Solve the following.−3nn+2+2n4n+8=68n+16-\frac{3n}{n+2}+\frac{2n}{4n+8}=\frac{6}{8n+16}

n=−310n=-\frac{3}{10}n=−103

Solve the following. − 3 n n + 2 + 2 n 4 n + 8 = 6 8 n + 16 -\frac{3n}{n+2}+\frac{2n}{4n+8}=\frac{6}{8n+16}

this problem, we are asked to solve the following equation for n. And for part a of this problem, what we need to do is figure out how to find a solution for what this n is. Now the strategy for solving problems where you have these rational expressions that make up the equation is to first find the least common denominator. This is going to help you a lot. Now the way I can find the least common denominator is by factoring the denominator in each of these terms that I see. Now n plus two is already factored. So negative three over n plus two is going to stay the same. Now for this two n over four n plus eight, I can factor a four out of this denominator. So that's going to be four. And then dividing each of these terms by four, we're going to get n plus two. Now all of this is going to equal six over eight n plus 16. And I can pull an eight out of each of these terms, giving me eight times n plus 16 divided by eight, which is two. So notice that this is the expression that we end up with. Now it turns out I can make this expression even more simple, because four is the same thing as two times two, and eight is the same thing as two times two times two. Two times two is four, times another two would give us eight. And writing it like this allows me to simplify these by putting exponents on them. So two times two is two squared, and two times two times two is two cubed. Now we have everything as factored as it can get. Now I noticed that we have an n plus two in common with each of these denominators. So listing the unique prime factors, we're going to have n plus two. Now I can also see we have a two squared, but we also have a two cubed. When you run into situations where you have the same base, you wanna look for the highest exponent, which in this case is two cubed. So now we have found the least common denominator. Now from here, what I'm gonna do is take my least common denominator, and I'm going to multiply it on both sides of our equation. So it's gonna be two cubed times n plus two that I'm multiplying on both sides. Now what I need to do from here is distribute this n cubed times n plus two into the parentheses for each of these terms on the left and for this term on the right. Now as I do this, I'm going to cancel the common terms that I see. So if I multiply two cubed times n plus two, I take that whole thing and I multiply it by three n over n plus two, notice the n plus two will cancel completely. So all we're going to end up with is two cubed times negative three n. Now if we go ahead and distribute this into the next term here, we'll notice once again the n plus two cancels completely. So all we're gonna end up with is two cubed, two cubed times two n over two squared, so times two n over two squared. And this is all going to be equal to and then in this case, multiplying this into the parentheses here, notice that the two cubed times n plus two entirely cancels. Those denominators completely go away, so all we end up with is six. So this is what we get. Now there's one more thing that I can do because notice we have a two cubed and we have a two squared. What I can do is I can cancel the two twos with two of these twos, leaving me with just a two. So, ultimately, we're going to get two cubed times negative three n plus two times two n equals six. And notice how this is a lot more simple than where we started with since we've cleared all the denominators. Now I'm not at my final solution yet because I still need to solve for this n. But to solve for it, well, two cubed, that's going to give me eight. So we'll have eight times negative three n, and we're going to have plus, and then we have two times two, which is four. N is equal to six. And then eight times negative three, that's gonna give me negative 24 n plus four n equals six. Negative 24 plus four is going to give me negative 20, and that's gonna be negative 20 n equals six. And I divide that negative 20 on both sides of the equation. The negative twenties cancel there, giving me that n is equal to negative six over 20. Now what I can do is break this down further because six is the same thing as two times three, and twenty is the same thing as two times 10. So if I cancel those twos right about there, that's going to give me that n is equal to negative three over 10. So this right here is our solution for n, and that's the answer to this problem. So this is how we can solve for a variable in a rational equation. The main step that you need to know is first finding the LCD and then multiplying it by both sides of the equation. That will significantly simplify the equation that you are dealing with and give you something straightforward that you can solve. So this is how you can find the solution. Hope you found this video helpful, and let's move on and get some more practice. See you in the next one.

