3.2.1

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Introduction
Your previous experiences have helped you understand
how to represent situations using algebraic equations,
and how solving those equations can reveal something
specific about those situations. Most of these equations
have involved integer coefficients. However, coefficients
can also be expressed as a ratio. A ratio compares two
quantities, and is often written as a fraction. Numbers
expressed as a ratio are called rational numbers.
1
3.2.1: Creating and Solving Rational Equations
Introduction, continued
More formally, a rational number is any number that can
be written as
m
n
, where m and n are integers and n ≠ 0.
In this lesson, we will explore ways to solve rational
equations, and to assess and validate the resulting
solutions.
2
3.2.1: Creating and Solving Rational Equations
Key Concepts
• A rational equation is an algebraic equation that
includes at least one term that is expressed as a ratio.
• A rational equation defines a specific relationship
between numbers and variables; this relationship can
be analyzed to find a solution to the equation.
• Finding the solution to a rational equation may be
accomplished by simplifying the rational terms. To
simplify rational terms, convert the rational equation
into an equivalent equation with integer coefficients.
3
3.2.1: Creating and Solving Rational Equations
Key Concepts, continued
• Finding the solution often requires using a common
denominator—a quantity that is a shared multiple
of the denominators of two or more fractions—to
eliminate any rational terms.
• Recall that the least common denominator is the least
common multiple of the denominators of the fractions
in the expression.
• Like other algebraic equations, a rational equation
may have no solutions.
4
3.2.1: Creating and Solving Rational Equations
Key Concepts, continued
• An extraneous solution can occur when the solution
process results in a value that satisfies the original
equation, but creates an invalid equation by creating a
denominator that is equal to 0.
• An extraneous solution can also occur when a
solution results in a value that is irrelevant in the
context of the original equation.
5
3.2.1: Creating and Solving Rational Equations
Key Concepts, continued
• Recall that a proportion is a statement of equality
between two ratios. Techniques for solving
proportions can often be used to simplify rational
equations.
• The process of solving rational equations can be
applied to real-world examples in which a specific task
that may be done alone is to be completed
cooperatively.
6
3.2.1: Creating and Solving Rational Equations
Common Errors/Misconceptions
• multiplying only some terms of the rational equation
when eliminating rational numbers
• incorrectly or prematurely reducing rational numbers
• eliminating terms instead of factors
• neglecting to check for extraneous solutions
7
3.2.1: Creating and Solving Rational Equations
Guided Practice
Example 2
Solve the rational equation
5
t
-
3
12
=
9
3t
for t.
8
3.2.1: Creating and Solving Rational Equations
Guided Practice: Example 2, continued
1. Determine the least common denominator.
The denominators in the equation are t, 12, and 3t.
The smallest multiple having all three of these values
as a factor is 12t ; therefore, the least common
denominator is 12t.
9
3.2.1: Creating and Solving Rational Equations
Guided Practice: Example 2, continued
2. Multiply each term of the equation by the
common denominator and simplify.
5
3
9
(12t) - (12t) = (12t)
t
12
3t
5
3
9
(12) - (t) = (4)
1
1
1
Multiply each term
by the LCD, 12t.
Simplify each
fraction.
(12)5 – (t)3 = (4)9
Continue to simplify.
60 – 3t = 36
10
3.2.1: Creating and Solving Rational Equations
Guided Practice: Example 2, continued
The result, 60 – 3t = 36, is an algebraic equation
without any remaining rational numbers that can be
solved using methods you have already learned.
11
3.2.1: Creating and Solving Rational Equations
Guided Practice: Example 2, continued
3. Solve the resulting equation.
60 – 3t = 36
Simplified equation from the
previous step
60 = 36 + 3t
Add 3t to both sides.
24 = 3t
Subtract 36 from both sides.
t=8
Divide both sides by 3.
12
3.2.1: Creating and Solving Rational Equations
Guided Practice: Example 2, continued
4. Verify the solution(s).
While it appears that we have a solution, t = 8, we
have to return to the original equation to confirm it.
Because there are variables in the denominators of
the first and last terms, we must determine what
value(s) would make the denominator equal 0, and
thereby invalidate the equation (since division by 0
yields an undefined result and is not acceptable).
