Transcript MA 128: Lecture 03 * 6/26/02

MA 1128: Lecture 03 – 5/28/2015
Solving Linear Equations
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Expressions
We will use the equal sign in two slightly different ways.
When we are simplifying expressions, we’ll start with an expression like
x(x + 22)
And change the form of the expression somehow.
For example, since 22 = 4, replacing 22 with 4 results in another expression
that is equal
x(x + 22) = x(x + 4)
Going one step further, we can distribute the first x over the x + 4,
x(x + 22) = x(x + 4) = x2 + 4x
and we get a third expression that is equal to the first two.
When changing expressions, the equal sign indicates that the expressions are
equivalent.
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Equivalent Expressions
Specifically, we will say that two expressions are equivalent, if they are equal for
all values of x (or whatever the variable letter is).
For example, if we put 2 in for the x, the first expression x(x + 22) is equal to
(2)((2) + 22) = 2(2 + 4) = 2(6) = 12
Putting 2 into the last expression x2 + 4x gives us the same result
(2)2 + 4(2) = 4 + 8 = 12.
If we were to put a 5 in for the x, we would get 45 from each expression.
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Practice Problems for Equivalent Expressions
1.
2.
3.
4.
5.
What do the first and last expressions equal if you put a 3 in for the x?
Are the numbers you get the same?
The distributive property applied to x(x + 1) results in what? Note that the
distributive property will give you an equivalent expression.
Put x = 2 into the expressions x(x + 1) and x2 + x. What do you get?
Put x = 1 into both expressions from problem 4. What do you get?
Click for answers:
1) 3 [[ (3)((3) + 22) = (3)(1) = 3 and (3)2 + 4(3) = 9  12 = 3]];
2) Yes; 3) x2 + x; 4) 6; 5) 0.
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Equations
We will also use the equal sign in an equation.
Here we will put an equal sign between two non-equivalent expressions,
and we’ll be asking the question, for what values of x are the expressions equal.
In solving an equation, we will start with an equation like
x + 4 = 2x
Here, the expressions x + 4 and 2x are not equivalent.
For most of the values we might put in for the x, the results won’t be equal.
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Practice Problems
1.
2.
Put a 3 in for the x in both expressions. What do you get?
Put a 4 in for the x in both expressions. What do you get?
Click for answers
1) 7 for one and 6 for the other.
2) 8 for both of them. When solving this equation, x = 8 is what we’ll be
looking for.
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Solving Equations
When we solve an equation, we are looking for those values for x that make the
two expressions equal.
These values are called solutions.
The solutions are obvious in some equations.
For example, in the equation
x = 4,
The only way x and 4 can be equal is if x is 4.
In the equation, as another example,
x – 1 = 10,
We can see pretty easily that x must be 11.
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Continued
For the equation
x2 = 4
We can see that x can be 2 or 2.
In this case, we’ll usually write something like
x = 2,2
to indicate that these are the solutions.
Again, look at how the two expressions x2 and 4 are in no way equivalent.
The expression 4 is always equal to 4,
and x2 can be any non-negative number.
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Practice Problems
Find all of the solutions for the following equations.
1.
x = 10.
2.
x2 = 9.
3.
x + 3 = 5.
4.
x – 4 = 2.
Click for answers.
1) x = 10; 2) x = 3,3; 3) x = 2; 4) x = 6.
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Special cases.
In this class, most of our equations will only have one or two solutions.
BUT there are a few special cases. For example, in the equation
3x = x + 2x
The two expressions are equal for any value of x (the two expressions are
equivalent).
For the solution, we’ll say that x can be any real number.
We’ll also see equations like
x = x + 2.
The expression on the right is always 2 larger than the one on the left,
So in this case, there are no solutions.
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Basic Techniques
Getting back to how we solve equations, we will follow the following Basic
Principle
Given an equation to solve, we will try to transform the equation into
one where the solutions are obvious.
Example. Consider the equation
3x + 7 = 1.
You may be able to see the solution, but let us try to transform this into a simpler
equation (where the solution is even more obvious).
We can add 7 to both sides (or subtract 7 from both sides).
3x + 7 – 7 = 1 – 7
3x + 0 = 6
3x = 6
This last equation is simpler than the original, but we can get simpler.
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More on techniques
We can restate our Basic Principle like this.
A mystery value for x will make both expressions equal.
Therefore, if we do the same thing to both sides, then the same x will
still make both sides equal.
In particular, by adding or multiplying both sides of an equation by the same
number, we can produce simpler equations with the same solutions.
In this example, we can simplify our equation further by multiplying both sides by
1/3.
1
3
1
3 x     6 
1x 
3
6
3
x = 2
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Continued
The last equation is so simple, we’ll use it to say what the solutions are, and we’ll
actually call an equation like x = 2, a solution.
Here’s another example.
7x – 3 = 5
7x – 3 + 3 = 5 + 3
7x = 8
1
7
1
7 x   8 
7
x
8
7
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Practice Problems
1.
2.
Solve 5x + 3 = 23
Solve 3x + 5 = 14
Click for answers.
1) x = 4; 2) x = 3.
Next Slide
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Terms
Consider the equation
2x – 3 + x = 9
In the order of operations,
we go in the order exponents  multiplication  addition
Exponents form the strongest bonds, and addition the weakest.
In this equation, the weakest bonds, the additions, separate the expressions into
2x, 3, and x on the left side
And just the 9 on the right.
These four things are called terms.
The 2x and x are called x-terms,
And the 3 and the 9 are called constant terms.
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Continued
One Basic Principle we’ll use to determine what’s simpler is the following.
Simpler equations have fewer terms.
On the left of the equation we’re considering, we have 2x + x.
That is, two x’s plus one x.
Altogether, there are three x’s.
This is the distributive property in action.
2x + x = (2 + 1)x = 3x.
In the equation we have
2x – 3 + x = 9
3x – 3 = 9.
We can always combine like terms to reduce the number of terms.
And this makes the equation simpler.
Next Slide
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Continuing Example
We’re trying to reduce the number of terms, and we also know that we want to end
up with an equation of the form x equals something.
We can get rid of the constant term 3 on the left by adding 3 to both sides.
3x – 3 + 3 = 9 + 3
3x = 12
From here we can divide both sides by 3, and end up with x = 4.
An equation with only x-terms and constant terms is called a linear equation.
Given a linear equation, we can always find its solutions (except for the special
cases mentioned earlier, there will be just one solution).
Let’s finish off by looking at a few examples.
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Example
Consider the equation
4x – 2(3x – 7) = 2x – 6.
We first want to make this look like a linear equation.
The 2(3x – 7) is not an x-term or a constant term.
The distributive property says multiply the 2 times the 3x and the 7.
That is, 2(3x) and 2(7).
If we do this, we get a linear equation.
4x – 6x + 14 = 2x – 6
Now combine like terms.
2x + 14 = 2x – 6
(continued)
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Continued
To finish this problem, we would like only an x-term on one side and only a
constant term on the other.
We can get this by adding terms to both sides.
2x + 14 – 2x = 2x – 6 – 2x
4x + 14 = 6
4x + 14 – 14 = 6 – 14
4x = 20
Now divide both sides by 4 (or multiply by 1/4).

