Solve for x X- 3x+2/x+1 = 1/x+1, №18010627, 17.06.2021 17:38

Mathematics

Step-by-step answer

P Answered by Master

Answer is x= 1/2. I have showed the steps in the picture
Solve for x
X- 3x+2/x+1 = 1/x+1

Mathematics

1. the width w of a rectangular swimming pool is x+4. the area a of the pool is 2x^3-29+12. what is an expression for the length of the pool? a. 2x^2+8x+3 b. 2x^2-8x-3 c. 2x^2-8x+3 d. 2x^2+8-3 2. simplify x/6x-x^2 a. 1/6-x; where x = 0,6 b. 1/6-x; where x=6 c. 1/6; where x=0 d. 1/6 3. simplify -12x4/x4+8x^5 a. -12/1+8x; where x= -1/8 b. -12/1+8x; where x= -1/8,0 4. simplify x+5/ x^2+6x+5 a. 1/x+1; where x= -1 b. 1/x+1; where x=-1, -5 5. simplify x^2-3x-18/x+3 a. x-3 b. x-6; where x= -3 c. x-6; where x= 6 d. 1/x+3; where x= -3 6. simplify 2/3a .

Step-by-step answer

P Answered by PhD

20

Question 1

To find the width of the rectangle, we divide the area by the length
$2x^{3}-29x+12$ ÷ x+4

We use the method of long division to get the answer. The method is shown in the first diagram below

$2x^{2}-8x+3$

Question 2:
$\frac{x}{6x-x^{2} } = \frac{x}{x(6-x)} = \frac{1}{6-x}$

Question 3:
$\frac{-12 x^{4} }{x^{4}+8 x^{5} }= \frac{-12 x^{4} }{ x^{4}(1+8x)}= \frac{-12}{1+8x}$

Question 4:
$\frac{x+5}{x^{2}+6x+5}= \frac{x+5}{(x+1)(x+5)}= \frac{1}{x+1}$

Question 5:
$\frac{x^{2}-3x-18} {x+3}= \frac{(x-6)(x+3)}{x+3}= \frac{x-6}{1}=x-6$

Question 6:
$\frac{2}{3a}$ × $\frac{2}{a^{2}}$ = $\frac{4}{3a^{3} }$ where $a \neq 0$

Question 7: (Question is not written well)
$\frac{x-5}{4x+8}$ × $(12x^{2}+32x+8)$
$\frac{12 x^{3}-28 x^{2} -152x-40 }{4x+8}$
By performing long division we get an answer $3 x^{2} -x-36$ with remainder of 248

Question 8:
$( \frac{x^{2}-16} {x-1})$ ÷ (x+4)

$( \frac{ x^{2}-16 }{x-1})$ × $\frac{1}{x+4}$
$\frac{(x+4)(x-1)}{x-1}$ × $\frac{1}{x+4}$
Cancelling out x+4

we obtain $\frac{x+1}{x-1}$

Question 9:
$\frac{x^{2}+2x+1} {x-2}$ ÷ $\frac{x^{2-1} }{x^{2}-4 }$
$\frac{ x^{2}+2x+1 }{x-2}$ × $\frac{x^{2}-4 }{x^{2}-1}$
Factorise all the quadratic expression gives
$\frac{(x+1)(x+1)}{x-2}$ × $\frac{(x-2)(x+2)}{(x+1)(x-1)}$
Cancelling out (x+1)

and

gives a simplest form
$\frac{(x+1)(x+2)}{x-1}$

Question 10:

$\frac{24 w^{10}+8w^{12} }{4 x^{4} }= \frac{24w^{10} }{4 x^{4} } + \frac{8 w^{12} }{4 x^{4} }$
Cancelling out the constants of each fraction
$\frac{6w^{10} }{x^{4} }+ \frac{2w^{12} }{x^{4}}= \frac{6w^{10}+2w^{12} }{ x^{4}}$

Question 11:

$\frac{-6m^{9}-6m^{8}-16m^{6} }{2m^{3} } = \frac{-2m^{6}(3m^{3}-3m^{2}-8)}{2m^{3} }$
Cancelling $2m^{3}$ gives us the simplified form
$-m^{3}(3m^{3}-3m^{2}-8) = -3m^{6}+3m^{5}+8m^{3}$

