A rational equation is a type of equation where it involves at least one rational expression, a fancy name for a fraction. The best approach to address this type of equation is to eliminate all the denominators using the idea of LCD (least common denominator). By doing so, the leftover equation to deal with is usually either linear or quadratic.
In this lesson, I want to go over ten (10) worked examples with various levels of difficulty. I believe that most of us learn math by looking at many examples. Here we go!
Example 1: Solve the rational equation below and make sure you check your answers for extraneous values.
Would it be nice if the denominators are not there? Well, we can’t simply vanish them without any valid algebraic step. The approach is to find the Least Common Denominator (also known Least Common Multiple) and use that to multiply both sides of the rational equation. It results in the removal of the denominators, leaving us with regular equations that we already know how to solve such as linear and quadratic. That is the essence of solving rational equations.
Always check your “solved answers” back into the original equation to exclude extraneous solutions. This is a critical aspect of the overall approach when dealing with problems like Rational Equations and Radical Equations.
Example 2: Solve the rational equation below and make sure you check your answers for extraneous values.
The first step in solving a rational equation is always to find the “silver bullet” known as LCD. So for this problem, finding the LCD is simple.
Try to express each denominator as unique powers of prime numbers, variables and/or terms.
Multiply together the ones with the highest exponents for each unique prime number, variable and/or terms to get the required LCD.
Example 3: Solve the rational equation below and make sure you check your answers for extraneous values.
It looks like the LCD is already given. We have a unique and common term [latex]\left( \right)[/latex] for both of the denominators. The number [latex]9[/latex] has the trivial denominator of [latex]1[/latex] so I will disregard it. Therefore the LCD must be [latex]\left( \right)[/latex].
= [-3/(x-3)](x-3)" width="291" height="65" />
Distribute the constant [latex]9[/latex] into [latex]\left( \right)[/latex].
It should work so yes, [latex]x = 2[/latex] is the final answer.
Example 4: Solve the rational equation below and make sure you check your answers for extraneous values.
I hope that you can tell now what’s the LCD for this problem by inspection. If not, you’ll be fine. Just keep going over a few examples and it will make more sense as you go along.
Try to express each denominator as unique powers of prime numbers, variables and/or terms.
Multiply together the ones with the highest exponents for each unique prime number, variable and/or terms to get the required LCD.
= " width="348" height="63" />
Quick note: If ever you’re faced with leftovers in the denominator after multiplication, that means you have an incorrect LCD.
Now, distribute the constants into the parenthesis on both sides.
Example 5: Solve the rational equation below and make sure you check your answers for extraneous values.
Focusing on the denominators, the LCD should be [latex]6x[/latex]. Why?
Remember, multiply together “each copy” of the prime numbers or variables with the highest powers.
= (1/2)(6x)" width="242" height="61" />
Distribute the constant into the parenthesis.
Example 6: Solve the rational equation below and make sure you check your answers for extraneous values.
Whenever you see a trinomial in the denominator, always factor it out to identify the unique terms. By simple factorization, I found that [latex] + 4x – 5 = \left( \right)\left( \right)[/latex]. Not too bad?
Finding the LCD just like in previous problems.
Try to express each denominator as unique powers of prime numbers, variables and/or terms. In this case, we have terms in the form of binomials.
Multiply together the ones with the highest exponents for each unique copy of a prime number, variable and/or terms to get the required LCD.
This aids in the cancellations of the commons terms later.
= + >(x+5)(x-1)>" width="488" height="51" />
You should end up with a very simple equation to solve.
Example 7: Solve the rational equation below and make sure you check your answers for extraneous values.
Since the denominators are two unique binomials, it makes sense that the LCD is just their product.
It makes a lot of sense to perform the FOIL method. Does that ring a bell?
Example 8: Solve the rational equation below and make sure you check your answers for extraneous values.
This one looks a bit intimidating. But if we stick to the basics, like finding the LCD correctly, and multiplying it across the equation carefully, we should realize that we can control this “beast” quite easily.
Expressing each denominator as unique powers of terms
Multiply each unique terms with the highest power to obtain the LCD
+
Be careful now with your cancellations.
= + >(6)(x+1)(x-1)" width="509" height="50" />
This is getting simpler in each step!
I would combine like terms on both sides also to simplify further.
Example 9: Solve the rational equation below and make sure you check your answers for extraneous values.
Let’s find the LCD for this problem, and use it to get rid of all the denominators.
Express each denominator as unique powers of terms.
Multiply each unique term with the highest power to determine the LCD.
+ = 1/(x-5)" width="256" height="51" />
+ > = (x+5)(x-5)" width="510" height="74" />
Multiply the constants into the parenthesis.
Example 10: Solve the rational equation below and make sure you check your answers for extraneous values.
Start by determining the LCD. Express each denominator as powers of unique terms. Then multiply together the expressions with the highest exponents for each unique term to get the required LCD.
So then we have,
= < - >(x^2)(x-2)" width="416" height="76" />
Always start with the simplest method before trying anything else. I will utilize the factoring method of the form [latex]x^2+bx+c=0[/latex] since the trinomial is easily factorable by inspection.
Set each factor equal to zero, then solve each simple one-step equation.
Again, always check the solved answers back into the original equations to make sure they are valid.
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