Finite Math Examples

(5, 6)

$(5, 6)$ ,

x + 6 y = 5

$x + 6 y = 5$

Step 1

Solve the equation for

y

$y$ in terms of

x

$x$ .

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Step 1.1

Subtract

x

$x$ from both sides of the equation.

6 y = 5 - x

$6 y = 5 - x$

Step 1.2

Divide each term in

6 y = 5 - x

$6 y = 5 - x$ by

6

$6$ and simplify.

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Step 1.2.1

Divide each term in

6 y = 5 - x

$6 y = 5 - x$ by

6

$6$ .

\frac{6 y}{6} = \frac{5}{6} + \frac{- x}{6}

$\frac{6 y}{6} = \frac{5}{6} + \frac{- x}{6}$

Step 1.2.2

Simplify the left side.

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Step 1.2.2.1

Cancel the common factor of

6

$6$ .

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Step 1.2.2.1.1

Cancel the common factor.

$\frac{6 y}{6} = \frac{5}{6} + \frac{- x}{6}$

Step 1.2.2.1.2

Divide

$y$ by

$1$ .

$y = \frac{5}{6} + \frac{- x}{6}$

Step 1.2.3

Simplify the right side.

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Step 1.2.3.1

Move the negative in front of the fraction.

$y = \frac{5}{6} - \frac{x}{6}$

Step 2

The Intermediate Value Theorem states that, if

$f$ is a real-valued continuous function on the interval

$[a, b]$ , and

$u$ is a number between

$f (a)$ and

$f (b)$ , then there is a

$c$ contained in the interval

$[a, b]$ such that

$f (c) = u$ .

$u = f (c) = 0$

Step 3

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Step 4

Calculate

$f (a) = f (5) = \frac{5}{6} - \frac{5}{6}$ .

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Step 4.1

Combine the numerators over the common denominator.

$f (5) = \frac{5 - 5}{6}$

Step 4.2

Simplify the expression.

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Step 4.2.1

Subtract

$5$ from

$5$ .

$f (5) = \frac{0}{6}$

Step 4.2.2

Divide

$0$ by

$6$ .

$f (5) = 0$

Step 5

Calculate

$f (b) = f (6) = \frac{5}{6} - \frac{6}{6}$ .

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Step 5.1

Combine the numerators over the common denominator.

$f (6) = \frac{5 - 6}{6}$

Step 5.2

Simplify the expression.

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Step 5.2.1

Subtract

$6$ from

$5$ .

$f (6) = \frac{- 1}{6}$

Step 5.2.2

Move the negative in front of the fraction.

$f (6) = - \frac{1}{6}$

Step 6

Since

$0$ is on the interval

$[- \frac{1}{6}, 0]$ , solve the equation for

$x$ at the root by setting

$y$ to

$0$ in

$y = \frac{5}{6} - \frac{x}{6}$ .

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Step 6.1

Rewrite the equation as

$\frac{5}{6} - \frac{x}{6} = 0$ .

$\frac{5}{6} - \frac{x}{6} = 0$

Step 6.2

Subtract

$\frac{5}{6}$ from both sides of the equation.

$- \frac{x}{6} = - \frac{5}{6}$

Step 6.3

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

$- x = - 5$

Step 6.4

Divide each term in

$- x = - 5$ by

$- 1$ and simplify.

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Step 6.4.1

Divide each term in

$- x = - 5$ by

$- 1$ .

$\frac{- x}{- 1} = \frac{- 5}{- 1}$

Step 6.4.2

Simplify the left side.

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Step 6.4.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 5}{- 1}$

Step 6.4.2.2

Divide

$x$ by

$1$ .

$x = \frac{- 5}{- 1}$

Step 6.4.3

Simplify the right side.

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Step 6.4.3.1

Divide

$- 5$ by

$- 1$ .

$x = 5$

Step 7

The Intermediate Value Theorem states that there is a root

$f (c) = 0$ on the interval

$[- \frac{1}{6}, 0]$ because

$f$ is a continuous function on

$[5, 6]$ .

The roots on the interval

$[5, 6]$ are located at

Step 8