Question 12:

$\frac{-4x}{x+7} - \frac{8}{x-7} = \frac{-4x(x-7)-8(x+7)}{(x+7)(x-7)}$
$\frac{-4 x^{2} +28x-8x-56}{(x+7)(X-7)}= \frac{-4 x^{2} +20x-56}{(x+7)(x-7)}$
Factorising the numerator expression
$\frac{(-4x+28)(x-2)}{(x+7)(x-7)} = \frac{-4(x-7)(x-2)}{(x+7)(x-7)}$
Cancelling out x-7

gives the simplified form
$\frac{-4x+8}{x-7}$

Question 13:

$\frac{3}{x-3} - \frac{5}{x-2}= \frac{x3(x-2)-5(x-2)}{y(x-3)(x-2)}$
$\frac{3x-6-5x+15}{(x-3)(x-2)}= \frac{-2x+9}{(x-3)(x-2)}$

Question 14:

$\frac{9}{x-1}- \frac{5}{x+4}= \frac{9(x+4)-5(x-1)}{(x-1)(x+4)}$ $\frac{9x+36-5x+5}{(x-1)(x+4)}= \frac{4x+41}{(x-1)(x+4)}$

Question 15:

$\frac{-3}{x+2}- \frac{(-5)}{x+3}= \frac{-3(x+3)-(-5)(x+2)}{(x+2)(x+3)}$
$\frac{-3x-9+5x+10}{(x+2)(x+3)}= \frac{2x+1}{(x+2)(x+3)}$

Question 16:

$\frac{4}{x}+ \frac{5}{x}=-3$
$\frac{9}{x}=-3$
x=-3

Question 17:

$\frac{1}{3x-6}- \frac{5}{x-2}=12$
$\frac{(x-2)-5(3x-6)}{(3x-6)(x-2)} = \frac{x-2-15x+30}{(3x-6)(x-2)}= \frac{-14x+28}{(3x-6)(x-2)}$

Question 18

1. the width w of a rectangular swimming pool is x+4. the area a of the pool is 2x^3-29+12. what is

Mathematics

1. the width w of a rectangular swimming pool is x+4. the area a of the pool is 2x^3-29+12. what is an expression for the length of the pool? a. 2x^2+8x+3 b. 2x^2-8x-3 c. 2x^2-8x+3 d. 2x^2+8-3 2. simplify x/6x-x^2 a. 1/6-x; where x = 0,6 b. 1/6-x; where x=6 c. 1/6; where x=0 d. 1/6 3. simplify -12x4/x4+8x^5 a. -12/1+8x; where x= -1/8 b. -12/1+8x; where x= -1/8,0 4. simplify x+5/ x^2+6x+5 a. 1/x+1; where x= -1 b. 1/x+1; where x=-1, -5 5. simplify x^2-3x-18/x+3 a. x-3 b. x-6; where x= -3 c. x-6; where x= 6 d. 1/x+3; where x= -3 6. simplify 2/3a .

Step-by-step answer

P Answered by PhD

20

Question 1

To find the width of the rectangle, we divide the area by the length
$2x^{3}-29x+12$ ÷ x+4

We use the method of long division to get the answer. The method is shown in the first diagram below

$2x^{2}-8x+3$

Question 2:
$\frac{x}{6x-x^{2} } = \frac{x}{x(6-x)} = \frac{1}{6-x}$

Question 3:
$\frac{-12 x^{4} }{x^{4}+8 x^{5} }= \frac{-12 x^{4} }{ x^{4}(1+8x)}= \frac{-12}{1+8x}$

Question 4:
$\frac{x+5}{x^{2}+6x+5}= \frac{x+5}{(x+1)(x+5)}= \frac{1}{x+1}$

Question 5:
$\frac{x^{2}-3x-18} {x+3}= \frac{(x-6)(x+3)}{x+3}= \frac{x-6}{1}=x-6$

Question 6:
$\frac{2}{3a}$ × $\frac{2}{a^{2}}$ = $\frac{4}{3a^{3} }$ where $a \neq 0$

Question 7: (Question is not written well)
$\frac{x-5}{4x+8}$ × $(12x^{2}+32x+8)$
$\frac{12 x^{3}-28 x^{2} -152x-40 }{4x+8}$
By performing long division we get an answer $3 x^{2} -x-36$ with remainder of 248