13
3.2.1: Creating and Solving Rational Equations
Guided Practice: Example 2, continued
The expression
t ≠ 0.
5 is undefined when t = 0; therefore,
t
Similarly, the expression
9
is also undefined when
3t
t = 0; once again, t ≠ 0. Our solution, t = 8, does not
violate this condition; therefore, it is an acceptable
result.
For the equation
5
t
-
3
12
=
9
3t
, t = 8.
✔
14
3.2.1: Creating and Solving Rational Equations
Guided Practice: Example 2, continued
15
3.2.1: Creating and Solving Rational Equations
Guided Practice
Example 3
Solve the rational equation
2
y
-
1
y +2
=
1
3
for y.
16
3.2.1: Creating and Solving Rational Equations
Guided Practice: Example 3, continued
1. Determine the least common
denominator.
The denominators are y, y + 2, and 3.
The smallest multiple having all three of these values
as a factor is 3y(y + 2), the irreducible product of all
three denominators.
The least common denominator is 3y(y + 2).
17
3.2.1: Creating and Solving Rational Equations
Guided Practice: Example 3, continued
2. Multiply each term by the common
denominator.
3y(y + 2) ·
2
y
- 3y(y + 2) ·
1
y +2
= 3y(y + 2) ·
1
3
18
3.2.1: Creating and Solving Rational Equations
Guided Practice: Example 3, continued
3. Simplify the resulting equation.
Start by rewriting each occurrence of the common
denominator in the equation as a ratio:
3y(y + 2) =
3y(y + 2)
1
Common denominator
rewritten as a ratio
3y(y + 2) 2 3y(y + 2)
1
· ·
=
1
y
1
y +2
3y(y + 2) 1
·
1
3
3.2.1: Creating and Solving Rational Equations
Rewritten
equation
19
Guided Practice: Example 3, continued
3(y + 2) • 2 – 3y • 1 =
y(y + 2) • 1
Simplify the ratios.
6y + 12 – 3y = y2 + 2y
Carry out the
multiplication.
3y + 12 = y2 + 2y
Subtract 3y from 6y.
20
3.2.1: Creating and Solving Rational Equations
Guided Practice: Example 3, continued
The algebraic equation that results this time,
3y + 12 = y2 + 2y, still has only integer coefficients.
However, it is no longer linear—this is a quadratic
equation because y is raised to a power of 2. We will
need to either factor or use the quadratic formula to
determine the value of y.
21
3.2.1: Creating and Solving Rational Equations
Guided Practice: Example 3, continued
4. Solve the quadratic equation by factoring
or using the quadratic formula.
Start by setting the equation equal to 0 so that the
equation is in the standard form of a quadratic
equation. Then, factor the quadratic.
22
3.2.1: Creating and Solving Rational Equations
Guided Practice: Example 3, continued
3y + 12 = y2 + 2y
Equation found in the
previous step
12 = y2 + (– y)
Subtract 3y from both sides.
0 = y2 – y – 12
Subtract 12 from both sides.
0 = (y – 4)(y + 3)
Factor the resulting quadratic.
y = 4 or y = –3
23
3.2.1: Creating and Solving Rational Equations
Guided Practice: Example 3, continued
5. Verify the solution(s).
Now that we have potential solutions, y = 4 or y = –3,
we must return to the original equation. We must
determine what value(s) would make the
denominators equal 0, and thereby invalidate the
equation.
24
3.2.1: Creating and Solving Rational Equations
Guided Practice: Example 3, continued
The expression
therefore, y ≠ 0.
The expression
2
y
is undefined when y = 0;
1
is also undefined when
y +2
y = –2; therefore, y ≠ –2.
There is no value for y that invalidates the last term,
1
, because the denominator does not contain
3
a variable.
25
3.2.1: Creating and Solving Rational Equations
Guided Practice: Example 3, continued
Our solutions, y = 4 or y = –3, do not violate the
condition that the value of y cannot make the
denominators equal 0; therefore, they are acceptable
results.
If
2
y
-
1
1
= , then y = 4 or y = –3.
y +2 3
✔
26
3.2.1: Creating and Solving Rational Equations
Guided Practice: Example 3, continued
27
3.2.1: Creating and Solving Rational Equations
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