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Rational Equations: Proportions Solutions

To solve rational equations, determine the value of the variable .
When the degree of the variable is one , there will be at most one solution.
When the degree of the variable is greater than one , that degree is the maximum number of possible solutions.
However, because there can be restrictions on the denominator, it is possible to have fewer solutions than the degree.
Therefore, it is important to check for restrictions on the denominator as well as extraneous solutions.
When one rational expression is set equal to another, this creates a proportion .
Solve proportions by finding the cross-product .

Example 1

Implement

$\frac{h + 2}{h} = \frac{2 h + 1}{h + 2}$

$h \neq - 2, 0 (h + 2) (h + 2) = h (2 h + 1)$

$\begin{array}{rcl} h^{2} + 4 h + 4 & = & 2 h^{2} + h \\ 0 & = & h^{2} - 3 h - 4 \\ (h + 1) (h - 4) & = & 0 \\ h & = & - 1, 4 \end{array}$

$\begin{array}{rcl} Check \\ \begin{matrix} h = - 1 & h = 4 \\ \frac{(- 1) + 2}{(- 1)} = \frac{2 (- 1) + 1}{(- 1) + 2} ✓ & \frac{(4) + 2}{(4)} = \frac{2 (4) + 1}{(4) + 2} ✓ \end{matrix} \end{array}$

Explain

Restrictions
Cross-product
Distribute
Move terms to one side 
Factor 
Solve for h

Because the solutions do not include values that would make the denominator zero, there are no extraneous solutions.

Note

Remember that you cannot simplify the numerator and denominator of expressions on opposite sides of the equation. However, it is critical to understand you can simplify expressions (no equal sign).

Example 2

Implement

$\frac{3}{x - 2} = \frac{5}{x + 2}$

$\begin{array}{rcl} \begin{matrix} x \neq - 2, 2 \end{matrix} \\ 3 (x + 2) & = & 5 (x - 2) \\ 3 x + 6 & = & 5 x - 10 \\ 16 & = & 2 x \\ x & = & 8 \\ Check \\ \frac{3}{(8) - 2} & = & \frac{5}{(8) + 2} ✓ \end{array}$

Explain

Restrictions 
Cross-product 
Distribute 
Isolate the variable