Solving Rational Equations
Summary
TLDRThis lesson focuses on solving rational equations by eliminating fractions and finding the least common multiple. The instructor demonstrates step-by-step solutions for various examples, including simplifying equations, factoring, and applying cross-multiplication. Techniques for finding x values are explored, with a final example involving factoring a quadratic expression and solving for x, yielding multiple solutions.
Takeaways
- 🔍 The lesson focuses on solving rational equations by eliminating fractions to simplify the problem.
- 📚 The first example demonstrates finding the least common multiple (LCM) of 8, 5, and 10, which is 40, to clear fractions from the equation.
- 🧩 After clearing fractions, the solution involves basic arithmetic operations to solve for x, resulting in x = 1/4 for the initial example.
- 📉 In the second problem, multiplying both sides by x eliminates the denominators, leading to a quadratic equation which factors to find x = 4 and x = 2.
- ✅ The third example uses cross-multiplication to transform the equation into a linear one, solving for x = 5.
- 🤔 The fourth problem involves taking the square root of both sides after cross-multiplication, yielding two potential solutions, x = 6 and x = -6.
- 🔢 For the fifth example, cross-multiplication and simplification lead to a solution of x = 7 after combining like terms.
- 📈 The sixth example uses the LCM of 2 and 3, which is 6, to eliminate fractions and solve for x = 1.
- 📉 The seventh problem involves finding a common denominator and simplifying to form a quadratic equation, which factors to x = 2 and x = -1.
- 🔗 The last example involves factoring a difference of squares and clearing fractions to form a quadratic equation, solving for x = 13 and x = -3.
Q & A
What is the first step to solve a rational equation involving fractions?
-The first step is to find the least common multiple (LCM) of the denominators and then multiply every fraction by that LCM to eliminate the fractions.
How do you find the least common multiple (LCM) of 8, 5, and 10 from the transcript?
-You list the multiples of each number and identify the smallest number that appears in all lists. In this case, the multiples of 5 are 5, 10, 15, etc., multiples of 8 are 8, 16, 24, 32, 40, and multiples of 10 include 10, 20, 30, 40. The LCM is 40.
What is the value of x in the equation 5/8 - 3/5 = x/10?
-After clearing the fractions by multiplying by the LCM (40), you get 25 - 24 = 4x/10, which simplifies to 1 = 4x/10. Solving for x gives x = 1/4.
How do you handle the equation x + 8/x = 6?
-You multiply both sides by x to eliminate the fraction, which gives x^2 + 8 = 6x. Then, you rearrange the equation to x^2 - 6x + 8 = 0 and factor it to (x - 4)(x - 2) = 0, giving x = 4 and x = 2.
What is the process for solving the equation (x + 3)/(x - 3) = 12/3?
-You cross-multiply to get 12(x - 3) = 3(x + 3). Simplifying gives 12x - 36 = 3x + 9. Then, you combine like terms and solve for x, which results in x = 5.
How do you solve the equation 9/x = x/4?
-You cross-multiply to get 9 * 4 = x^2, which simplifies to 36 = x^2. Taking the square root of both sides gives x = ±6.
In the equation 4/(x - 3) = 9/(x + 2), what is the step after cross-multiplying?
-After cross-multiplying, you get 4(x + 2) = 9(x - 3). Expanding and simplifying leads to 4x + 8 = 9x - 27, and then you solve for x, which results in x = 7.
What is the least common multiple (LCM) of 2 and 3, and how is it used in the equation (x + 2)/3 = (x + 9)/2?
-The LCM of 2 and 3 is 6. Multiplying both sides of the equation by 6 eliminates the fractions, leading to 2x + 4 = 3x + 27/2, which simplifies to x = 1.
How do you solve the equation 4/x + 8/(x + 2) = 4?
-You multiply both sides by the common denominator x(x + 2), which gives 4(x + 2) + 8x = 4x^2 + 8x. Simplifying and solving the quadratic equation gives x = 2 and x = -1.
In the final example of the transcript, how do you simplify the equation (x + 5)/(x - 5) - 5/(x + 5) = 14/(x^2 - 25)?
-You factor x^2 - 25 as (x + 5)(x - 5) and multiply both sides by this expression to clear the fractions. Simplifying leads to x^2 - 10x - 39 = 0, which factors to (x - 13)(x + 3) = 0, giving x = 13 and x = -3.
