Finite Math Examples

$3 x \cdot 1 - 4 x \cdot 2 = - 7$ , $- 6 x \cdot 1 + 8 x \cdot 2 = 14$

Step 1

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

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Step 1.1

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

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

$10 x = 14$

Step 1.2

Simplify the left side.

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Step 1.2.1

Cancel the common factor of $- 5$ .

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Step 1.2.1.1

Cancel the common factor.

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

$10 x = 14$

Step 1.2.1.2

Divide $x$ by $1$ .

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

$10 x = 14$

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

$10 x = 14$

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

$10 x = 14$

Step 1.3

Simplify the right side.

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Step 1.3.1

Dividing two negative values results in a positive value.

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

$10 x = 14$

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

$10 x = 14$

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

$10 x = 14$

Step 2

Replace all occurrences of $x$ with $\frac{7}{5}$ in each equation.

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Step 2.1

Replace all occurrences of $x$ in $10 x = 14$ with $\frac{7}{5}$ .

$10 (\frac{7}{5}) = 14$

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

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify $10 (\frac{7}{5})$ .

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Step 2.2.1.1

Cancel the common factor of $5$ .

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Step 2.2.1.1.1

Factor $5$ out of $10$ .

$5 (2) (\frac{7}{5}) = 14$

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

Step 2.2.1.1.2

Cancel the common factor.

$5 \cdot (2 (\frac{7}{5})) = 14$

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

Step 2.2.1.1.3

Rewrite the expression.

$2 \cdot 7 = 14$

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

$2 \cdot 7 = 14$

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

Step 2.2.1.2

Multiply $2$ by $7$ .

$14 = 14$

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

$14 = 14$

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

$14 = 14$

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

$14 = 14$

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

Step 3

Remove any equations from the system that are always true.

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

Step 4