Intermediate Algebra Tutorial 7: Linear Equations in One Variable 
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Learning Objectives
After completing this tutorial, you should be able to:

Introduction
This is where we start getting into the heart of what algebra is about  solving equations. In this tutorial we will be looking specifically at linear equations and their solutions. We will start off slow and solve equations that use only one property to make sure you have the individual concepts down. Then we will pick up the pace and mix 'em up where you need to use several properties and steps to get the job done. Equations can be used to help us solve a variety of problems. In later tutorials, we will put them to use to solve word problems. Then you can answer those tricky math questions. 
Tutorial
Two expressions set equal to each other 
An equation that can be written in the form 
The following is an example of a linear equation: 3x  4 = 5 
A value, such that, when you replace the variable with it, (the left side comes out equal to the right side) 
Set of all solutions 
Example 1: Determine if any of the following values for x are solutions to the given equation. 3x  4 = 5; x = 3, 5. 
Checking 3  Checking 5 
in General Get the variable you are solving for alone on one side and everything else on the other side using INVERSE operations. 
The following will give us the tools that we need to solve linear equations. 
If a = b, then a + c = b + c If a = b, then a  c = b  c 
In other words, if two expressions are equal to each other and you add or subtract the exact same thing to both sides, the two sides will remain equal. Note that addition and subtraction are inverse operations of each other. For example, if you have a number that is being added that you need to move to the other side of the equation, then you would subtract it from both sides of that equation. Example 2: Solve for the variable. x  5 = 2. 
x  5 = 2  *Inverse of sub. 5 is add. 5 
Note that if you put 7 back in for x in the original problem you will see that 7 is the solution to our problem. 
Example 3: Solve for the variable. y + 4 = 7. 
y + 4 = 7  *Inverse of add. 4 is sub. 4 
Note that if you put 11 back in for y in the original problem you will see that 11 is the solution we are looking for. 
If a = b, then a(c) = b(c) If a = b, then a/c = b/c where c is not equal to 0. 
In other words, if two expressions are equal to each other and you multiply or divide (except for 0) the exact same constant to both sides, the two sides will remain equal. Note that multiplication and division are inverse operations of each other. For example, if you have a number that is being multiplied that you need to move to the other side of the equation, then you would divide it from both sides of that equation. Note, for multiplication and division, it is not guaranteed that if you multiply by the variable you are solving for that the two sides are going to be equal. But is guaranteed that the two sides are going to be equal if you are multiplying or dividing by a constant or another variable that you are not solving for. We will talk more about this in a later tutorial. For this tutorial just note you can use this property with constants and variables you are not solving for. Example 4: Solve for the variable. x/2 = 5. 
*Inverse of div. by 2 is mult. by 2 
If you put 10 back in for x in the original problem, you will see that 10 is the solution we are looking for. 
Example 5: Solve for the variable. 5x = 7. 
*Inverse of mult. by 5 is div. by 5 
If you put 7/5 back in for x in the original problem, you will see that 7/5 is the solution we are looking for. 
The examples above were using only one property at a time to help you understand the different properties that we use to solve equations. However, most times, we have to use several properties to get the job done. The following is a strategy that you can use to help you solve linear equations that are a little bit more involved. 
Note that your teacher or the book you are using may have worded these steps a little differently than I do, but it all boils down to the same concept  get your variable on one side and everything else on the other using inverse operations. Step 1: Simplify each side, if needed.
Step 2: Use Add./Sub. Properties to move the variable term to one side and all other terms to the other side. Step 3: Use Mult./Div. Properties to remove any values that are in front of the variable. Step 4: Check your answer. I find this is the quickest and easiest way to approach linear equations. Example 6: Solve for the variable. 10  3x = 7. 
*Inverse of add. 10 is sub. 10 *Inverse of mult. by 3 is div. by 3 
Be careful going from line 4 to line 5. Yes, there is a negative sign. But, the operation between the 3 and x is multiplication not subtraction. So if you were to add 3 to both sides you would have ended up with 3x + 3 instead of the desired x. If you put 1 back in for x in the original problem you will see that 1 is the solution we are looking for. 
