Solve the equation 2x-1=4x+5

Mathematics

2 answers:

Feliz [49]4 years ago

8 0

Answer:

x = -3

Step-by-step explanation:

Solve for x:

2 x - 1 = 4 x + 5

Subtract 4 x from both sides:

(2 x - 4 x) - 1 = (4 x - 4 x) + 5

2 x - 4 x = -2 x:

-2 x - 1 = (4 x - 4 x) + 5

4 x - 4 x = 0:

-2 x - 1 = 5

Add 1 to both sides:

(1 - 1) - 2 x = 1 + 5

1 - 1 = 0:

-2 x = 5 + 1

5 + 1 = 6:

-2 x = 6

Divide both sides of -2 x = 6 by -2:

(-2 x)/(-2) = 6/(-2)

(-2)/(-2) = 1:

x = 6/(-2)

The gcd of 6 and -2 is 2, so 6/(-2) = (2×3)/(2 (-1)) = 2/2×3/(-1) = 3/(-1):

x = 3/(-1)

Multiply numerator and denominator of 3/(-1) by -1:

Answer: x = -3

timurjin [86]4 years ago

7 0

Answer:

-3

Step-by-step explanation:

2x-1=4x+5

-1=2x+5

-6=2x

x=-3

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Rasek [7]

With ϕ ≈ 1.61803 the golden ratio, we have 1/ϕ = ϕ - 1, so that

$I = \displaystyle \int_0^\infty \frac{\sqrt[\phi]{x} \tan^{-1}(x)}{(1+x^\phi)^2} \, dx = \int_0^\infty \frac{x^{\phi-1} \tan^{-1}(x)}{x (1+x^\phi)^2} \, dx$

Replace $x \to x^{\frac1\phi} = x^{\phi-1}$ :

$I = \displaystyle \frac1\phi \int_0^\infty \frac{\tan^{-1}(x^{\phi-1})}{(1+x)^2} \, dx$

Split the integral at x = 1. For the integral over [1, ∞), substitute $x \to \frac1x$ :

$\displaystyle \int_1^\infty \frac{\tan^{-1}(x^{\phi-1})}{(1+x)^2} \, dx = \int_0^1 \frac{\tan^{-1}(x^{1-\phi})}{\left(1+\frac1x\right)^2} \frac{dx}{x^2} = \int_0^1 \frac{\pi2 - \tan^{-1}(x^{\phi-1})}{(1+x)^2} \, dx$