Download Chapter 1 Section 1.1

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Linear Equations
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Warm-up
A linear equation in one variable is an equation that can be written in
ax  b  0 where a and b are constants and a ≠ 0.
the form __________
Solve for x in the following equations.
3x  6
2 x  9
0  2 x  5
 x  3  1 x
 x  12
2  5 x  3( x  1)
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A simple linear equation
Starting Equation
3x  6
Divide both sides of
the equation by 3
3x 6

3 3
Solution
x2
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General method for solving
equations of the form ax  c
Starting Equation
ax  c
eg.
Divide both sides of
the equation by a
ax c

a a
3
x
6
eg.

3 3
c
x
a
eg.
Solution
3x  6
x2
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Example 1
Starting Equation
2x  9
Divide both sides of
the equation by 2
2x  9

2
2
Solution
9
x
2
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Example 2
Starting Equation
 x  12
Divide both sides of
the equation by −1
 x 12

1
1
Solution
x  12
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Grouping like terms
Starting Equation
0  2x  5
 2x
 2x
2x  5
Divide both sides of
the equation by 2
Solution
2x 5

2
2
5
x 
2
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Grouping like terms
The idea behind “grouping like terms” is to get all of the
terms involving x on one side of the equation and all of
the constants on the other side of the equation.
Take the previous problem as an example.
Starting Equation
Grouping like terms
Term involving
0  2x  5
 2x
 2x
2x  5
x
Constant
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Example: Grouping like terms
Starting Equation
 x  3  1 x
3
Group like terms
3
xx  4  xx
 2x  4
Divide both sides of
the equation by −2
Solution
 2x
4

2
2
x  2
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Solving Linear Equations Using Distribution
Starting Equation
2  5 x  3( x  1)
Distribute
2  5x  3x  3
Group like terms
Divide both sides of
the equation by −8
Solution
2
2
 5x  3x  1
 3x
 3x
 8x  1
8 8
x1
8
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1.1.2(A)
Starting Equation
Distribute
4( x  1)  2  3(2 x  1)  4
4x  4  2  6x  3  4
4x  2  6x  7
2
Groups like terms
Divide both sides by −2
Solution
2
4x  6x  5
 6x
 6x
 2x  5
2
2
x5
2
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Overview
 Our method of solving linear equations involves repeated
simplification. Each step is designed to reduce the
problem to a form that we already now how to solve.
Steps:
1. Distributing: Results in only needing to group like terms
2. Grouping like terms: Results in the form
3. Dividing: Results in a solution
ax  c
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Reducing to a previously solved problem
 TELL JOKE NOW!!
Next, we will learn how to solve linear equations with
• Decimals
• Fractions
• Variables in the denominator.
Rather than solving such problems, I will teach you how to
reduce them to previously solved problems. That is, I will
show you how to convert such problems into the form we
just learned how to solve.
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Linear equations with Decimals
 If a linear equation has decimals, we multiply both sides
of the equation by a power of 10 (i.e., 10, 100, 1000, etc)
to get rid of the decimals. This is not necessary, but it can
help if you don’t like working with decimals.
Example:
 If our problem has the decimals 0.1, 0.2, and −0.8, we
multiply both sides of the equation by 10. Our decimals
then become 10×0.1=1, 10 ×0.2=2, and 10 ×(−0.8)= −8
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1.1.2(C) Linear equation with decimals
Starting Equation
0.1(4 x  8)  0.2( x  4)  0.8 x
Multiply by 10 to 100.1(4x  8)  0.2( x  4)  10 0.8x
get rid of decimals
Distribute the 10 100.1(4x  8) 100.2( x  4)  10 0.8x
1(4 x  8)  2( x  4)  8 x
The problem is now in a form you can solve.
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Example: Linear equation with decimals
Starting Equation
Multiply by 100 to
get rid of decimals
0.03( x  1)  0.17 x  0.01x
1000.03( x  1)  0.17 x  1000.01x
Distribute the 100 1000.03( x 1)1000.17 x  1000.01x
3( x  1)  17 x  x
The problem is now in a form you can solve.
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Linear equations with Fractions
 If a linear equation has fractions, we multiply both sides
of the equation by a common denominator to get rid of
the fractions. Ideally, we should multiply by the least
common denominator. This is not necessary, but it can
help if you don’t like working with fractions.
Example:
 If our problem has the fractions 1/4, 1/2, and 1/8, we
multiply both sides of the equation by 8. Our fractions
then become 8×1/4=2, 8 ×1/2=4, and 8 ×1/8= 1.
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1.1.2(B) Linear equation with fractions
Starting Equation
Multiply by 8 to
get rid of fractions
Distribute the 8
1
1
1
( p  2)  (3 p  1)  ( p  1)
4
2
8
1
1

