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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WTAMU> Virtual Math Lab > Intermediate Algebra
Last revised on July 1, 2011 by Kim Seward.
All contents copyright (C) 2002  2011, WTAMU and Kim Seward. All rights reserved.
The following steps provide a good method to use when solving linear equations.
 Simplify each side of the equation by removing parentheses and combining like terms.
 Use addition or subtraction to isolate the variable term on one side of the equation.
 Use multiplication or division to solve for the variable.
Note: Fractions may be removed by multiplying each side of the equation by the common denominator.
Example:
Solve for z: 7z – (3z – 4) = 12
Solution:
Step 1. Simplify the left side of the equation by removing parentheses and combining like terms.
Distribute through by 1.
7z– 3z+ 4 = 12
Combine like terms on the left side of the equation.
4z + 4 = 12
Step 2. Use subtraction to isolate the variable term on the left side of the equation.
Subtract 4 from each side of the equation.
4z + 4 – 4 = 12 – 4
4z = 8
Step 3. Use division to solve for the variable.
Divide each side of the equation by 4.
The solution to 7z – (3z – 4) = 12 is z = 2.
Solve for y: (y  11)  (y + 8) = 6y
Solve for x: 0.8(5x + 15) = 2.6  (x + 3)
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Isolate the term with the absolute value by adding 5 to both sides.
[latex]\begin{array}{r}3\left4w1\right5=10\\\underline{\,\,\,\,\,\,\,\,\,\,\,\,\,+5\,\,\,+5}\\ 3\left4w1\right=15\end{array}[/latex]
Divide both sides by 3. Now the absolute value is isolated.
[latex]\begin{array}{r} \underline{3\left4w1\right}=\underline{15}\\3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\,\,\\\left4w1\right=\,\,5\end{array}[/latex]
Write the two equations that will give an absolute value of 5 and solve them.
[latex] \displaystyle \begin{array}{r}4w1=5\,\,\,\,\,\,\,\,\,\,\,\,\text{or}\,\,\,\,\,\,\,\,\,\,4w1=5\\\underline{\,\,\,\,\,\,\,+1\,\,+1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{\,\,\,\,\,\,\,\,\,+1\,\,\,\,\,+1}\\\,\,\,\,\,\underline{4w}=\underline{6}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{4w}\,\,\,\,\,\,\,=\underline{4}\\4\,\,\,\,\,\,\,\,\,\,\,4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4\,\,\\\,\,\,\,\,\,\,\,w=\frac{3}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,w=1\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,w=\frac{3}{2}\,\,\,\,\,\text{or}\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]
Check the solutions in the original equation.
[latex] \displaystyle \begin{array}{r}\,\,\,\,\,3\left 4w1\, \right5=10\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\left 4w1\, \right5=10\\\\3\left 4\left( \frac{3}{2} \right)1\, \right5=10\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\left 4w1\, \right5=10\\\\\,\,\,\,\,\,3\left \frac{12}{2}1\, \right5=10\,\,\,\,\,\,\,3\left 4(1)1\, \right5=10\\\\\,\,\,\,\,\,\,\,3\left 61\, \right5=10\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\left 41\, \right5=10\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\left(5\right)5=10\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3\left 5 \right5=10\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,155=10\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,155=10\\10=10\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,10=10\end{array}[/latex]
Both solutions check
Answer
[latex]w=1\,\,\,\,\text{or}\,\,\,\,w=\frac{3}{2}[/latex]
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Steps linear equations
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