Example 7: Solve for the variable. 2(x + 5)  7 = 3(x  2). 
*Remove ( ) by using dist. prop. *Get all x terms on one side *Inverse of add. 3 is sub. 3 *Inverse of mult. by 1 is div. by 1 
If you put 9 back in for x in the original problem you will see that 9 is the solution we are looking for. 
Example 8: Solve for the variable: . 
*To get rid of the fractions, *Get all the x terms on one side *Inverse of add. 2 is sub. 2 *Inverse of mult. by 3 is div. by 3

If you put 4/3 back in for x in the original problem you will see that 4/3 is the solution we are looking for. 
A contradiction is an equation with one variable that has no solution. 
Example 9: Solve for the variable. 4x 1 = 4(x+ 3). 
*Get all the x terms on one side 
Where did our variable, x, go??? It disappeared on us. Also note how we ended up with a FALSE statement, 1 is not equal to 12. This does not mean that x = 12 or x = 1. Whenever your variable drops out AND you end up with a false statement, then after all of your hard work, there is NO SOLUTION. So, the answer is no solution. 
An identity is an equation with one variable 
Example 10: Solve for the variable. 5x+ 10 = 5(x+ 2). 
*Get all the x terms on one side 
This time when our variable dropped out, we ended up with a TRUE statement. Whenever that happens your answer is ALL REAL NUMBERS. So, the answer is all real numbers. 
Practice Problems
These are practice problems to help bring you to the next level. It will allow you to check and see if you have an understanding of these types of problems. Math works just like anything else, if you want to get good at it, then you need to practice it. Even the best athletes and musicians had help along the way and lots of practice, practice, practice, to get good at their sport or instrument. In fact there is no such thing as too much practice. To get the most out of these, you should work the problem out on your own and then check your answer by clicking on the link for the answer/discussion for that problem. At the link you will find the answer as well as any steps that went into finding that answer. 
Practice Problems 1a  1e: Solve for the variable.
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Old school equations with Sal
Video transcript
Linear Equations
6.2 Solving Linear Equations
Equations of the form ax+b=0 are called linear equations in the variable x. In this section we will be concerned with the problem of solving linear equations, and equations that reduce to linear equations.
We deﬁne two equations as equivalent if they have the same solution set. The following two operations on an equation always result in a new equation which is equivalent to the original one. These operations, sometimes called elementary transformations, are:
T.1 The same expression representing a real number may be added to both sides of an equation.
T.2 The same expression representing a nonzero real number may be multiplied into both sides of an equation.
Using these operations we may transform an equation whose solution set is not obvious through a series of equivalent equations to an equation that has an obvious solution set.
Example 1. Solve the equation
(a) 2x3=4+x
Add x to both sides to obtain
x+2x3=x+4+x (T.1)
or x3=4
Add 3 to both sides to obtain
x3+3=4+3 (T.1)
or x=7
Since 2x3=4+x is equivalent to x3=4, which, in turn, is equivalent to x=7, whose solution set is obviously {7}, we know that the solution set of (a) is {7}.
Let’s see how our Linear equation solver solves this and similar problems. Click on "Solve Similar" button to see more examples.
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Example 2. Solve the equation
(b) 1/2x+2/3=5/2x1
Add (1/2)x to both sides to obtain
2/3=5/2x1/2x1 (T.1)
or 2/3=2x1
Add 1 to both sides to obtain
1+2/3=2x (T.1)
or 5/3=2x
Multiply both sides by 1/2 to obtain
5/6=x (T.2)
Thus, the solution set of (b) is {5/6}.
Every linear equation can be solved in the same way as in the above examples. In fact, let us consider the general linear equation
ax+b=0
Add b to both sides to obtain
ax=b
Multiply both sides by 1/a to obtain
x=(b/a)
if a a!=0. The general linear equation, therefore, has as its solution set {b/a}, if a!=0. Thus each linear equation has at most one solution.
