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Algebra 1 Solving Rational Equations

Name: Algebra 1: Solving Rational Equations
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Section 10: Rational Equations: Lecture 4 | 16:24 min

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Solving Rational Equations

To solve a rational expression first find the LCD of all the terms, and multiply by it. This will transform the equation so that it no longer contains fractions.
Always check the solutions you find when working rational equations. A possible solution may not work in the original because it makes us divide by zero.
Some formulas contain rational expressions. When solving these for a specific variable, work to isolate it on one side.

Solving Rational Equations

Solve:
[x/(x + 1)] = [(x − 4)/(x − 5)]

x( x − 5 ) = ( x + 1 )( x − 4 )
x² − 5x = x² − 3x − 4
− 2x = − 4

x = 2

Solve:
[y/(y + 8)] = [(y + 6)/(y − 7)]

y( y − 7 ) = ( y + 8 )( y + 6 )
y² − 7y = y² + 14y + 48
− 21y = 48

y = − 2[2/7]

Solve:
[4/(x − 2)] = [(x + 6)/(x − 1)]

4( x − 1 ) = ( x − 2 )( x + 6 )
4x − 4 = x² + 4x − 12
8 = x²

x = ±√8

Solve:
[y/(2y − 1)] + [3y/(y − 5)] = 8

( 2y − 1 )( y − 5 )g[y/(2y − 1)] + [3y/(y − 5)]( 2y − 1 )( y − 5 ) = 8( 2y − 1 )( y − 5 )
y( y − 5 ) + 3y( 2y − 1 ) = 8( 2y² − 11 + 5 )
y² − 5y + 6y² − 3y = 16y² − 88y + 40
0 = 9y² − 80y + 40
x = [( − b ±√{b² − 4ac} )/2a]
y = [( − ( − 80 ) ±√{( − 80 )² − 4( 9 )( 40 )} )/2( 9 )]
y = [(80 ±√{4960} )/18]
y = [(80 ±4√{310} )/18]
y = [(40 ±2√{310} )/9]
Extraneous Solutions: 2y − 1 = 0
2y = 1
y = [1/2]
y − 5 = 0

y = 5

[n/(n + 3)] + [n/(n − 4)] = 7

( n + 3 )( n − 4 )[n/(n + 3)] + ( n + 3 )( n − 4 )[n/(n − 4)] = 7( n + 3 )( n − 4 )
n(n − 4) + n(n + 3) = 7(n² − n − 12)
n² − 4n + n² + 3n = 7n² − 7n − 84
5n² − 6n − 84 = 0
n = [( − ( − 6) ±√{( − 6)² − 4(5)( − 84)} )/2(5)]
n = [(6 ±√{1716} )/10]
n = [(6 ±2√{429} )/10]
n = [(3 ±√{429} )/5]
Extraneous Solutions: n + 3 = 0
n = − 3
n − 4 = 0

n = 4

[3r/(r − 2)] + [10r/(2r + 1)] = 1

3r(2r + 1) + 10r(r − 2) = (r − 2)(2r + 1)
6r² + 3r + 10r² − 20r = 2r² − 3r − 2
14r² − 14r + 2 = 0
r = [( − ( − 14) ±√{( − 14)² − 4(14)(2)} )/2(14)]
r = [(14 ±√{84} )/28]
r = [(14 ±2√{21} )/28]
r = [(7 ±√{21} )/14]
Extraneous Solutions: r − 2 = 0
r = 2
2r + 1 = 0

r = − [1/2]

Eric finishes a puzzle in 20 minutes. Duy finished the same puzzle in 15 minutes. If they work together, how long does it take them to finish the same puzzle?

[1/20] + [1/15] = [1/x]
[1/20](20)(15)(x) + [1/15](20)(15)(x) = [1/x](20)(15)(x)
15x + 20x = 300
35x = 300

x = 8[4/7] minutes

Irene makes a costume in three hours. Michelle makes the same costume in five hours. If they work together how long does it take them to make the costume?