Question 8:
$( \frac{x^{2}-16} {x-1})$ ÷ (x+4)

$( \frac{ x^{2}-16 }{x-1})$ × $\frac{1}{x+4}$
$\frac{(x+4)(x-1)}{x-1}$ × $\frac{1}{x+4}$
Cancelling out x+4

we obtain $\frac{x+1}{x-1}$

Question 9:
$\frac{x^{2}+2x+1} {x-2}$ ÷ $\frac{x^{2-1} }{x^{2}-4 }$
$\frac{ x^{2}+2x+1 }{x-2}$ × $\frac{x^{2}-4 }{x^{2}-1}$
Factorise all the quadratic expression gives
$\frac{(x+1)(x+1)}{x-2}$ × $\frac{(x-2)(x+2)}{(x+1)(x-1)}$
Cancelling out (x+1)

and

gives a simplest form
$\frac{(x+1)(x+2)}{x-1}$

Question 10:

$\frac{24 w^{10}+8w^{12} }{4 x^{4} }= \frac{24w^{10} }{4 x^{4} } + \frac{8 w^{12} }{4 x^{4} }$
Cancelling out the constants of each fraction
$\frac{6w^{10} }{x^{4} }+ \frac{2w^{12} }{x^{4}}= \frac{6w^{10}+2w^{12} }{ x^{4}}$

Question 11:

$\frac{-6m^{9}-6m^{8}-16m^{6} }{2m^{3} } = \frac{-2m^{6}(3m^{3}-3m^{2}-8)}{2m^{3} }$
Cancelling $2m^{3}$ gives us the simplified form
$-m^{3}(3m^{3}-3m^{2}-8) = -3m^{6}+3m^{5}+8m^{3}$

Question 12:

$\frac{-4x}{x+7} - \frac{8}{x-7} = \frac{-4x(x-7)-8(x+7)}{(x+7)(x-7)}$
$\frac{-4 x^{2} +28x-8x-56}{(x+7)(X-7)}= \frac{-4 x^{2} +20x-56}{(x+7)(x-7)}$
Factorising the numerator expression
$\frac{(-4x+28)(x-2)}{(x+7)(x-7)} = \frac{-4(x-7)(x-2)}{(x+7)(x-7)}$
Cancelling out x-7

gives the simplified form
$\frac{-4x+8}{x-7}$

Question 13:

$\frac{3}{x-3} - \frac{5}{x-2}= \frac{x3(x-2)-5(x-2)}{y(x-3)(x-2)}$
$\frac{3x-6-5x+15}{(x-3)(x-2)}= \frac{-2x+9}{(x-3)(x-2)}$

Question 14:

$\frac{9}{x-1}- \frac{5}{x+4}= \frac{9(x+4)-5(x-1)}{(x-1)(x+4)}$ $\frac{9x+36-5x+5}{(x-1)(x+4)}= \frac{4x+41}{(x-1)(x+4)}$

Question 15:

$\frac{-3}{x+2}- \frac{(-5)}{x+3}= \frac{-3(x+3)-(-5)(x+2)}{(x+2)(x+3)}$
$\frac{-3x-9+5x+10}{(x+2)(x+3)}= \frac{2x+1}{(x+2)(x+3)}$

Question 16:

$\frac{4}{x}+ \frac{5}{x}=-3$
$\frac{9}{x}=-3$
x=-3

Question 17:

$\frac{1}{3x-6}- \frac{5}{x-2}=12$
$\frac{(x-2)-5(3x-6)}{(3x-6)(x-2)} = \frac{x-2-15x+30}{(3x-6)(x-2)}= \frac{-14x+28}{(3x-6)(x-2)}$

Question 18

1. the width w of a rectangular swimming pool is x+4. the area a of the pool is 2x^3-29+12. what is

Mathematics

What is the equation of the oblique asymptote? h(x)= x^2-3x-4/x+1 ○a. y=x+4 ○b. y= x ○c. y= x^2-3 ○d. y=x-4

Step-by-step answer

P Answered by PhD

21

y=x-4

Step-by-step explanation:

What you are looking for is also known as the slant asymptote. The slant asymptote occurs when the degree of the numerator is one degree more than the denominator which is what you have.

