Rational Equations and Inequalities

Rational Equations

Rational Equations are equations containing rational expressions. Rational Equations only require one rational expression (usually displayed on the right). The left side can either contain another rational expression, a zero, or a non-zero constant.

Rational Inequalities

Rational Inequalities are inequalities that contain rational expressions. Rational Inequalities only require one rational expression (usually displayed on the right). The left side can either contain another rational expression, a zero, or a non-zero constant.

The General Form of a rational inequality indicates that the main expression is on the left side while the zero (or secondary expression) stays on the right side. They can be expressed in the following ways:

  • \(f(x) \gt 0\)
  • \(f(x) \geq 0\)
  • \(f(x) \lt 0 \)
  • \(f(x) \leq 0\)

It should be noted that even though a rational inequality can have one side equal to or greater/less than the other, both sides can NEVER be fully equivalent to each other.

Solving Rational Equations and Inequalities

Rational equations and inequalities can be solved in a handful of different ways. These include:

  • Solving the Inequality Algebraically. This involves factoring the expressions in the numerator and denominator to find asymptotes and restrictions. Then, multiply both sides by the factored denominators and simplify to obtain a polynomial equation. This usually requires setting one side (normally the right) to zero
  • Graphing Using Key Features. Using the rational function, we are able to determine characteristics such as asymptotes and intercepts. From there, we can sketch a graph around these features to determine its solutions
  • Using a Number Line. We can use this to help determine the intervals outlined by the function’s zeroes/critical points. This will allow us to determine our test points
  • Using an Interval Table. This is a continuation of the previous technique. We can use the test point to help determine the signs of the factors on each interval and record them in a table

For solving problems, it is likely that you use several of these techniques in conjunction with each other.


Example

Solve the rational equation \(\cfrac{x+8}{x-4} = 5\).

First, we can multiply both sides by the denominator of the expression:

\(\left(\cfrac{x+8}{x-4}\right)(x-4) = 5(x-4)\)

\(x + 8 = 5x - 20\)

Next, we can collect like terms by moving them to their respective sides:

\(5x - x = 20 + 8\)

\(4x = 28\)

Finally, we can divide both sides by the coefficient of \(x\) to get our solution:

\(\cfrac{4x}{4} = \cfrac{28}{4}\)

\(x = 7\)

Therefore, we can determine our solution is \(\boldsymbol{x = 7}\).


Example

Solve the rational inequality \(\cfrac{2x - 13}{x + 6} \geq 7\).

First, we can move all terms onto one side:

\(\cfrac{2x - 13}{x + 6} - 7 \geq 0\)

Next, we can combine terms under a common denominator:

\(\cfrac{2x - 13}{x + 6} - 7(x+6) \geq 0\)

\(\cfrac{2x - 13 - 7x - 42}{x + 6}\geq 0\)

Finally, we can collect like terms in the numerator:

\(\cfrac{-5x - 55}{x + 6}\geq 0\)

In order to help solve this inequality, we can draw a graph with the expression's key features. We can determine these features as such:

  • \(\text{VA}: x = -6\)
  • \(\text{HA}: y = -5\)
  • \(x\)-int: \(x = -11\)
  • \(y\)-int: \(y = -\cfrac{55}{6}\)

Using these key features, we are able to sketch a graph of the function. This will help determine the intervals to satisfy the inequality:

Graph of Reciprocal Function expressed as 2x-13/(x+6).

From looking at the graph, we can determine the interval(s) where \(\cfrac{2x - 13}{x + 6} \geq 7\) is \(\boldsymbol{-11 \geq x \gt -6}\).

This can be expressed in interval notation as \(\boldsymbol{x \in [-11, -6)}\).


Solve the rational equation \(\cfrac{-2x-10}{5x} - \cfrac{3}{5} = 15 + \cfrac{6}{x}\).

First, we can multiply all terms by the Lowest Common Denominator, \(10x\), to remove the denominators:

\(\left(\cfrac{-2x-10}{\cancel{5x}}\right)(2\cancel{10x}) - \left(\cfrac{3}{\cancel{5}}\right)(2\cancel{10}x) = (15)(10x) + \left(\cfrac{6}{\cancel{x}}\right)(10\cancel{x})\)

\((-2x-10)(2) - (3)(2x) = 150x + (6)(10)\)

\(-4x-20 - 6x = 150x + 60\)

\(-10x-20 = 150x + 60\)

Next, we can collect like terms by moving them to their respective sides:

\(-80 = 150x + 10x\)

\(-80 = 160x\)

Finally, we can divide both sides by the coefficient of \(x\) to get our solution:

\(\cfrac{160x}{160} = \cfrac{80}{160}\)

\(x = \cfrac{1}{2} = 0.5\)

Therefore, we can determine that our solution is \(\boldsymbol{x = 0.5}\).


Solve the rational inequality \(\cfrac{x + 1}{x^2 - 16} \lt 0\)

First, we can factor the denominator to help determine all restrictions:

\(\cfrac{x+ 1}{(x - 4)(x+ 4)}\)

From the numerator, we can determine that zeroes occur at \(x = -1\).

From the denominator, we can determine that restrictions occur at \(x = 4\) and \(x = -4\).

In order to solve this inequality, a number line can be used to help determine its intervals:

Number line outlining the different inervals of the Reciprocal Function.

As shown in the number line, the numbers with dotted lines correspond to the expressions restrictions whereas the dot corresponds to its zero.

We can now use a table to consider the signs of the factors on each interval. In order to test, we need to pick an arbitrary number in each interval:

Interval \(x\)-value Signs of Factors of Expression Sign of Expression
\((- \infty, -4)\) \(x = -5\) \(\cfrac{(-)}{(-)(-)}\) \(-\)
\((-4, -1)\) \(x = -2.5\) \(\cfrac{(-)}{(-)(+)}\) \(+\)
\(-1\) \(x = -1\) \(\cfrac{0}{(+)(-)}\) \(0\)
\((-1, 4)\) \(1.5\) \(\cfrac{+}{(+)(-)}\) \(-\)
\((4, +\infty)\) \(5\) \(\cfrac{+}{(+)(-)}\) \(+\)

Therefore, we can determine that the inequality\(\cfrac{x + 1}{x^2 - 16} \lt 0\) is true for the intervals \(\boldsymbol{x > 4}\) and \(\boldsymbol{-1 < x 4}\).

In Interval Notation, the solution set is \(\boldsymbol{x \in (-\infty, -4) \cup (-1, -4)}\).


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