Outlines
📚 Solving Rational Equations
This paragraph introduces the process of solving rational equations with examples. The first example involves simplifying a complex fraction and finding the least common multiple (LCM) of 8, 5, and 10, which is 40. The fractions are then multiplied by 40 to eliminate them, leading to the solution x = 1/4. The second example shows how to solve an equation with a variable in the denominator by multiplying both sides by x to eliminate the fraction and then factoring to find x = 4 and x = 2. The third example uses cross-multiplication to solve an equation, resulting in x = 5. The fourth example involves taking the square root of both sides to find two possible solutions, x = ±6. The final example in this paragraph demonstrates cross-multiplication and solving for x = 7 after simplifying the equation.
🔍 Advanced Rational Equation Techniques
This paragraph delves into more complex rational equation problems. The first example involves finding the LCM of 2 and 3, which is 6, and multiplying through to eliminate fractions, leading to the solution x = 1. The second example requires finding a common denominator of x(x + 2) and simplifying to find two solutions, x = 2 and x = -1. The third example involves factoring a difference of squares and multiplying through to eliminate fractions, resulting in two potential solutions, x = 13 and x = -3. The paragraph demonstrates a step-by-step approach to solving rational equations, emphasizing the importance of finding common denominators, factoring, and simplifying equations to isolate the variable.
📘 Final Rational Equation Challenges
The final paragraph presents a challenging rational equation that requires factoring and careful manipulation to solve. The equation involves a difference of squares, which is factored into (x + 5)(x - 5). The fractions are eliminated by multiplying through by the common denominator, and the equation is simplified to x^2 - 5x - 39 = 0. Factoring this quadratic equation yields (x - 13)(x + 3) = 0, leading to the solutions x = 13 and x = -3. This example showcases the application of algebraic techniques to solve more intricate rational equations.
Mindmap
Keywords
💡Rational Equations
💡Least Common Multiple (LCM)
💡Cross Multiply
💡Factoring
💡Distributing
💡Combining Like Terms
💡Solving for x
💡Variable Cancellation
💡Equation Transformation
💡Roots of an Equation
💡Simplifying Fractions
Highlights
Introduction to solving rational equations.
Method to clear fractions by finding the least common multiple (LCM).
Example of solving 5/8 - 3/5 = x/10 by using LCM of 40.
Technique of multiplying each fraction by the LCM to eliminate denominators.
Solving the equation by simplifying and finding x = 1/4.
Approach to solving the equation x + 8/x = 6 by multiplying both sides by x.
Factoring and solving for x in the equation x^2 - 6x + 8 = 0.
Finding the solutions x = 4 and x = 2 for the equation.
Cross-multiplication method for equations involving two fractions.
Solving x + 3/(x - 3) = 12/3 by cross-multiplication and simplification.
Determining x = 5 for the given equation.
Cross-multiplication for equations with fractions separated by an equal sign.
Solving 9/x = x/4 by taking the square root of both sides.
Finding two solutions, x = 6 and x = -6, for the equation.
Solving complex rational equations by cross-multiplication and factoring.
Example of solving 4/(x - 3) = 9/(x + 2) by cross-multiplication and simplification.
Determining x = 7 for the equation after simplification.
Using LCM to eliminate fractions in complex rational equations.
Solving x + 2/3 + 4/(x + 2) = x + 9/2 by multiplying by the LCM.
Finding x = 1 by simplifying and solving the resulting equation.
Solving rational equations with a common denominator by factoring and simplification.
Example of solving 4/x + 8/(x + 2) = 4 by finding a common denominator and simplifying.
Determining the solutions x = 2 and x = -1 for the equation.
Advanced technique of factoring and solving rational equations with quadratic denominators.
Solving i(x)/(x + 5) - 5/(x - 5) = 14/(x^2 - 25) by factoring and clearing fractions.
Finding the solutions x = 13 and x = -3 for the equation.
Transcripts
now in this lesson we're going to focus
on solving
rational equations
so let's start with our first example
5 over 8
minus 3 over 5
and let's set that equal to
x over 10.
what do we need to do
in order to find the value of x
what would you do
the best thing we can do is clear away
all fractions
we have an eight a five and a ten
what is the least common multiple of
eight five and ten
well we can make a list
multiples of five are five ten fifteen
and so forth
multiples of 8
are 8 16 24
32 and 40.
multiples of 10
also include 40. so 40 is the least
common multiple
let's multiply every fraction
by 40.
so what's 5
over 8 times 40
you can do 5 times 40 which is 200 and
then divide 200 by 8 or
you can do 40 divided by 8 which is 5
times the 5 on top
and that's going to be 25
now what about three fifths of 40
40 divided by five is eight eight times
three is twenty four
and the last one
forty divided by ten is four times x
that's four x
twenty five minus twenty four is one
and so x
is equal to one fourth so that's the
answer
here's the next problem x plus eight
over x is equal to 6.