The next two examples are of equations that reduce to linear equations.
Example 3. Solve the equation
23+4y(5y+4)=9+10y(2y+3)
We expand both sides to obtain
23+20y^2+16y=9+20y^2+30y
Add 20y^2 to both sides to obtain
23+16y=9+30y
We now solve as in the previous examples.
23+16y=9+30y
239=30y16y
14=14y
y=1
Thus the solution set is {1}.
Let’s see how our step by step math solver solves this and similar problems. Click on "Solve Similar" button to see more examples.
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Example 4. Solve the equation
(c) (2x)/(x1)=2/(x1)+1
The replacement set of (c) is all real numbers except 1. Assuming that x!=1, we multiply both sides of (c) by x1 to obtain
(d) 2x=2+x1,x!=1
Solving the equation 2x =2+x 1, we obtain 1 as the only solution Since 1 is not in the replacement of (d), (d) has no solution. Furthermore, (c) is equivalent to (d), therefore (c) has no solution.
6.3 Solving Literal Equations
An equation containing more than one variable, or containing symbols representing constants such as a,b, and c, can be solved for one of the symbols in terms of the remaining symbols by applications of the operations T.1 and T.2 in the preceding section. The student will encounter such problems in other courses.
Example 1. Solve cx3a=b for x.
Add 3a to both sides.
cx=b+3a
Multiply both sides by 1/c.
x=(b+3a)/c
This last equation expresses x in terms of the other symbols.
Example 2. Solve 3ay2b=2cy for y.
Add 2b to both sides.
3ay=2cy+2b
Add 2cy to both sides.
3ay2cy=2b
Factor out y.
(3a2c)y=2b
Multiply both sides by 1/((3a2c))
y=(2b)/(3a2c)
Example 3. Solve a/x+b/(2x)=c for x.
Multiply both sides by 2x.
2a+b=2cx
2cx=2a+b
Multiply by 1/(2c).
x=(2a+b)/(2c)
We conclude this section by including two more examples similar to those that the student may encounter in other areas.
Let’s see how our math solver solves this and similar problems. Click on "Solve Similar" button to see more examples.
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Example 4. Solve A=P(1+rt) for r.
Apply the distributive law.
A=P+Prt
Add P to both sides.
AP=Prt
Multiply both sides by 1/(Pt).
(AP)/(Pt)=r
Example 5. Solve 1/R=1/r_1+1/r_2 for r_1.
Add the two terms on the right—hand side.
1/(R)=(r_2+r_1)/(r_1r_2)
Multiply by Rr_1r_2.
r_1r_2=R(r_2+r_1)
r_1r_2=Rr_2+Rr_1
Add Rr_1 to both sides.
r_1r_2Rr_1=Rr_2
Factor out r_1.
r_1(r_2R)=Rr_2
Multiply by 1/(r_2R).
r_1=(Rr_2)/(r_2R)
6.4 Solving Statement Problems
One of the fundamental applications of algebra is solving problems that have been stated in words. A statement problem is a word description of a situation that involves both known and unknown quantities. In this section each problem will be solvable by means of one equation involving one unknown.
Our problem is to choose the unknown and to determine the equation that it must satisfy. Although there is no single approach to all of the problems, the following suggestions are sometimes helpful:
1. Read the problem carefully until the situation is thoroughly understood.
2. Determine what quantities are asked for, then choose the one that seems to be the best to use as the unknown.
3. Establish the relationship between the unknown and the other quantities in the problem.
4. Find the information that tells which two quantities are equal.
5. Use the information in (4) to write the equation.
6. Solve the equation and check the solution to see that it satisﬁes the original problem.
At this stage, the emphasis will be on translating statement problems into equations. Although some of the problems can be solved almost by inspection, the practice that we obtain in setting up equations will prove helpful in working more difficult problems.
