Up

Equations

When faced with a rational equation, we encounter two difficulties.  The first difficulty we face is that we are dealing with equations with complication expressions in denominators.  Secondly, we face the possibility of what are known as extraneous solutions, that is, false solutions that appear.  In this article we will deal with both difficulties.

                Let’s try to solve for \(a\) in the following equation:

\(\Large \frac{{a - 3}}{{2{a^2} + 18a + 28}} + \Large \frac{1}{{2{a^2} + 18a + 28}} = \Large \frac{1}{{{a^2} + 9a + 14}}\)

We should try to multiply both sides of the equation by some expressions or numbers which would eliminate our denominators.  But what?  If we do some factoring, the answer will become clearer:

\(\Large \frac{{a - 3}}{{2{a^2} + 18a + 28}} + \Large \frac{1}{{2{a^2} + 18a + 28}} = \Large \frac{1}{{{a^2} + 9a + 14}}\)

\(\Large \frac{{a - 3}}{{2\left( {{a^2} + 9a + 14} \right)}} + \Large \frac{1}{{2\left( {{a^2} + 9a + 14} \right)}} = \Large \frac{1}{{{a^2} + 9a + 14}}\)

Hopefully you can see that all three terms in this equation contain the expression \({a^2} + 9a + 14\). If we multiply both sides of the equation by that expression, we will do some major cancelling in our denominators.

\(\Large \frac{{a - 3}}{{2\left( {{a^2} + 9a + 14} \right)}} + \Large \frac{1}{{2\left( {{a^2} + 9a + 14} \right)}} = \Large \frac{1}{{{a^2} + 9a + 14}}\)

\(\left( {{a^2} + 9a + 14} \right)\left( {\Large \frac{{a - 3}}{{2\left( {{a^2} + 9a + 14} \right)}} + \Large \frac{1}{{2\left( {{a^2} + 9a + 14} \right)}}} \right) = \left( {\Large \frac{1}{{{a^2} + 9a + 14}}} \right)\left( {{a^2} + 9a + 14} \right)\)

\(\Large \frac{{a - 3}}{2} + \Large \frac{1}{2} = 1\)

Wow!  That operation really cleaned up the equation.  Let’s clean up some more by multiplying both sides by 2, and then solving for \(a\)!. We have

\(\Large \frac{{a - 3}}{2} + \Large \frac{1}{2} = 1\)

\(\left( 2 \right)\left( {\Large \frac{{a - 3}}{2} + \Large \frac{1}{2}} \right) = \left( 1 \right)\left( 2 \right)\)

\(a - 3 + 1 = 2\)

\(a - 2 = 2\)

\(a = 4\)

Here we must be careful.  When we multiply both sides of an equation by an expression containing variables, we must check for extraneous, or false, solutions.  The reasoning behind this has to do with the fact that division by 0 is undefined.  You can do more research to learn about extraneous solutions if you wish.  But all it boils down to is that we must check the solution we got in our original equation.  We have

\(\Large \frac{{4 - 3}}{{2{{\left( 4 \right)}^2} + 18\left( 4 \right) + 28}} + \Large \frac{1}{{2{{\left( 4 \right)}^2} + 18\left( 4 \right) + 28}} = \Large \frac{1}{{{4^2} + 9\left( 4 \right) + 14}}\)

\(\Large \frac{1}{{32 + 72 + 28}} + \Large \frac{1}{{32 + 72 + 28}} = \Large \frac{1}{{16 + 36 + 14}}\)

\(\Large \frac{1}{{132}} + \Large \frac{1}{{132}} = \Large \frac{1}{{66}}\)

\(\Large \frac{2}{{132}} = \Large \frac{1}{{66}}\)

\(\Large \frac{1}{{66}} = \Large \frac{1}{{66}}\)

The solution checks.  Therefore \(a = 4\) is a solution to our original equation.  Let’s look at another example.  We want to solve for \(p\) in

\(\Large \frac{8}{{p - 5}} = 1 + \Large \frac{1}{{p - 5}}\)

First multiply both sides by \(p - 5\)

\(\left( {p - 5} \right)\left( {\Large \frac{8}{{p - 5}}} \right) = \left( {p - 5} \right)\left( {1 + \Large \frac{1}{{p - 5}}} \right)\)

\(8 = p - 5 + 1\)

\(8 = p - 4\)

\(p = 12\)

Again, check for extraneous solutions

\(\Large \frac{8}{{12 - 5}} = 1 + \Large \frac{1}{{12 - 5}}\)

\(\Large \frac{8}{7} = 1 + \Large \frac{1}{7}\)

\(\Large \frac{8}{7} = \Large \frac{8}{7}\)

The solution checks.  So \(p = 12\) is a solution to the original equation.

Below you can download some free math worksheets and practice.


Downloads:
1254 x

Solve each equation. Remember to check for extraneous solutions.

This free worksheet contains 10 assignments each with 24 questions with answers.

Example of one question:

Rational-Expressions-Equations-easy

Watch below how to solve this example:

 

Downloads:
1102 x

Solve each equation. Remember to check for extraneous solutions.

This free worksheet contains 10 assignments each with 24 questions with answers.

Example of one question:

Rational-Expressions-Equations-medium


Watch below how to solve this example:

 

Downloads:
1093 x

Solve each equation. Remember to check for extraneous solutions.

This free worksheet contains 10 assignments each with 24 questions with answers.

Example of one question:

Rational-Expressions-Equations-hard

Watch below how to solve this example:

 
 
 

Facebook PageGoogle PlusTwitterYouTube Channel

Algebra and Pre-Algebra

Beginning Algebra
Adding and subtracting integer numbers
Dividing integer numbers
Multiplying integer numbers
Sets of numbers
Order of operations
The Distributive Property
Verbal expressions
Beginning Trigonometry
Finding angles
Finding missing sides of triangles
Finding sine, cosine, tangent
Equations
Absolute value equations
Distance, rate, time word problems
Mixture word problems
Work word problems
One step equations
Multi step equations
Exponents
Graphing exponential functions
Operations and scientific notation
Properties of exponents
Writing scientific notation
Factoring
By grouping
Common factor only
Special cases
Linear Equations and Inequalities
Plotting points
Slope
Graphing absolute value equations
Percents
Percent of change
Markup, discount, and tax
Polynomials
Adding and subtracting
Dividing
Multiplying
Naming
Quadratic Functions
Completing the square by finding the constant
Graphing
Solving equations by completing the square
Solving equations by factoring
Solving equations by taking square roots
Solving equations with The Quadratic Formula
Understanding the discriminant
Inequalities
Absolute value inequalities
Graphing Single Variable Inequalities
Radical Expressions
Adding and subtracting
Dividing
Equations
Multiplying
Simplifying single radicals
The Distance Formula
The Midpoint Formula
Rational Expressions
Adding and subtracting
Equations
Multiplying and dividing
Simplifying and excluded values
Systems of Equations and Inequalities
Graphing systems of inequalities
Solving by elimination
Solving by graphing
Solving by substitution
Word problems