1

8 ( p  2)  (3 p  1)  8 ( p  1)
2
4

8

1
 1

1

8 ( p  2)  8 (3 p  1)  8 ( p  1)
4
 2

8

2( p  2)  4(3 p  1)  ( p  1)
The problem is now in a form you can solve.
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Example: Linear equation with fractions
Starting Equation
Multiply by 15 to
get rid of fractions
Distribute the 15
1
1
1
(2 y  1)  ( y  4)  ( y  1)
3
15
5
1
1

1

15 (2 y  1)  ( y  4)  15 ( y  1)
15
3

5

1

1

1

15 (2 y  1)  15 ( y  4)  15 ( y  1)
3

15

5

5(2 y  1)  1( y  4)  3( y  1)
The problem is now in a form you can solve.
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Variables in the denominators
 If an equation has variables in the denominators, it is NOT a
linear equation. Such equations, however, can lead to linear
equations.
 We treat such equations like those with fractions. That is, we
multiply both sides of the equation by a common denominator
to get rid of the variables in the dominators. Ideally, we should
multiply by the least common denominator.
Example:
 If our problem has x−3 in the denominator of one term, and
x−3 in the denominator of another term, we multiply both
sides of the equation by x−3. After the multiplication, the
terms will have no variables in the denominators.
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1.1.2(D) Variables in the denominators
Starting Equation
2x
5
1 
x 3
x3
Multiply both sides
by (x −3)
 2x

 5 
( x  3) 
 1  ( x  3) 
x 3 
 x  3 
Distribute the (x−3)
 2x 
 5 


( x  3) 

(
x

3
)
1

(
x

3
)
 x  3 
 x  3 
2 x  ( x  3)  5
The problem is now in a form you can solve.
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Example: With variables in in the denominators
Starting Equation
3
2
4


x  1 x  2 ( x  1)( x  2)
Multiply both sides by (x −1)(x+2)


2 
4
 3
( x  1)( x  2)

 ( x  1)( x  2)


x

1
x

2
(
x

1
)(
x

2
)




Distribute the (x−1)(x+2)


4
 3 
 2 
( x  1)( x  2)
 ( x  1)( x  2)
 ( x  1)( x  2)



x

1
x

2
(
x

1
)(
x

2
)






3( x  2)  2( x  1)  4
The problem is now in a form you can solve.
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Example: With variables in in the denominators
Starting Equation
3
2
4

 2
x 1 x  2 x  x  2
The denominator x2+x−2 on the right-hand-side of the equation
is not factored. That makes it difficult to find the least common
denominator. Therefore, we first factor x2+x−2.
Equation after factoring
3
2
4


x  1 x  2 ( x  1)( x  2)
That is the starting equation on the previous slide.
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Extraneous Solutions
Starting Equation
x
1
1 
x 1
x 1
Multiply both sides
by (x−1)
Distribute the (x−1)
 x

 1 
( x  1) 
 1  ( x  1) 
 x 1 
 x  1
 x 
 1 
( x  1) 
 ( x  1)1  ( x  1) 

