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3 years ago

6

What is the value of x? 3x−8=8x+4/2 Enter your answer in the box.

1 answer:

bekas [8.4K]3 years ago

3 0

Answer:

x = -2

Step-by-step explanation:

3x -8 =8x +4/2

3x -8 =8x +2

3x -3x -8 = 8x -3x +2 (subtract 3x both sides )

-8 = 5x +2

-8-2 =5x+2 -2 (subtract 2 both sides )

-10 =5x

-10/5 = 5x/5 (divide 5 both sides)

-2 = x

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What number fills in the blank to complete the factorization of 3x + 24?<br><br> (x + 8)

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Answer:

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Step-by-step explanation:

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2 years ago

Problem 4: Solve the initial value problem

pishuonlain [190]

Separate the variables:

$y' = \dfrac{dy}{dx} = (y+1)(y-2) \implies \dfrac1{(y+1)(y-2)} \, dy = dx$

Separate the left side into partial fractions. We want coefficients a and b such that

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$\implies \dfrac1{(y+1)(y-2)} = \dfrac{a(y-2)+b(y+1)}{(y+1)(y-2)}$

$\implies 1 = a(y-2)+b(y+1)$

$\implies 1 = (a+b)y - 2a+b$

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So we have

$\dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = dx$

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$\dfrac13 \ln\left|\dfrac{y-2}{y+1}\right| = x + C$

$\ln\left|\dfrac{y-2}{y+1}\right| = 3x + C$

$\dfrac{y-2}{y+1} = e^{3x + C}$

$\dfrac{y-2}{y+1} = Ce^{3x}$

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$\dfrac{1-2}{1+1} = Ce^{0} \implies C = -\dfrac12$

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$\boxed{\dfrac{y-2}{y+1} = -\dfrac12 e^{3x}}$

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$\dfrac{y-2}{y+1} = \dfrac{(y+1)-3}{y+1} = 1-\dfrac3{y+1}$

Then

$1 - \dfrac3{y+1} = -\dfrac12 e^{3x}$

$\dfrac3{y+1} = 1 + \dfrac12 e^{3x}$

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$y = \dfrac6{2+e^{3x}} - 1$

$\boxed{y = \dfrac{4-e^{3x}}{2+e^{3x}}}$

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3 years ago

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