \frac{1}{3} \cdot (12 x + 3) = 0$

$y = 12 x + 3$

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify $8 x - \frac{1}{3} \cdot (12 x + 3)$ .

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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.

$8 x - \frac{1}{3} \cdot (12 x) - \frac{1}{3} \cdot 3 = 0$

$y = 12 x + 3$

Step 2.2.1.1.2

Cancel the common factor of $3$ .

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Step 2.2.1.1.2.1

Move the leading negative in $- \frac{1}{3}$ into the numerator.

$8 x + \frac{- 1}{3} \cdot (12 x) - \frac{1}{3} \cdot 3 = 0$

$y = 12 x + 3$

Step 2.2.1.1.2.2

Factor $3$ out of $12 x$ .

$8 x + \frac{- 1}{3} \cdot (3 (4 x)) - \frac{1}{3} \cdot 3 = 0$

$y = 12 x + 3$

Step 2.2.1.1.2.3

Cancel the common factor.

$8 x + \frac{- 1}{3} \cdot (3 (4 x)) - \frac{1}{3} \cdot 3 = 0$

$y = 12 x + 3$

Step 2.2.1.1.2.4

Rewrite the expression.

$8 x - 1 (4 x) - \frac{1}{3} \cdot 3 = 0$

$y = 12 x + 3$

$8 x - 1 (4 x) - \frac{1}{3} \cdot 3 = 0$

$y = 12 x + 3$

Step 2.2.1.1.3

Multiply $4$ by $- 1$ .

$8 x - 4 x - \frac{1}{3} \cdot 3 = 0$

$y = 12 x + 3$

Step 2.2.1.1.4

Cancel the common factor of $3$ .

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Step 2.2.1.1.4.1

Move the leading negative in $- \frac{1}{3}$ into the numerator.

$8 x - 4 x + \frac{- 1}{3} \cdot 3 = 0$

$y = 12 x + 3$

Step 2.2.1.1.4.2

Cancel the common factor.

$8 x - 4 x + \frac{- 1}{3} \cdot 3 = 0$

$y = 12 x + 3$

Step 2.2.1.1.4.3

Rewrite the expression.

$8 x - 4 x - 1 = 0$

$y = 12 x + 3$

$8 x - 4 x - 1 = 0$

$y = 12 x + 3$

$8 x - 4 x - 1 = 0$

$y = 12 x + 3$

Step 2.2.1.2

Subtract $4 x$ from $8 x$ .

$4 x - 1 = 0$

$y = 12 x + 3$

$4 x - 1 = 0$

$y = 12 x + 3$

$4 x - 1 = 0$

$y = 12 x + 3$

$4 x - 1 = 0$

$y = 12 x + 3$

Step 3

Solve for $x$ in $4 x - 1 = 0$ .

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Step 3.1

Add $1$ to both sides of the equation.

$4 x = 1$

$y = 12 x + 3$

Step 3.2

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

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Step 3.2.1

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

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

$y = 12 x + 3$

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Cancel the common factor of $4$ .

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Step 3.2.2.1.1

Cancel the common factor.

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

$y = 12 x + 3$

Step 3.2.2.1.2

Divide $x$ by $1$ .

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

$y = 12 x + 3$

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

$y = 12 x + 3$

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

$y = 12 x + 3$

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

$y = 12 x + 3$

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

$y = 12 x + 3$

Step 4

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

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Step 4.1

Replace all occurrences of $x$ in $y = 12 x + 3$ with $\frac{1}{4}$ .

$y = 12 (\frac{1}{4}) + 3$

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

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify $12 (\frac{1}{4}) + 3$ .

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Step 4.2.1.1

Cancel the common factor of $4$ .

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Step 4.2.1.1.1

Factor $4$ out of $12$ .

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

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

Step 4.2.1.1.2

Cancel the common factor.

$y = 4 \cdot (3 (\frac{1}{4})) + 3$

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

Step 4.2.1.1.3

Rewrite the expression.

$y = 3 + 3$

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

$y = 3 + 3$

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

Step 4.2.1.2

Add $3$ and $3$ .

$y = 6$

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

$y = 6$

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

$y = 6$

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

$y = 6$

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

Step 5

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

$(\frac{1}{4}, 6)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(\frac{1}{4}, 6)$

Equation Form:

$x = \frac{1}{4}, y = 6$

Step 7