 x  1
 x  1
x  ( x  1)  1
Extraneous Solution
x 1
Plugging in 1 for x in the original equation
would result in dividing by zero. Thus, x=1 is not
a solution. It is called an extraneous solution.
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Variables in the denominators
There are two key differences between equations with
variables in the denominators and those with fractions.
1. If an equation has variables in the denominators, it is important
to first factor the expressions in the denominators. This makes it
easier to find a least common denominator.
2. An equation with variables in the denominator may have
extraneous solutions. Therefore, it is important to check that the
solution you get does not make any of the denominators zero.
Checking answers on a Calculator
1. Type in your answer. (e.g. 4)
2. Press STO, X,T,Ѳ,n , and press ENTER
3. Enter the original equation.
(e.g. 2x+1=9)
(Use X,T,Ѳ,n to type x.
Press 2nd, then MATH (TEST),
then ENTER to type =.)
4. Press ENTER.
If a 1 appears your answer is right.
If a 0 appears your answer is wrong.
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Identifying Linear equations
A linear equation in one variable is an equation that can be written in
ax  b  0 where a and b are constants and a ≠ 0.
the form __________
To decide if an equation is linear, we try and get it in this form.
To do this, simply try to solve it. But instead of actually solving it,
stop short to try and get it in this form. If it cannot be written in
this form, then the equation is not linear.
Note: When trying to get the equation into this form, you
cannot multiply or divide by an expression with a variable.
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Example: Identifying linear equations
Starting Equation
7 x  1  3x
 3x
Equation in the
form ax  b  0
a4
b 1
 3x
4x  1  0
ax  b  0
The equation is linear.
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1.1.1(B) Identifying linear equations
Starting Equation
4 x 3  x
2x  3  x
x
x
Equation in the
form ax  b  0
1 x 3  0
a 1
ax  b  0
b  3
The equation is linear.
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1.1.1(D) Identifying linear equations
Starting Equation
( x  1) x  7
Note: When trying to get the equation into the form ax+b=0,
you cannot multiply or divide by an expression with a variable.
We would have to multiply by x to get the equation into the from
ax+b=0. Therefore, the equation is non-linear.
It does, however, lead to the linear equation −6x−1=0.
The equation is non-linear.
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1.1.1(C) Identifying linear equations
Starting Equation
3x  1 x  2
Note: When trying to get the equation into the form ax+b=0,
you cannot multiply or divide by an expression with a variable.
We would have to multiply by x to get the equation into the from
ax+b=0. Therefore, the equation is non-linear.
It also does NOT lead to a linear equation.
The equation is non-linear.
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1.1.1(A) Identifying linear equations
Starting Equation
3x  2  4 x
 4x 2
2
 4x 2
 4 x  3x  2  0
2
The equation cannot be written in the form ax  b  0
The equation is non-linear.
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Quadratic equations
An equation in the form ax2+bx+c=0 where a≠0 is called a
quadratic equation. The equation −4x2+3x+2=0 from the
previous slide is an example of a quadratic equation.
All quadratic equations are non-linear. You may simply
assume this fact. Although a mathematician would want
you to prove it.
JOKE:
An argument is needed to convince a reasonable person.
A proof is needed to convince and unreasonable one.
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Systems of equations
So far we have learned to solve for one variable given a single
equation. Next we shall try and solve for multiple variables given
multiple equations.
Examples:
2x  5 y  7
x  2y  3
x  2y
14 x  14 y  1
There are two methods for solving systems of equations:
Substitution and Elimination. Both work by combining
the equations into a single equation with one variable.
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Substitution
Starting Equations
x  2y
14 x  14 y  1
Substitute 2y for x in
the second equation
14( 2 y )  14 y  1
Solution for y
Plug in 1/42 for y in
the first equation to
get the solution for x
1
y
42
1
 1 
x  2

21
 42 
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1.1.3 Substitution
Starting Equations
Solve for x in the
second equation
Substitute −2y−8 for
x in the first equation
Solution for y
Plug in −5 for y in the
second equation to
get an equation for x
4 x  2 y  18
x  2 y  8
x  2 y  8
4(2 y  8)  2 y  18
y  5
x  2(5)  8
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1.1.3 Elimination
Starting Equations
Add the two
equations together
Solution for x
Plug in 2 for x in one of
the original equations
to get an equation for y
4 x  2 y  18
 x  2 y  8
5x  10
x2
2  2 y  8
Now solve for y
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1.1.4 Elimination
Starting Equations
Multiply the first
equation by −2
Add the latter two
equations together
3A  5B  15
6 A  2B  12
 6 A 10B  30
12B  18
We chose to multiply the first equation by −2, so that when we
would add the two equations together the terms with A would
cancel leaving us with an equation that has just one variable.
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Substitution and elimination
Both substitution of elimination work by combining the equations into a
single equation with one variable.
 Substitution accomplishes this by substituting one variable by an
equivalent expression that is written in terms of another variable.
 Elimination works by multiplying the equations by appropriate
constants, so that when added together one of the variables will be
eliminated.
The resulting equation with one variable is then solved. The solution to
this equation is plugged into one of the original equations . That
produces an equation in the second variable, which may then be solved.