So to find the slant asymptote we can use polynomial division.

We have a choice to use synthetic division here because the denominator is linear.

-1 goes on the outside because we are dividing by (x+1).

-1 | 1 -3 -4

| -1 4

|

1 -4 0

The asymptote is the quotient part which is y=x-4.

So answer is y=x-4.

Mathematics

What is the equation of the oblique asymptote? h(x)= x^2-3x-4/x+1 ○a. y=x+4 ○b. y= x ○c. y= x^2-3 ○d. y=x-4

Step-by-step answer

P Answered by PhD

21

y=x-4

Step-by-step explanation:

What you are looking for is also known as the slant asymptote. The slant asymptote occurs when the degree of the numerator is one degree more than the denominator which is what you have.

So to find the slant asymptote we can use polynomial division.

We have a choice to use synthetic division here because the denominator is linear.

-1 goes on the outside because we are dividing by (x+1).

-1 | 1 -3 -4

| -1 4

|

1 -4 0

The asymptote is the quotient part which is y=x-4.

So answer is y=x-4.

Mathematics

Which solution shown below contains an error? 3 4 3x 4 3x+4 ? ਦੇਸ਼ ਦੇ 1 +2 2 x+2 1 x+1 +2 2 3 + 21-12 8x+8 (x+1)(x-6) * (x+1)(x-6) +1 - 8 1px-4 (+1)(-6)

Step-by-step answer

P Answered by Master

92

B

Step-by-step explanation:

1/x+2 + 1/x+2 = 2/x+2 = 1/x+1.

Mathematics

Which solution shown below contains an error? 3 4 3x 4 3x+4 ? ਦੇਸ਼ ਦੇ 1 +2 2 x+2 1 x+1 +2 2 3 + 21-12 8x+8 (x+1)(x-6) * (x+1)(x-6) +1 - 8 1px-4 (+1)(-6)

Step-by-step answer

P Answered by Master

92

B

Step-by-step explanation:

1/x+2 + 1/x+2 = 2/x+2 = 1/x+1.

Mathematics

hi if anyone is good with extraneous solutions pleas help m tessa solves the equation below by first squaring both sides of the equation√x^2-3x-6=x-1 what extraneous solution does tessa obtain x=

Step-by-step answer

P Answered by Specialist

1

x = -7/5

Step-by-step explanation:

If we square both sides of the equation, we get:

$\sqrt{x^2-3x-6}=x-1\\ (\sqrt{x^2-3x-6})^2=(x-1)^2\\x^2-3x-6=x^2-2x+1\\$

Then, solving for x, we get:

$x^2-3x-6=x^2-2x+1\\-3x-6=2x+1\\-6-1=2x+3x\\-7=5x\\\frac{-7}{5}=x$

So, x is equal to -7/5

Mathematics

Find f(g(x)). State the domain of the composite function. f(x) = 3x-2/x+1 g(x) = x+5/2x-3

Step-by-step answer

P Answered by Master

ℝ - {(-2/3),(3/2)}

Step-by-step explanation:

We want the domain of f(g(x)). So, firstly, we have to find the domain for g(x) and, then, for f(g(x)).

- Domain of g(x): Since the expression is a fracion, we must exclude the values of x that make null the denominator. Hence,

$g(x)=\dfrac{x+5}{2x-3}\Longrightarrow 2x-3\neq 0\iff \boxed{x\neq\dfrac{3}{2}}$

- Domain of f(g(x)): We'll find its expression:

$f(x) = \dfrac{3x-2}{x+1}\\\\f(g(x)) = \dfrac{3g(x)-2}{g(x)+1}\\\\f(g(x)) = \dfrac{3\cdot\dfrac{x+5}{2x-3}-2}{\dfrac{x+5}{2x-3}+1}=\dfrac{~~~\dfrac{3(x+5)-2(2x-3)}{2x-3}~~~}{\dfrac{(x+5)+(2x-3)}{2x-3}}\\\\f(g(x)) =\dfrac{3(x+5)-2(2x-3)}{(x+5)+(2x-3)}=\dfrac{3x+15-4x+6}{x+5+2x-3}\\\\\boxed{f(g(x)) =\dfrac{21-x}{3x+2}}$