1
4
1
  4 x      20 
4
x 5
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Example
Consider the equation
3(x – 4) = 6x – (4 – 5x)
3x – 12 = 6x – 4 + 5x
3x – 12 = 11x – 4
3x – 12 – 11x = 11x – 4 – 11x
8x – 12 = 4
8x – 12 + 12 = 4 + 12
8x = 8

1
8
1
  8 x    8 
8
x  1
Next Slide
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Practice Problems
1.
2.
Solve 8 – 3(2x – 4) = 5 + 3x – 4x
Solve 2 + 5(x + 1) = 3x + 3
Click for answers.
1) x = 3; 2) x = 2.
Next Slide
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Example
x
4

1
6

7
x
3
Given an equation with fractions, most people like to use the basic trick of
multiplying both sides of the equation by a common denominator.
For this equation, 12 works, since 4, 6, and 3 all divide 12.
24 works also, but you will get bigger numbers.
Remember: By the distributive property, you must multiply every term.
x 1
7

12     12   x 
4 6
3

3 x  2  28  12 x
This gets you back to stuff we’ve already seen.
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Example
Consider the equation
3.9x – 5.7(2x – 3) = 9.3
Here we have decimal numbers.
Do everything the same way as you would with whole numbers.
You may want to use your calculator, and you may need to round.
When you round, the more places you keep the better.
Keep at least a couple of places past the decimal point.
3.9x – 11.4x + 17.1 = 9.3
7.5x + 17.1 = 9.3
7.5x = 7.8
x
 7 .8
 7 .5
x  1 . 04
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Practice Problems
1. Solve the equation
3
5
x
5
 4x 
2
9
10
2. Solve the equation
0 . 05 ( 2000  2 x )  . 04 ( 2500  6 x )
Click for answers:
1) x = 1; 2) x = 0. Note: An answer of x = 0 is fine. Zero is a perfectly good
number, and can be a solution. This is very different from not having a solution,
which means that no number for x will work.
End
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