[1/3] + [1/5] = [1/x]
[1/3](5)(3)(x) + [1/5](5)(3)(x) = [1/x](5)(3)(x)
5x + 3x = 15
8x = 15

x = [15/8] = 1[7/8] hours

[3a/(a + 2)] + [(a − 6)/(a² − 4)] = 1

LCM = (a − 2)(a + 2)
[3a/(a + 2)](a − 2)(a + 2) + [(a − b)/(a² − 4)](a − 2)(a + 2) = (a − 2)(a + 2)
(3a)(a − 2) + (a − 6) = a² − 4
3a² − 6a + a − 6 = a² − 4
2a² − 5a − 2 = 0
a = [( − ( − 5 ) ±√{( − 5 )² − 4( 2 )( 2 )} )/2( 2 )]

a = [(5 ±√{41} )/4]

[p/(3p + 2)] − [5p/(p − 1)] = 5

LCM = (3p + 2)(p − 1)
[p/(3p + 2)](3p + 2)(p − 1) − [5p/(p − 1)](3p + 2)(p − 1) = 5(3p + 2)(p − 1)
p(p − 1) − 5p(3p + 2) = 5(3p² − p − 2)
p² − p − 15p² − 10p = 15p² − 5p − 10
29p² + 6p − 10 = 0
p = [( − 6 ±√{6² − 4( 29 )( − 10 )} )/2( 29 )]
p = [( − 6 ±√{1196} )/58]
p = [( − 6 ±2√{299} )/58]

p = [( − 3 ±√{299} )/29]

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Answer

Solving Rational Equations

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

Intro

Objectives

Solving Rational Equations

Isolate the Specified Variable

Example 1

Example 2

Example 3

Example 4

Intro 0:00
Objectives 0:07
Solving Rational Equations 0:23

Isolate the Specified Variable

Example 1 1:58
Example 2 5:00
Example 3 8:23
Example 4 13:25

Algebra 1 Online Course

Section 1: Properties of Real Numbers
	Basic Types of Numbers	30:41
	Operations on Numbers	19:26
	Order of Operations	12:06
	Properties of Real Numbers	18:52
Section 2: Linear Equations
	The Vocabulary of Linear Equations	12:22
	Solving Linear Equations in One Variable	28:52
	Solving Formulas	12:02
	Applications of Linear Equations	28:41
	Applications of Linear Equations, Motion & Mixtures	24:26
Section 3: Graphing
	Rectangular Coordinate System	22:55
	Slope & Graphing	27:58
	Linear Equations in Two Variables	20:36
Section 4: Functions
	Introduction to Functions	21:24
	Graphing Functions	16:12
Section 5: Systems of Linear Equations
	Systems of Linear Equations	25:54
	Solving a System Using Substitution	20:01
	Solving a System Using Elimination	19:40
	Applications of Systems of Equations	24:34
Section 6: Inequalities
	Solving Linear Inequalities in One Variable	17:13
	Compound Inequalities	16:13
	Solving Equations with Absolute Values	14:12
	Inequalities with Absolute Values	17:07
	Graphing Inequalities in Two Variables	15:33
	Systems of Inequalities	21:13
Section 7: Polynomials
	Integer Exponents	44:51
	Adding & Subtracting Polynomials	18:33
	Multiplying Polynomials	25:07
	Dividing Polynomials	44:56
Section 8: Factoring Polynomials
	Greatest Common Factor & Factor by Grouping	28:27
	Factoring Trinomials	21:44
	Factoring Trinomials Using the AC Method	30:09
	Special Factoring Techniques	30:14
Section 9: Quadratic Equations
	Solving Quadratic Equations by Factoring	23:38
	Solving Quadratic Equations	29:27
	Equations in Quadratic Form	16:47
	Quadratic Formulas & Applications	29:04
	Graphs of Quadratics	26:53
	Polynomial Inequalities	21:42
Section 10: Rational Equations
	Multiply & Divide Rational Expressions	26:41
	Adding & Subtracting Rational Expressions	20:24
	Complex Fractions	18:23
	Solving Rational Equations	16:24
	Rational Inequalities	18:54
	Applications of Rational Expressions	20:20
	Variation & Proportion	27:04
Section 11: Radical Equations
	Rational Exponents	14:32
	Simplify Rational Exponents	15:12
	Adding & Subtracting Radicals	17:22
	Multiply & Divide Radicals	19:24
	Solving Radical Equations	15:05
	Complex Numbers	29:16

Transcription: Solving Rational Equations

Welcome back to www.educator.com.0000

In this lesson we are going to work on solving rational equations.0003

Two things that we will look at is I can solve a equation that contains a rational expression.0009

I will throw in an example or we deal with a formula that also involves a rational expression.0016

Recall that when you are working with equations that have a lot of fractions in them we will often use the least common denominator0025

to go ahead and clear out all of those fractions.0032

These particular types of equations you definitely want to make sure that you check your solutions.0036

