Algebra Examples

$f (10) = 5$ $f (2) = - 3$

Step 1

Divide each term in $f (10) = 5$ by $10$ and simplify.

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Step 1.1

Divide each term in $f (10) = 5$ by $10$ .

$\frac{f \cdot 10}{10} = \frac{5}{10}$

Step 1.2

Simplify the left side.

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Step 1.2.1

Cancel the common factor of $10$ .

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Step 1.2.1.1

Cancel the common factor.

$\frac{f \cdot 10}{10} = \frac{5}{10}$

Step 1.2.1.2

Divide $f$ by $1$ .

$f = \frac{5}{10}$

Step 1.3

Simplify the right side.

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Step 1.3.1

Cancel the common factor of $5$ and $10$ .

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Step 1.3.1.1

Factor $5$ out of $5$ .

$f = \frac{5 (1)}{10}$

Step 1.3.1.2

Cancel the common factors.

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Step 1.3.1.2.1

Factor $5$ out of $10$ .

$f = \frac{5 \cdot 1}{5 \cdot 2}$

Step 1.3.1.2.2

Cancel the common factor.

$f = \frac{5 \cdot 1}{5 \cdot 2}$

Step 1.3.1.2.3

Rewrite the expression.

$f = \frac{1}{2}$

Step 2

Replace all occurrences of $f$ with $\frac{1}{2}$ in each equation.

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Step 2.1

Replace all occurrences of $f$ in $f (2) = - 3$ with $\frac{1}{2}$ .

$(\frac{1}{2}) (2) = - 3$

$f = \frac{1}{2}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Cancel the common factor of $2$ .

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Step 2.2.1.1

Cancel the common factor.

$\frac{1}{2} \cdot 2 = - 3$

$f = \frac{1}{2}$

Step 2.2.1.2

Rewrite the expression.

$1 = - 3$

$f = \frac{1}{2}$

$1 = - 3$

$f = \frac{1}{2}$

$1 = - 3$

$f = \frac{1}{2}$

$1 = - 3$

$f = \frac{1}{2}$

Step 3

Since $1 = - 3$ is not true, there is no solution.

No solution