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 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:
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.
Rational equations and inequalities can be solved in a handful of different ways. These include:
For solving problems, it is likely that you use several of these techniques in conjunction with each other.
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 \(\boldsymbol{x = 7}\).
Solve the rational inequality \(\cfrac{2x - 13}{x + 6} \geq 7\).
First, we can move all terms onto one side:
Next, we can combine terms under a common denominator:
Finally, we can collect like terms in the numerator:
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:
Using these key features, we are able to sketch a graph of the function. This will help determine the intervals to satisfy the inequality:
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 \(\boldsymbol{x \in [-11, -6)}\).
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 \(\boldsymbol{x = 0.5}\).
First, we can factor the denominator to help determine all restrictions:
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:
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 \(\boldsymbol{x \in (-\infty, -4) \cup (-1, -4)}\).