Example 1. If 2 times a certain integer is added to the next consecutive integer the result is 34. Find the integers.
Step 1. Reread!
Step 2. Let x be the ﬁrst integer.
Step 3. Then x+ 1 is the next consecutive integer.
Step 4. 2 times a certain integer plus the next consecutive integer is 34.
Step 5. 2x+(x+1)=34
Step 6. Solve.
2x+(x+1)=34
3x+1=34
3x=33
x=11
Check. 2*11+(11+1)=34
Example 2. Bob and Joe together earned $60. Both were paid at the same rate, but Bob worked three times as long as Joe. How much did each receive?
Step 1. Reread!
Step 2. Let x be the number of dollars that Joe received.
Step 3. Then 3x is the number of dollars that Bob received
Step 4. Bob and Joe together earned $60.
Step 5.3x+x=60
Step 6. Solve.
3x+x=60
4x=60
x=15
3x=45
Check 3*15+15=60
Example 3. The sum of the digits of a two digit number is 12. If the digits are reversed the number is decreased by 36. What is the number?
Step 1. Reread!
Step 2. Let x be the tens digit.
Step 3. Then 12  x is the units digit.
Step 4. If the digits are reversed then the number is decreased by 36
Step 5.10(12x)+x = 10x+ (12x) 36
Step 6. Solve.
10(12x)+x = 10x+ (12x) 36
=12010x+x=10x+12x36
=1209x=9x24
=144=18x
=x=8
=12x=4
Therefore the number is 84.
Check. 8436=48
Example 4. How many pounds of candy valued at 48¢ per pound should be added to 50 pounds of candy valued at 80¢ per pound in order for the store owner to be able to sell the candy at 60¢ per pound?
Step 1. Reread!
Step 2. Let x be the number of pounds of 48¢ per pound candy.
Step 3. Then 50+x would be the pounds of candy he would have at 60¢ per pound.
Step 4. The amount of candy at 48¢ per pound times 48¢, plus the amount of candy at 80¢ per pound times 80¢, must be equal to the amount of candy at 60¢ per pound times 60¢.
Step 5. (48¢/lb)(x lbs) + (80¢/lb) (50 lbs) = (60¢/lb) [(50+x)lbs]
Step 6. Solve.
48x+80*50=60(50+x)
48x+4000=3000+60x
1000=12x
x=(83(1)/3)lbs
Check. (83+1/3)48+80*50=60(50+83+1/3)
Problems involving velocities (or speeds) will use the formula
d=rt
where d is the distance traveled, r is the rate, and t is the time. When the formula is used, d and r must be expressed in the same unit of distance, whiler and t must be expressed in the same unit of time.
Example 5. A group of students drove to a lake in the north woods to ﬁsh. They traveled 380 miles in 7 hours, of which 4 hours were on a paved highway and the remaining time was on a dirt road. If the average speed on the dirt road was 25 miles per hour less than the average speed on the highway, then ﬁnd for each part of the trip the average speed and the distance traveled.
Step 1. Reread!
Step 2. Let x be the speed on the dirt road.
Step 3. Then x+25 is the speed on the highway.
Step 4. The distance traveled on the highway plus the distance traveled on the dirt road is equal to 380 miles.
Step 5. Since d=rt, we have
[(x+25)(mi)/(hr)](4hrs)+[x(mi)/(hr)](3hrs)=380mi
Step6. Solve.
(x+25)4+3x=380
4x+100+3x=380
7x=280
x=40 miles per hour
x+25=65miles per hour
Check. (40+25)4+40*3=380
Work problems which involve the rate of performance can often be solved by ﬁrst ﬁnding the fractional part of the task done by each person or machine in one unit of time, and then ﬁnding an equation that relates these various fractional parts.
Example 6. A boy can cut a lawn in 4 hours while the father can cut it in 3 hours. How long would it take them to cut the same lawn working together?
