Solving Rational Equations | General Mathematics
Summary
TLDRThis educational video tutorial guides viewers through solving rational equations, which are equations containing rational expressions, or quotients of two polynomials. The presenter methodically demonstrates the process using three examples, starting with finding the least common denominator (LCD), multiplying through by the LCD, simplifying, and solving for the variable. Each solution is checked for extraneous roots. The video is interactive, encouraging viewers to ask questions in the comments section, and the presenter ensures clarity by checking solutions in the original equations.
Takeaways
- 📘 The video is a tutorial on solving rational equations, which are equations involving rational expressions.
- 🔢 A rational expression is defined as a quotient of two polynomials.
- ❌ The instructor emphasizes that the denominator should not be equal to zero.
- 📝 The first example involves solving the equation 2/x - 3/2x = 1/5 by finding the least common denominator (LCD).
- 🔄 The process of solving involves multiplying each term by the LCD and simplifying the equation.
- 🔍 After simplifying, the example leads to solving for x, which results in x = 5/2.
- 🔄 The solution x = 5/2 is checked by substituting it back into the original equation to confirm its validity.
- 📐 The second example demonstrates solving a rational equation with binomials in the denominator.
- 🔢 The LCD for the second example is the product of the binomials (x + 1)(x - 3), and the solution is found to be x = 11.
- 🔄 The solution x = 11 is verified by substitution, confirming it is a valid solution to the equation.
- 📘 The third example is a more complex rational equation that simplifies into a quadratic equation, leading to two potential solutions, x = -7 and x = 2.
- 🔄 The solutions are checked, and it's found that x = -7 is valid while x = 2 is an extraneous solution due to making the denominator zero.
Q & A
What is a rational equation according to the video?
-A rational equation is an equation involving rational expressions, which are expressions that can be written as a quotient of two polynomials.
Why is it important to not set the LCD to zero when solving rational equations?
-Setting the LCD (Least Common Denominator) to zero would eliminate the denominators, which is not allowed in mathematics as it can lead to undefined expressions and loss of potential solutions.
What is the first step in solving the example equation 2/(x-3) + 3/2 = 1/5?
-The first step is to find the LCD, which in this case is 10x, by multiplying the denominators x, 2, and 5.
How is the equation 2/(x-3) + 3/2 = 1/5 transformed after multiplying by the LCD?
-After multiplying by the LCD (10x), the equation becomes 20x/x + 30x/2x = 10x/5.
What is the simplified form of the equation after the first example's transformation?
-The simplified form is 20x = 15, which simplifies further to x = 5/2 after combining like terms and solving for x.
How do you check if the solution x = 5/2 is valid for the first example?
-You substitute x = 5/2 back into the original equation and simplify to ensure both sides of the equation are equal.
What is the LCD for the second example equation 3/x + 1 = 2/(x-3)?
-The LCD for the second example is the product of the two binomials (x + 1) and (x - 3).
How is the equation 3/x + 1 = 2/(x-3) simplified after multiplying by the LCD?
-After multiplying by the LCD, the equation simplifies to 3(x - 3) = 2(x + 1), which further simplifies to 3x - 9 = 2x + 2.
What is the solution for x in the second example after simplification?
-After simplifying and solving for x, the solution is x = 11.
How do you verify the solution x = 11 in the second example?
-You substitute x = 11 back into the original equation and check if both sides are equal, confirming it is a valid solution.
What is the LCD for the third example equation x^2 - 4x / (x - 2) = 14 - 9x / (x - 2)?
-The LCD for the third example is x - 2, as it is the common denominator in the equation.
What type of equation results after multiplying the third example by the LCD?
-After multiplying by the LCD, the resulting equation is a quadratic equation: x^2 - 4x = 14 - 9x.
What are the solutions for x in the third example after solving the quadratic equation?
-The solutions for x are x = -7 and x = 2, found by factoring the quadratic equation.
How do you determine if x = 2 is an extraneous solution in the third example?
-By substituting x = 2 into the original equation, you find that the denominator becomes zero, making it an extraneous solution.