This will be especially important for these types of equations this is because we are dealing with expressions that involve fractions,0041

we will have some restricted values, values that we cannot have.0051

Some of these values may try and show up as our solution.0056

We simply have to get rid of them.0059

They are not valid and they will make our denominator 0.0061

When working with these expressions that have a lot of fractions we like to keep the least common denominator throughout the entire process0068

or that way we can go back in and make sure that makes sense in the original.0076

Let us go ahead.0081

When you see my example for formulas here again and know that the same process plays out.0086

We like to use that least common denominator to go ahead and try and clear things out0092

Of course, our goal is to isolate which ever variable we are trying to solve for.0096

When we are all done with that formula we still have an equation of these words0103

and for more variables still floating around in there.0107

That is okay as long as we get that variable we are looking for completely isolated.0109

That is how we will know it is solved.0114

Let us take a look at one of these.0119

I have 1-2 / x + 1 = 2x / x +1 and we can see that this is a rational equation because looking at all these rational expressions I have,0121

think of those fractions.0131

To help out I’m going to work on getting a common denominator.0134

Already these 2 parts of it have an x + 1 in the bottom.0139

We are just going to focus on trying to give an x + 1 to the one over there on the left.0144

I can do that by taking 1 / 1 and then multiplying the top and bottom of that by x + 1.0151

(x + 1)(x +1) - 2 / (x + 1) = 2x /(x +1)0162

On the left side that allows us to finally combine things now that we have a common denominator.0175

In doing so, we just put the top together.0182

x + 1 - 2 / x +10184

Doing a little bit of simplifying let us go ahead and subtract 2 from 1, x - 1 now x + 1 = 2x / x +1.0196

At this stage notice how we have two fractions that are set equal to one another and the bottoms are exactly the same.0211

Since the bottoms are the same then I know that the tops of these must also be the same.0219

Let us just focus on the top part for a little bit and see if we can find a solution out of that.0225

x - 1 = 2x this one is not so bad to solve.0230

Let us just go ahead and subtract an x from both sides and I think we will be able to get a possible solution.0236

It looks like a possible solution is that -1 = x.0245

Be careful this can sometimes happen with these rational equations.0249

It looks like we have done all of our steps correctly and it looks like we have found a solution, but this guy does not work.0253

To see why it does not work, take this possible solution all the way back to one of our original denominators0260

and you can see that if you try and plug it in and it makes the bottom 0.0267

We have to get rid of that possible solution.0272

Since we do not have any other possibilities left, we can say that this particular equation has no solution.0275

With these types of equations, even when you find something that looks like it is a solution you have to go back0284

and check to make sure that it make sense in the original.0291

Oftentimes you may end up getting rid of things that will simply make that denominator 0.0294

Let us try a different one.0301

This one is 2 /(p² - 2p) = 3 /(p² – p)0302

Before working with this, I need to see what is in the denominator.0311

To get a better idea we are going to go ahead and factor.0316

In this side I can see that they both have a p in common, we will go ahead and take that out.0321

I get p × p -2.0326

On the right side here, it looks like it has a p in common as well, p × p -1.0330

They almost have the same denominator, but I need to essentially give each one its missing factor.0339

Let us give the left side p -1 and we will give the right side p -2.0346

Just copy this down and see where the extra pieces are coming from.0353

On the left I will put in p -1 on the bottom and on the top.0365

The one on the right we will give p - 2 in the bottom and p -2 on top.0374

Distributing this will give us (2p -2) / and now we have this nice common denominator equals (3p – 6) / (p) (p -1) (p -2)0385

This is right back to that situation we had before.0406

The bottoms of each of these fractions are exactly the same I know that the tops of the fractions will also end up being the same.0409

Let us extract out just the tops and focus on those.0418

If we have to solve this what can we do?0425

I will go ahead and add 2 to both sides 2p = 3p - 4 and let us go ahead and subtract 3p from both sides, -p = -4.0428

One last step, let us go ahead and multiply both sides by -1.0448

My possible solution is that p = 4.0455

We cannot necessarily just assume that that is the solution, not until we go back to the original0459

and make sure that it is not going to give us 0 in the bottom.0464

Let us check to see what factors are in the bottom.0468

Some of our restricted values that we simply cannot have is 1p = 0, we cannot have that one.0473

We also cannot have p = 2 and we cannot have p = 1.0480

Those three values are restricted.0487

Fortunately when we look down at what we got, it is not any of these restricted values.0490

We will go ahead and keep it as our solution.0497

The solution to this one is p = 4.0499

Let us look at a little bit large one, one that has a few more things in the denominator.0505

