Finite Math Examples

$(- 10, 8)$ , $7 x - 5 y = 2$

Step 1

Solve the equation for $y$ in terms of $x$ .

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Step 1.1

Subtract $7 x$ from both sides of the equation.

$- 5 y = 2 - 7 x$

Step 1.2

Divide each term in $- 5 y = 2 - 7 x$ by $- 5$ and simplify.

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Step 1.2.1

Divide each term in $- 5 y = 2 - 7 x$ by $- 5$ .

$\frac{- 5 y}{- 5} = \frac{2}{- 5} + \frac{- 7 x}{- 5}$

Step 1.2.2

Simplify the left side.

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Step 1.2.2.1

Cancel the common factor of $- 5$ .

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Step 1.2.2.1.1

Cancel the common factor.

$\frac{- 5 y}{- 5} = \frac{2}{- 5} + \frac{- 7 x}{- 5}$

Step 1.2.2.1.2

Divide $y$ by $1$ .

$y = \frac{2}{- 5} + \frac{- 7 x}{- 5}$

Step 1.2.3

Simplify the right side.

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Step 1.2.3.1

Simplify each term.

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Step 1.2.3.1.1

Move the negative in front of the fraction.

$y = - \frac{2}{5} + \frac{- 7 x}{- 5}$

Step 1.2.3.1.2

Dividing two negative values results in a positive value.

$y = - \frac{2}{5} + \frac{7 x}{5}$

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 $f (a) = f (- 10) = - \frac{2}{5} + \frac{7 (- 10)}{5}$ .

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Step 4.1

Combine the numerators over the common denominator.

$f (- 10) = \frac{- 2 + 7 (- 10)}{5}$

Step 4.2

Simplify the expression.

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Step 4.2.1

Multiply $7$ by $- 10$ .

$f (- 10) = \frac{- 2 - 70}{5}$

Step 4.2.2

Subtract $70$ from $- 2$ .

$f (- 10) = \frac{- 72}{5}$

Step 4.2.3

Move the negative in front of the fraction.

$f (- 10) = - \frac{72}{5}$

Step 5

Calculate $f (b) = f (8) = - \frac{2}{5} + \frac{7 (8)}{5}$ .

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Step 5.1

Combine the numerators over the common denominator.

$f (8) = \frac{- 2 + 7 (8)}{5}$

Step 5.2

Simplify the expression.

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Step 5.2.1

Multiply $7$ by $8$ .

$f (8) = \frac{- 2 + 56}{5}$

Step 5.2.2

Add $- 2$ and $56$ .

$f (8) = \frac{54}{5}$

Step 6

Since $0$ is on the interval $[- \frac{72}{5}, \frac{54}{5}]$ , solve the equation for $x$ at the root by setting $y$ to $0$ in $y = - \frac{2}{5} + \frac{7 x}{5}$ .

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Step 6.1

Rewrite the equation as $- \frac{2}{5} + \frac{7 x}{5} = 0$ .

$- \frac{2}{5} + \frac{7 x}{5} = 0$

Step 6.2

Add $\frac{2}{5}$ to both sides of the equation.

$\frac{7 x}{5} = \frac{2}{5}$

Step 6.3

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

$7 x = 2$

Step 6.4

Divide each term in $7 x = 2$ by $7$ and simplify.

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Step 6.4.1

Divide each term in $7 x = 2$ by $7$ .

$\frac{7 x}{7} = \frac{2}{7}$

Step 6.4.2

Simplify the left side.

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Step 6.4.2.1

Cancel the common factor of $7$ .

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Step 6.4.2.1.1

Cancel the common factor.

$\frac{7 x}{7} = \frac{2}{7}$

Step 6.4.2.1.2

Divide $x$ by $1$ .

$x = \frac{2}{7}$

Step 7

The Intermediate Value Theorem states that there is a root $f (c) = 0$ on the interval $[- \frac{72}{5}, \frac{54}{5}]$ because $f$ is a continuous function on $[- 10, 8]$ .

The roots on the interval $[- 10, 8]$ are located at $x = \frac{2}{7}$ .

Step 8