Step 1. Reread!
Step 2. Let x be the number of hours that it would take them to cut the lawn Working together.
Step 3. Choose one hour as our unit of time. Now the boy can cut 1/4 of the lawn in one hour, the father can cut 1/3 of the lawn in one hour, and togeﬂner they can cut 1/x of the lawn in one hour.
Step 4. The amount cut by the boy in one hour plus the amount cut by the father in one hour is equal to the amount they can cut together in one hour.
Step 5.1/3+1/4=1/x
Step 6. Solve.
1/3+1/4=1/x
7/12=1/x
x=12/7 hours
Linear Equations
A linear equation is an equation for a straight line
These are all linear equations:
y = 2x + 1  
5x = 6 + 3y  
y/2 = 3 − x 
Let us look more closely at one example:
Example: y = 2x + 1 is a linear equation:
The graph of y = 2x+1 is a straight line
 When x increases, y increases twice as fast, so we need 2x
 When x is 0, y is already 1. So +1 is also needed
 And so: y = 2x + 1
Here are some example values:
x  y = 2x + 1 

1  y = 2 × (1) + 1 = 1 
0  y = 2 × 0 + 1 = 1 
1  y = 2 × 1 + 1 = 3 
2  y = 2 × 2 + 1 = 5 
Check for yourself that those points are part of the line above!
Different Forms
There are many ways of writing linear equations, but they usually have constants (like "2" or "c") and must have simple variables (like "x" or "y").
Examples: These are linear equations:
But the variables (like "x" or "y") in Linear Equations do NOT have:
Examples: These are NOT linear equations:
y^{2} − 2 = 0  
3√x − y = 6  
x^{3}/2 = 16 
SlopeIntercept Form
The most common form is the slopeintercept equation of a straight line:
Example: y = 2x + 1
 Slope: m = 2
 Intercept: b = 1
PointSlope Form
Another common one is the PointSlope Form of the equation of a straight line:
y − y_{1} = m(x − x_{1}) 
Example: y − 3 = (¼)(x − 2)
It is in the form y − y_{1} = m(x − x_{1}) where:
 y_{1} = 3
 m = ¼
 x_{1} = 2
General Form
And there is also the General Form of the equation of a straight line:
Ax + By + C = 0 
(A and B cannot both be 0) 
Example: 3x + 2y − 4 = 0
It is in the form Ax + By + C = 0 where:
There are other, less common forms as well.
As a Function
Sometimes a linear equation is written as a function, with f(x) instead of y:
y = 2x − 3 
f(x) = 2x − 3 
These are the same! 
And functions are not always written using f(x):
y = 2x − 3 
w(u) = 2u − 3 
h(z) = 2z − 3 
These are also the same! 
The Identity Function
There is a special linear function called the "Identity Function":
f(x) = x
And here is its graph:
It makes a 45° (its slope is 1)
It is called "Identity" because what comes out is identical to what goes in:
In  Out 

0  0 
5  5 
−2  −2 
...etc  ...etc 
Constant Functions
Another special type of linear function is the Constant Function ... it is a horizontal line:
f(x) = C
No matter what value of "x", f(x) is always equal to some constant value.
Using Linear Equations
You may like to read some of the things you can do with lines:
Gradient (Slope) of a Straight LineY Intercept of a Straight LineDistance Between 2 PointsFinding Intercepts From an EquationGraph MenuAlgebra Menu
Copyright © 2017 MathsIsFun.com
Sours: https://www.mathsisfun.com/algebra/linearequations.htmlEquations help solving linear
Solving a System of Equations
Systems of linear equations take place when there is more than one related math expression. For example, in \(y = 3x + 7\), there is only one line with all the points on that line representing the solution set for the above equation.