This one is 1 / x -2 + 1/5 =2/5 × (x² – 4)0511

Quite a bit going on the bottom and it looks like we definitely need to factor one of our pieces0518

just so we can see what total factors we have in here.0523

1/x – 2 + 1/5 = 2/ I have 5 and then the x² – 4 that happens to be the difference of squares.0528

(x + 2)(x – 2)0545

It looks like this will be our LCD and we are going to give it to these other fractions here with missing pieces.0550

The 1/x – 2 and see what we can put into that one.0560

It needs to have a 5, we will put that down below and up top and it also needs to have an x + 2 and below and up top.0569

Onto the next fraction here, this is 1/5 and it looks like it is missing an x - 2 piece so we will throw that in there.0585

We are missing x + 2 missing both of those pieces.0597

On the other side of our equal sign it already has the LCD, so we just leave it just as it is.0607

We spread out quite a bit but now we are going to focus on the tops of everything since all the bottoms are now exactly the same.0616

We will also use our distribution property a little bit to help us out by multiplying 5 × 20626

and we will take the two binomials over here and we will foil them so we can put those together.0633

5x + 10 would be that first piece on the top of the first fraction.0641

We have our first terms x², outside terms 2x, inside terms – 2x and last terms -4.0649

We want that all equal to 2.0658

You can notice this is just the tops of all of our fractions.0661

Let us go ahead and combine some things.0666

My + 2x and my - 2x will go ahead and cancel each other out and we will have x², I will take care of that one.0669

5x took care of that one and combining the 10 and -4, +6 = 2.0680

We are almost there. Let us just go ahead and subtract 2 from both sides and see that we need to solve a quadratic.0689

We are going to solve x² + 5x + 4 is all equal to 0.0698

This quadratic is not too bad I can end up factoring it using reverse foil without too much problem.0703

Our first terms x and x and two terms that need to multiply to give us 4 but add to give us 5.0710

I am thinking of 1 and 4, those will do it, both positive.0719

This will give us two possible solutions.0725

When I take x + 1 = 0 and x + 4 = 0, x could equal -1 or x could equal -4.0727

There are two possible things that I'm worried about.0739

Before we put our seal of approval on them,0744

let us go back to some of those original fractions and see what some of our restricted values are.0746

Looking at this first one, the bottom would be 0 when x is equal to 2 so I know one of my restricted values x can not equal 2.0752

The bottom of this one is a number 5, so that one is not is not going to be 0.0762

Over here I can see that when x =-2 I will run into a 0 on the bottom, x cannot equal -2.0768

This factor here is exactly the same as the first one.0776

We already have that list as one of our restricted values.0780

As long as I know my possible solutions are 2 and -2, it looks like we are okay.0783

Since I'm looking at -1 and -4, both of these are going to be valid solutions, none of them are restricted.0789

The process of solving these rational equations you just have to work on combining them and focus on the tops for a bit.0796

Let us look at that formula as I promised.0807

To see how you could solve the formula involving some these rational expressions.0809

To make things interesting, let us go ahead and solve for y.0814

In order to get rid of a lot of these things in the bottom we will identify our lowest common denominator.0822

One has an x another one has y and the other one has z.0830

It looks like we will need all three of those parts, x, y and z in our new denominator.0833

We will go through and give each of them their missing pieces.0840

Here we have our originals, let us throw in the missing.0846

The first one it already has an x in the bottom, we will give it y and z on the top and the bottom.0852

The next one already has a y, so give it the x and z.0860

For the last one we have z so xy.0868

This will ensure that all of the bottoms are exactly the same.0878

We will go ahead and turn your attention to just the tops of all of these and we will continue moving forward.0882

2yz = 1 × x × z that will be just xz + xy.0889

We are trying to get these y isolated all by himself and looks like we have two of them, one on the left side and one on the right side.0899

Let us get them on the same side by subtracting an xy from both sides.0907

2yz – xy0914

We can factor out that common y that way we will have 1y to deal with.0922

Let us go ahead and factor out the common y out front.0928

y and then left over we will have 2z – x.0931

We are almost free with this one.0941

To get the y completely isolated, let us finally divide both sides by that 2z – x.0946

y = xz / 2z - x and this one is solved because we have completely isolated the variable we are looking for.0952

In a sense we have created a new type of formula.0970

Be careful in solving many of these different types of rational equations.0973

Make sure you always check your solutions and use that least common denominator to help yourself out.0976

Thank you for watching www.educator.com.0982

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Solving Rational Equations

Slide Duration:

Table of Contents

Section 1: Properties of Real Numbers

Basic Types of Numbers

30m 41s

Intro