Outlines
📘 Introduction to Solving Rational Equations
The video begins with a welcome to the channel and an introduction to rational equations. The instructor explains that a rational equation involves a rational expression, which is a quotient of two polynomials. The importance of not having a zero denominator is emphasized. The first example involves solving the equation 2/(x-3) + 3/2x = 1/5. The process includes finding the least common denominator (LCD), multiplying each term by the LCD, simplifying, and solving for x. The solution x = 5/2 is found and verified by substituting back into the original equation. The instructor also cautions about the possibility of extraneous roots.
🔍 Detailed Walkthrough of Rational Equation Example
The second paragraph delves deeper into solving a rational equation with the example 3/(x-3) + 1 = 2/(x+1) - 3. The instructor demonstrates the process of finding the LCD, which in this case is (x+1)(x-3), and then multiplying each term by the LCD. The equation is then simplified to 3(x+1) - 3(x-3) = 2(x+1) - 6(x-3). After simplifying, the equation is solved for x, yielding x = 11. The solution is checked by substituting x = 11 back into the original equation, confirming that it is indeed a solution.
📐 Advanced Rational Equation Solving Techniques
The final paragraph introduces a more complex rational equation, x^2 - 4x / (x-2) = (14 - 9x) / (x-2). The LCD is identified as (x-2), and the equation is multiplied by this LCD, resulting in a quadratic equation. The instructor then rearranges and simplifies the equation to x^2 - 4x + 5x - 14 = 0, which simplifies further to x^2 + x - 14 = 0. Factoring the quadratic equation yields (x+7)(x-2) = 0, leading to two potential solutions: x = -7 and x = 2. The instructor checks both solutions and confirms that x = -7 is a valid solution while x = 2 is an extraneous solution due to it making the denominator zero. The video concludes with an invitation for questions and a farewell.
Mindmap
Keywords
💡Rational Equation
💡Rational Expression
💡LCD (Least Common Denominator)
💡Quotient
💡Multiplying by LCD
💡Simplifying
💡Like Terms
💡Extraneous Root
💡Substitution
💡Quadratic Equation
Highlights
Introduction to solving rational equations.
Definition of a rational expression as a quotient of two polynomials.
Explanation of the importance of not setting the denominator to zero.
Step-by-step guide to solving the first example equation.
Finding the least common denominator (LCD) for a rational equation.
Multiplying each term by the LCD to eliminate denominators.
Simplification of the equation after multiplying by the LCD.
Combining like terms to simplify the equation further.
Solving for x by isolating the variable.
Checking the solution by substituting back into the original equation.
Verification that the solution does not make any denominator zero.
Introduction to the second example with binomials in the denominator.
Multiplying each term by the LCD to simplify the equation.
Simplification and solving for x in the second example.
Checking the solution for the second example.
Introduction to the third example with a quadratic equation.
Identifying the LCD for the third example and multiplying through.
Transforming the resulting equation into a standard quadratic form.
Solving the quadratic equation for x.
Checking the solutions for the third example and identifying extraneous roots.
Conclusion of the lesson and invitation for questions or clarifications.
Transcripts
hello class welcome to my channel
in this video i will show you how to
solve rational
equations so rational equation is an
equation involving rational expression
so rational expression is an expression
that can be written
as a quotient of two polynomials
so okay so the starting uh quotient
you queued up but not in detail should
not be equal to
zero let's try example number one
solve for x two over x minus three over
two
x equals one over five
so one first step not in detail is we
need to find the
lcd okay so dito
merlin denominators we have x
two x and five so two and five alumni
lcd
okay then step two we need to multiply
each term of the equation
by the lcd so nito
distribute nothing on 10x don't starting
equation okay
so we have 10x times 2 that
is 20x over x
minus 10x times 3
that is 30x over 2x
equals 10 x times one
that is ten x over five
next is we need to simplify
so 20 x over x omega0
is 20 minus 30 x divided by 2x we have
15
then young adding x
equals 10x divided by 5
that is 2x then combining like terms
20 minus 15 we have 5 equals
2x then solving for x we need to divide
both sides of the equation
by two so that we can cancel this one
then x is equals to five
over two okay so
after nothing must muslim cx we need to
check
out or not indeed
extraneous root okay
so to check this is a substitute not in
your 5 over to the unsatin
original equation so we have 2 over
x which is 5 over 2
minus 3 over two