When you are given 2 equations in the same question, and asked to solve for a unique answer, you can visualize the problem as be two lines on the same xyplane. The following two equations are graphed on the same xyplane:
$$ y = 3x + 5 $$ $$ y =  x $$The solution set to any equation is the place where BOTH equations meet on the xyplane. This meeting place is called the Point of Intersection. If you have a linear equation and a quadratic equation on the same xyplane, there may be TWO POINTS where the graph of each equation will meet or intersect. Here's a geometric view:
Here is a sample of two equations with two unknown variables:
Example
$$ x + y = 10 $$ $$ 3x + 2y = 20 $$There are three methods to solve our sample question.
 1) We can solve it graphically
 2) We can solve it algebraically
 3) We can also solve it through algebraic elimination
I will solve the question using all 3 methods. Method 1: Solve Graphically:
To solve graphically, it is best to write BOTH equations in the slopeintercept form or in the form: \(y = mx + b\) where m = the slope and b = the yintercept as your first step. So, \(x + y = 10\) becomes \(y =  x + 10\) (slopeintercept form). Next, \(3x + 2y = 20\) becomes \(y = \frac{3x}{2} + 10\) when written in slopeintercept form.
Then, graph the two lines, leading to the point of intersection. After graphing these lines, you'll find that BOTH equations meet at point (0,10). Point (0,10) means that if you plug x = 0 and y = 10 into BOTH original equations, you will find that it solves both equations. Here's what these two equations look like on the xyplane:
Method 2: Solve algebraically
Steps:
1) Solve for eaither x or y in the first equation (\(x + y = 10\)). I will solve for y. So, \(x + y = 10\) becomes \(y = x + 10\).
2) Plug the value of y (that is, x + 10) in the second equation to find x. Our second equation was \(3x + 2y = 20\) and, after substituting, becomes \(3x + 2(x + 10 ) = 20\)
Next: Solve for x.
$$ 3x 2x + 20 = 20 $$ $$ x + 20 = 20 $$ $$ x = 0 $$3) Plug x = 0 into EITHER original equations to find the value of y. I will use our second equation.
$$ 3x + 2y = 20 $$ $$ 3(0) + 2y = 20 $$ $$ 0 + 2y = 20 $$ $$ y = 10 $$So, our point of intersection is once again (0,10).
Method 3: Algebraic Elimination
This method deals with matching the variables to ELIMINATE or do away with one. Keep in mind that it is your choice which variable you want to eliminate first.
GOAL: Eliminate x and solve for y or viceversa. Let's go back to our original equations.In our second 3x + 2y = 20, you can eliminate 3x by multiplying 3 by EVERY term in our first equation (x + y = 10).
x + y = 10
3x + 2y = 20
3(x) + 3(y) = 3(10)
3x + 2y = 20
3x + 3y = 30
3x + 2y = 20
NOTICE that 3x and 3x are eliminated. See it? See why? Here's why: A negative PLUS a positive = ZERO.
We now have this:
3y = 30
2y = 20
3y + 2y = 30 + 20
y = 10
y = 10.
Next: To find x, we plug y = 10 into EITHER of the original equations. By now you should see that our answer for x will be ZERO.
Here it is:
I will use x + y = 10
x + 10 = 10
x = 0.
Do you see what I see? Yes, I found the SAME point of intersection again, which is (0,10).
