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3 years ago

(x/x+3)-(8/x+2)=(-13x-6)/(x^2+5x+6)

Mathematics

1 answer:

Fed [463]3 years ago

3 0

Answer:

x = -9

x = 2

Step-by-step explanation:

Given:

$\dfrac{x}{x+3}-\dfrac{8}{x+2}=\dfrac{-13x-6}{x^2+5x+6}$

First, note that

$x^2+5x+6=x^2+2x+3x+6=x(x+2)+3(x+2)=(x+2)(x+3)$

Since $x+2$ and $x+3$ stay in the denominator, then

$x\neq -2\ \text{and}\ x\neq -3$

Now add two fractions which stay in the left part. The common denominator is $(x+2)(x+3),$ so multiply the numerator of the first fraction by $(x+2)$ and the numerator of the second fraction by $(x+3)$ and subtract them in the numerator:

$\dfrac{x}{x+3}-\dfrac{8}{x+2}=\dfrac{-13x-6}{(x+2)(x+3)}\\ \\\dfrac{x(x+2)-8(x+3)}{(x+2)(x+3)}=\dfrac{-13x-6}{(x+2)(x+3)}\\ \\x(x+2)-8(x+3)=-13x-6\\ \\x^2+2x-8x-24=-13x-6\\ \\x^2+2x-8x+13x-24+6=0\\ \\x^2+7x-18=0\\ \\x^2+9x-2x-18=0\\ \\x(x+9)-2(x+9)=0\\ \\(x+9)(x-2)=0\\ \\x+9=0\ \text{or}\ x-2=0\\ \\x=-9\ \text{or}\ x=2$

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4 years ago

Hi, how do we do this question?

Nutka1998 [239]

Answer:

$\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{-2(ln|3x + 1| - 3x)}{9} + C$

General Formulas and Concepts:

<u>Algebra I</u>

Terms/Coefficients
Factoring

<u>Algebra II</u>

Polynomial Long Division

<u>Calculus</u>

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Integration

Integrals
Integration Constant C
Indefinite Integrals

Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Integration Property [Addition/Subtraction]: $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

Logarithmic Integration

U-Substitution

Step-by-step explanation:

*Note:

You could use u-solve instead of rewriting the integrand to integrate this integral.

<u>Step 1: Define</u>

<em>Identify</em>

$\displaystyle \int {\frac{2x}{3x + 1}} \, dx$

<u>Step 2: Integrate Pt. 1</u>

[Integrand] Rewrite [Polynomial Long Division (See Attachment)]: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \int {\bigg( \frac{2}{3} - \frac{2}{3(3x + 1)} \bigg)} \, dx$
[Integral] Rewrite [Integration Property - Addition/Subtraction]: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \int {\frac{2}{3}} \, dx - \int {\frac{2}{3(3x + 1)}} \, dx$
[Integrals] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}\int {} \, dx - \frac{2}{3}\int {\frac{1}{3x + 1}} \, dx$
[1st Integral] Reverse Power Rule: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{3}\int {\frac{1}{3x + 1}} \, dx$

<u>Step 3: Integrate Pt. 2</u>

<em>Identify variables for u-substitution.</em>

Set <em>u</em>: $\displaystyle u = 3x + 1$
[<em>u</em>] Differentiate [Basic Power Rule]: $\displaystyle du = 3 \ dx$

<u>Step 4: Integrate Pt. 3</u>

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}\int {\frac{3}{3x + 1}} \, dx$
[Integral] U-Substitution: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}\int {\frac{1}{u}} \, du$
[Integral] Logarithmic Integration: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}ln|u| + C$
Back-Substitute: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}ln|3x + 1| + C$
Factor: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = -2 \bigg( \frac{1}{9}ln|3x + 1| - \frac{x}{3} \bigg) + C$
Rewrite: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{-2(ln|3x + 1| - 3x)}{9} + C$