Solve the following.−3nn+2+2n4n+8=68n+16-\frac{3n}{n+2}+\frac{2n}{4n+8}=\frac{6}{8n+16}

n=−310n=-\frac{3}{10}n=−103

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1. Review of Real Numbers2h 43m

Simplifying Fractions
30m

Evaluating Exponents
29m

Evaluating Expressions
15m

Translating Phrases to Expressions
16m

Translating Sentences to Equations
9m

The Distributive Property
17m

Simplifying Expressions
44m

2. Linear Equations and Inequalities5h 35m

Solving Linear Equations
2h 23m

Linear Inequalities in One Variable
46m

Set Operations and Compound Inequalities
1h 7m

Absolute Value Equations
38m

Absolute Value Inequalities
39m

3. Solving Word Problems2h 46m

Introduction to Problem Solving
37m

Formulas
25m

Percent Problem Solving
59m

Mixture Problem Solving
43m

4. Graphs and Functions5h 12m

The Rectangular Coordinate System
44m

Graph Linear Equations in Two Variables
24m

Graph Linear Equations Using Intercepts
23m

Slope of a Line
44m

Slope-Intercept Form
38m

Point Slope Form
22m

Linear Inequalities in Two Variables
28m

Introduction to Relations and Functions
53m

Function Notation
15m

Composition of Functions
17m

5. Systems of Linear Equations1h 53m

Solving Systems of Linear Equations by Graphing
29m

Solving Systems of Linear Equations by Substitution
15m

Solving Systems of Linear Equations by Elimination
45m

Systems of Linear Inequalities
23m

6. Exponents, Polynomials, and Polynomial Functions3h 17m

The Product Rule
9m

The Quotient Rule
10m

Intro to the Power Rules
17m

The Power of a Quotient Rule
13m

Negative Exponents
24m

Simplifying Exponential Expressions Using All Exponent Rules
6m

Intro to Polynomials
21m

Adding and Subtracting Polynomials
20m

Multiplying Polynomials
38m

Special Products
34m

7. Factoring2h 49m

Factoring by Greatest Common Factor & Grouping
37m

Factoring Trinomials of the Form x² + bx + c
11m

Factoring Trinomials of the Form ax² + bx + c
37m

Factoring Special Products
41m

Quadratic Equations & Applications
41m

8. Rational Expressions and Functions3h 44m

Simplifying Rational Expressions
42m

Multiplying and Dividing Rational Expressions
25m

Adding and Subtracting Rational Expressions with Common Denominators
19m

Least Common Denominators
32m

Adding and Subtracting Rational Expressions with Different Denominators
32m

Rational Equations
44m

Direct & Inverse Variation
27m

9. Roots, Radicals, and Complex Numbers3h 56m

Introduction to Radical Expressions
46m

Simplifying Radical Expressions
47m

Adding and Subtracting Radicals
19m

Multiplying, Dividing, and Rationalizing Radicals
53m

Introduction to Complex Numbers
18m

Multiplying and Dividing Complex Numbers
51m

10. Quadratic Equations and Functions3h 1m

The Square Root Property
25m

Completing the Square
26m

The Quadratic Formula
40m

Graphing Quadratic Equations
1h 28m

11. Inverse, Exponential, & Logarithmic Functions2h 31m

Introduction to Inverse Functions
25m

Exponential Functions
35m

Introduction to Logarithmic Functions
33m

Properties of Logarithms
55m

12. Conic Sections & Systems of Nonlinear Equations58m

Graphing Circles
32m

Ellipses
15m

Parabolas
10m

13. Sequences, Series, and the Binomial Theorem1h 51m

Introduction to Sequences
40m

Arithmetic Sequences
27m

Geometric Sequences
29m

Factorials
13m

8. Rational Expressions and Functions

Rational Equations

Intermediate Algebra

8. Rational Expressions and Functions

Rational Equations

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Multiple Choice

Solve the following.
$- \frac{3 n}{n + 2} + \frac{2 n}{4 n + 8} = \frac{6}{8 n + 16}$

$n=-$\frac$12$

$n=$\frac{3}{10}$$

$n=-$\frac{3}{10}$$

$n=1$

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Verified step by step guidance

Start by identifying the given equation: $-\frac{3n}{n+2} + \frac{2n}{4n+8} = \frac{6}{8n+16}$.

Notice that the denominators \$4n+8$ and $8n+16$ can be factored to simplify the equation. Factor them as $4(n+2)$ and $8(n+2)$ respectively.

Rewrite the equation using the factored denominators: $-\frac{3n}{n+2} + \frac{2n}{4(n+2)} = \frac{6}{8(n+2)}$.

To eliminate the denominators, multiply every term in the equation by the least common denominator (LCD), which is \$8(n+2)$, to clear the fractions.

After clearing denominators, simplify the resulting equation and solve the linear equation for $n$ by isolating the variable on one side.

5:21m

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Struggling with Intermediate Algebra?

Solve the following.−3nn+2+2n4n+8=68n+16-\(\frac{3n}{n+2}\)+\(\frac{2n}{4n+8}\)=\(\frac{6}{8n+16}\)

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Solve the following.
$- \frac{3 n}{n + 2} + \frac{2 n}{4 n + 8} = \frac{6}{8 n + 16}$