By Mr. Feliz
(c) 2005
Solving Equations: (lesson 1 of 4)
Solving Linear Equations
Equations with no parentheses
Example 1
Solve 5x  4  2x + 3 = 7  3x + 5 + 2x
Solution 1
Step 1: Combine the similar terms  $5x  4  2x + 3 =  7  3x + 5 + 2x$ 
$3x  1 =  x  2$  
Step 2: Add x to both sides.  $3x  1 + x =  x  2 + x$ 
$4x  1 =  2$  
Step 3: Add 1 to both sides.  $4x  1 + 1 =  2 + 1$ 
$4x =  1$  
Step 4: Divide both sides by 4:  $\frac{{4x}}{4} = \frac{{  1}}{4}$ 
Solution is:  $x =  \frac{1}{4}$ 
Exercise 1: Solve equations
Equations with parentheses
Example 2
Solve $2 (x  4) + 4 (2  x) = 5x  4 (x + 1)$
Solution 2
Step 1: Simplify both sides:  $2 (x  4) + 4 (2  x) = 5x  4 (x + 1)$ 
$2x  8 + 8  4x = 5x  4x  4$  
$ 2x = x  4$  
Step 2: Subtract x from both sides.  $ 2x  x = x  4  x$ 
$ 3x =  4$  
Step 3: Divide both sides by 3:  $\frac{{  3x}}{{  3}} = \frac{{  4}}{{  3}}$ 
Solution is:  $x = \frac{4}{3}$ 
Exercise 2: Solve equations
Equations containing fractions
Example 3
Solve $x + \frac{1}{2} = \frac{x}{2}  \frac{2}{3}$
Solution 3
Step 1: Multiply both sides by the LCD. Lowest common multiple of 2 and 3 is 6. So, we multiply both sides by 6.  $6 \cdot \left( {x + \frac{1}{2}} \right) = 6 \cdot \left( {\frac{x}{2}  \frac{2}{3}} \right)$ 
Step 2: Remove brackets:  $6 \cdot x + 6 \cdot \frac{1}{2} = 6 \cdot \frac{x}{2}  6 \cdot \frac{2}{3}$ 
Step3: This problem is similar to the previous  $6x + 3 = 3x  4$ 
$6x + 3  3x = 3x  4  3x$  
$3x + 3 =  4$  
$3x + 3  3 = 4  3$  
$3x =  7$  
$\frac{{3x}}{3} = \frac{{  7}}{3}$  
Solution is:  $x =  \frac{7}{3}$ 
Exercise 3: Solve equations
More Complicated Example
Example 4
$\frac{{2x}}{{x + 2}} = \frac{4}{{x  10}} + 2$
Solution 4
In this case the LCD is (x + 2)(x  10). Here is the complete solution to this problem.
$$\frac{{2x}}{{x + 2}} = \frac{4}{{x  10}} + 2$$ $$\cancel{{(x + 2)}}(x  10)\frac{{2x}}{{\cancel{{x + 2}}}} = (x + 2)\cancel{{(x  10)}}\frac{4}{{\cancel{{x  10}}}} + (x + 2)(x  10) \cdot 2$$ $$(x  10) \cdot 2x = (x + 2) \cdot 4 + 2({x^2} + x  10x  20)$$ $$\cancel{{2{x^2}}}  20x = 4x + 8 + \cancel{{2{x^2}}} + 2x  20x  40$$ $$  20x = 4x + 8 + 2x  20x  40$$ $$  20x =  14x  32$$ $$  20x + 14x =  14x  32 + 14x$$ $$  6x =  32$$ $$x = \frac{{  32}}{{  6}}$$ $$x = \frac{{16}}{3}$$
Much more Complicated Example
Example 5
$\frac{1}{{x + 3}} = \frac{{  2x}}{{{x^2} + 5x + 6}}$
Solution 4
The first step is to factor the denominators
$\frac{1}{{x + 3}} = \frac{{  2x}}{{(x + 3)(x + 2)}}$
The LCD is (x + 3)(x + 2). The solution is:
$$\frac{1}{{x + 3}} = \frac{{  2x}}{{(x + 3)(x + 2)}}$$ $$(x + 3)(x + 2)\frac{1}{{x + 3}} = (x + 3)(x + 2)\frac{{  2x}}{{(x + 3)(x + 2)}}$$ $$x + 2 =  2x$$ $$x + 2x =  2$$ $$3x =  2$$ $$x =  \frac{2}{3}$$
Exercise 4: Solve equations
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Give me a place to stand, and I will move the earth.
Archimedes of Syracus
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I like mathematics because it is not human and has nothing particular to do with this planet or with the whole accidental universe  because like Spinoza's God, it won't love us in return.
Bertrand Russell
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