Finite Math Examples

$3 x - y = 0$ , $6 x + 7 y = 1$

Step 1

Solve for $x$ in $3 x - y = 0$ .

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Step 1.1

Add $y$ to both sides of the equation.

$3 x = y$

$6 x + 7 y = 1$

Step 1.2

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

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Step 1.2.1

Divide each term in $3 x = y$ by $3$ .

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

$6 x + 7 y = 1$

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.

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

$6 x + 7 y = 1$

Step 1.2.2.1.2

Divide $x$ by $1$ .

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

$6 x + 7 y = 1$

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

$6 x + 7 y = 1$

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

$6 x + 7 y = 1$

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

$6 x + 7 y = 1$

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

$6 x + 7 y = 1$

Step 2

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

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Step 2.1

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

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

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

Step 2.2

Simplify the left side.

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Step 2.2.1

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

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Step 2.2.1.1

Cancel the common factor of $3$ .

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Step 2.2.1.1.1

Factor $3$ out of $6$ .

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

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

Step 2.2.1.1.2

Cancel the common factor.

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

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

Step 2.2.1.1.3

Rewrite the expression.

$2 y + 7 y = 1$

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

$2 y + 7 y = 1$

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

Step 2.2.1.2

Add $2 y$ and $7 y$ .

$9 y = 1$

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

$9 y = 1$

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

$9 y = 1$

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

$9 y = 1$

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

Step 3

Divide each term in $9 y = 1$ by $9$ and simplify.

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Step 3.1

Divide each term in $9 y = 1$ by $9$ .

$\frac{9 y}{9} = \frac{1}{9}$

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

Step 3.2

Simplify the left side.

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Step 3.2.1

Cancel the common factor of $9$ .

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Step 3.2.1.1

Cancel the common factor.

$\frac{9 y}{9} = \frac{1}{9}$

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

Step 3.2.1.2

Divide $y$ by $1$ .

$y = \frac{1}{9}$

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

$y = \frac{1}{9}$

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

$y = \frac{1}{9}$

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

$y = \frac{1}{9}$

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

Step 4

Replace all occurrences of $y$ with $\frac{1}{9}$ in each equation.

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Step 4.1

Replace all occurrences of $y$ in $x = \frac{y}{3}$ with $\frac{1}{9}$ .

$x = \frac{\frac{1}{9}}{3}$

$y = \frac{1}{9}$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify $\frac{\frac{1}{9}}{3}$ .

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Step 4.2.1.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{1}{9} \cdot \frac{1}{3}$

$y = \frac{1}{9}$

Step 4.2.1.2

Multiply $\frac{1}{9} \cdot \frac{1}{3}$ .

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Step 4.2.1.2.1

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

$x = \frac{1}{9 \cdot 3}$

$y = \frac{1}{9}$

Step 4.2.1.2.2

Multiply $9$ by $3$ .

$x = \frac{1}{27}$

$y = \frac{1}{9}$

$x = \frac{1}{27}$

$y = \frac{1}{9}$

$x = \frac{1}{27}$

$y = \frac{1}{9}$

$x = \frac{1}{27}$

$y = \frac{1}{9}$

$x = \frac{1}{27}$

$y = \frac{1}{9}$

Step 5

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

$(\frac{1}{27}, \frac{1}{9})$

Step 6

The result can be shown in multiple forms.

Point Form:

$(\frac{1}{27}, \frac{1}{9})$

Equation Form:

$x = \frac{1}{27}, y = \frac{1}{9}$

Step 7