times five over two
equals one over five
okay so we have two over five over two
so d two paramount simplifying
which is two over five
minus determine three
over hands so we have three over
five equals one over five
so two times two that is four over five
minus three over five equals
one over five so four minus three
that is one then capital nothing in the
denominator
then we have one over five
equals one over five so therefore
five over two is the solution of two
over x
minus three over two x equals
one over five so next let's have
example number two three over x
plus one equals two over x
minus three so our first step is we need
to find
that lcd so in our case we have two
binomials
x plus one and x minus three
so normally pagoda you adding
denominators
i'm adding lcd i x plus one and
x minus three and then okay
so next we need to multiply each term
of our equation by the lcd
so 2 over x minus 3. so multiply that
into
by x plus one
and x minus three
so first multiplication as you can see
x plus one so um
3 and x minus 3
okay so we have 3 times
x minus 3 equals
it's the second multiplication not in
this right side
america hansel namant is the x minus 3
then my a1 is 2 times
x plus 1 next after that after now
multiplication
we need to simplify our equation so by
distribution
we have three times x that is three x
then three times negative three we have
negative nine
then for two x number one f four two
multiply nothing and we have
two x plus two
okay the next letting my x
really but not in the left side then lat
no
constants are right side so we have 3x
so in 2x pagoda but we have negative 2x
then say negative 9 transpose nothing
done and
that is positive 9. combining like terms
we have x
equals 11.
okay then after nothing what is cx
we need to check kunta lagabang solution
c11
then rational
equation so by substitution we have 3
over 11 plus 1
equals 2 over x which is eleven
minus three so we have
three over twelve
equals two over eight
then simplifying that n so three over 12
that is equivalent
to 1 4 then 2 over 8 is also equivalent
to
1 4. so therefore
11 is the solution
duns equation okay
next let's have example number three
number three x squared minus four x all
over
x minus 2 equals 14 minus 9 x
all over x minus 2. so again i'm adding
first
first step is we need to identify the
lcd
so as you notice on denominators nothing
ipad
so meaning our lcd is x
minus two okay then we need to multiply
each term of our equation by
the lcd so dito
see 14 minus nine x over x minus two
is multiply not in guys by
x minus two okay
lcd so we have x squared minus 4x
equals 14 minus 9x
then i'm acting resulting equation is a
quadratic equation
x squared this is a left side so paramus
also we need to transpose
everything the unselect
left side okay so we have x squared
minus four x
minus four x so alpha but nothing though
we have negative 14
then your negative nine x figure but we
have positive
nine x equals zero
then combining like terms we have x
squared
then negative four x plus nine x that is
five x
minus fourteen equals zero
so power of magnifying sulfide i think
quadratic equation
is if a factor or nothing since factor
volume
quadratic equation so we have x
and x so negative 14
factorial negative 14 napa peanut at nut
and we have positive five so
in this case and i and i is equal
numbers i
positive seven and negative two
dial seven times negative two we have
negative fourteen
then seven minus two that is
positive five so tamayo atom factor
after nothing macho in factors equate
not
in both factors by zero
so we have x plus seven equals zero
and x minus two equals zero
okay then solving for x we have two
values of x
x equals negative seven and x
equals positive two okay
so parenthesis that we need to check
your negative seven bar and two i put a
solution
equation so checking dial
so checking unit is negative seven
so negative seven squared
minus 4 times negative 7
over negative 7
minus 2 equals
14 minus nine times negative seven
all over negative seven minus two
so simplifying nothing in the numerator
negative seven squared that
is 49 then negative 4 times negative 2
negative 7 is positive 28
all over negative 9
equals 14
then negative 9 times negative 7 that is
positive 63 over
negative 9 okay
then 49 plus 28 that is
77 over negative 9
equals 14 plus 63 is
77 over negative 9.
so therefore c negative 7 is a
solution non-acting equation
okay next trinomial not in c
positive two so we have
two squared minus 4
times 2 over 2
minus 2 equals
14 minus 9 times 2
over two minus two so these
guys as you notice in denominator
nothing
not two minus two is making zero so
meaning
undefined so therefore
hindi kung
is a extraneous
solution
okay so ditonic data posts and
in lesson guys if you have questions
or clarification by
section below so thank you guys for
watching
see you in our next video bye
you
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