feel free to pause the video and work on
this example
so what we're going to do in this
problem we're going to multiply both
sides by x
so x times x is x squared
and then 8 over x times x the x
variables will cancel it's just going to
be 8 and 6 times x is 6x
now let's move the 6x from the right
side
to the left side on the right side is
positive 6x but on the left side it's
going to be negative
now we can factor it
what two numbers multiply to eight but
add up to negative six
this is negative four and two
so it's x minus four times x minus two
and so we can clearly see that x is
equal to positive four
and positive two
and so that's it
here's the next one
x plus 3
divided by
x minus 3 let's say that's equal to 12
over 3.
whenever you have two fractions
separated by an equal sign what you want
to do is you want to cross multiply
so 12
times x minus 3 is 12x
minus 36
and 3 times x plus 3
that's 3x
plus 9.
now let's subtract both sides by 3x
and let's add 36 to both
sides 12x minus 3x is 9x
9 plus 36 is 45
and 45
divided by 9 is 5.
so x
is equal to 5.
try this one
nine divided by x is x over four
so once again we have two fractions
separated by an equal sign
let's cross multiply
x times x is x squared
and 9 times 4 is 36
so all we need to do is take the square
root of both sides
the square root of 36
is plus or minus 6. so there's two
answers positive six and negative six
now what about this one four
divided by x minus three
and let's say that's equal to nine
over x plus two
so for this problem as well cross
multiply
so four times x plus two that's
four x plus eight
and then nine times x minus three that's
nine x minus twenty seven
so let's subtract both sides by 4x
and let's add 28 i mean not 28 but
rather 27 to both sides
8 plus 27 that's 35
9 minus 4 is 5.
so all we need to do now is divide by 5.
35 divided by 5 is 7.
so x
is equal to 7.
now let's say that we have x
plus two
divided by three
plus
four
and let's say that's equal to x plus
nine
divided by two
find the value of x
the least common multiple of two and
three is six
so let's multiply everything by six to
get rid of the fractions
six divided by three is two now let's
multiply two by x plus two
and that's going to be two x plus four
now four times six
is twenty four
and six divided by two is three and
three times
x plus nine that's going to be 3x
plus 27
so now let's combine 4 and 24 which is
28
now let's subtract both sides by 2x
and also by 27
28 minus 27 is one
3x minus 2x is x
so therefore x is equal to one
here's the next problem
four divided by x
plus
eight divided by x plus two
let's set that equal to four find the
value of x
so in this case the common denominator
is x
times x plus two
if we multiply 4 over x by
x x plus 2
the x variables will cancel
and so that's going to leave behind 4
times x plus 2
which if we distribute 4 is going to be
4x
plus 8.
now x plus 2 will cancel leaving behind
x times 8 or simply 8x
and then here we'll have 4 times x times
x plus 2
which is 4x
x plus 2.
now we can add 4x and 8x
that's going to be 12x
and now let's distribute the 4x 4x times
x is 4x squared 4x times 2 is 8x
everything on the left side let's move
it to the right side
so instead of having positive 12x on the
left side it's going to be negative 12x
on the right side
and 8 is going to change to negative 8.
now let's combine like terms
8x minus 12x is negative 4x
so now what we need to do is factor
we can take out a four
and this will leave us with x squared
minus x instead of plus x
minus two now two numbers that multiply
to negative two but at negative one
is going to be a negative two and
positive one
so it's gonna be x minus two times x
plus one
so if we set each factor equal to zero
we can see that x
is equal to two and x is equal to
negative one
and so that's going to be the answer
to the problem
now let's try the last example
5
or rather
i'll take that back not 5 x
over x plus 5
minus
5
over x minus 5.
let's say that's equal to 14
over x squared minus 25.
go ahead and find the value of x now
what we should do first is factor x
squared minus 25
and that's going to be x plus five
times x minus five
now we need to clear away all fractions
so let's multiply the top well let's
multiply both sides the left side and
the right side
by x plus five
times x minus five the common
denominator
so if we take this fraction and multiply
it by these two we can see that x plus
five will cancel
leaving behind x
times x minus five
now if we take the second fraction
multiply by those two
the x minus five term will cancel
leaving behind five times x plus five
and then x plus 5 will cancel and x
minus 5 will cancel
leaving 14.
so now let's distribute x times x minus
5.
that's x squared minus 5x
and if we distribute the negative 5
it's going to be negative 5x
minus 25.
now let's subtract both sides by 14 and
let's combine like terms
negative 5x and negative 5x that's
negative 10x
negative 25 minus 14
that's negative 39.
what two numbers
multiplied to negative 39
but add to negative 10.
i'm thinking of negative 13 and 3
so this is going to be x
minus 13
x plus 3.
so therefore x is equal to 13
and negative 3.
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