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3 years ago

What is the value of x in the equation 3x − 2 = x + 8?

Mathematics

1 answer:

N76 [4]3 years ago

4 0

Answer:

x=5

Step-by-step explanation:

We need to find the value of x the given equation: 3x-2=x+8

the goal when solving for equations is to isolate the variable onto one side, and have the number (value of what x is) on the other side

here's the equation:

3x-2=x+8

first, add 2 to both sides

3x-2=x+8

+2 +2

___________

3x=x+10 (2 goes away on the left side, as -2+2=0)

now, subtract x from both sides

3x=x+10

-x -x

______

2x=10 (x goes away on the right side, as x-x=0)

we have the variable isolated on one side, with the number on the other, but we need the value of x by itself (1x)

divide both sides by 2

x=5 (because 2x/2=1x, or x. 10/2=5).

therefore, the value of x is 5

Hope this helps!

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1. Write the form of the partial fraction decomposition of the rational expression. Do not solve for the constants.

german

Answer:

Step-by-step explanation:

To write the form of the partial fraction decomposition of the rational expression:

We have:

$\mathbf{\dfrac{8x-4}{x(x^2+1)^2}= \dfrac{A}{x}+\dfrac{Bx+C}{x^2+1}+\dfrac{Dx+E}{(x^2+1)^2}}$

Using partial fraction decomposition to find the definite integral of:

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By using the long division method; we have:

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So;

$\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}= 2x+\dfrac{x+20}{x^2-8x-20}$

By using partial fraction decomposition:

$\dfrac{x+20}{(x-10)(x+2)}= \dfrac{A}{x-10}+\dfrac{B}{x+2}$

$= \dfrac{A(x+2)+B(x-10)}{(x-10)(x+2)}$

x + 20 = A(x + 2) + B(x - 10)

x + 20 = (A + B)x + (2A - 10B)

Now; we have to relate like terms on both sides; we have:

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By solvong the expressions above; we have:

$A = \dfrac{5}{2}$ $B = \dfrac{3}{2}$

Now;

$\dfrac{x+20}{(x-10)(x+2)} = \dfrac{5}{2(x-10)} + \dfrac{3}{2(x+2)}$

Thus;

$\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}= 2x + \dfrac{5}{2(x-10)}+ \dfrac{3}{2(x+2)}$

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$\int \dfrac{2x^3-16x^2-39x+20}{x^2-8x-20} \ dx = \int \begin {bmatrix} 2x + \dfrac{5}{2(x-10)}+ \dfrac{3}{2(x+2)} \end {bmatrix} \ dx$

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