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podryga [215]
3 years ago
12

2(x + 4)= -5 + 1 what is this answer

Mathematics
2 answers:
fiasKO [112]3 years ago
6 0

Answer:

x=-6

Step-by-step explanation:

Given Equation:

                    2(x + 4)= -5 + 1

Solving the Equation for 'x':

                  2(x + 4)= -4

              2(x) + 2(4)= -4\\

                      2x+8=-4

Subtracting '8' both sides:

                        2x=-4-8

                        2x=-12

Dividing by '2' both sides:

                         x=-12/2\\

                           x=-6

The Solution for the equation is:  x=-6

Nitella [24]3 years ago
6 0
Answer: X = -6

Explanation:
2(x+4) = -5 + 1
2x + 8 = -4
2x = -4 - 8
2x = -12 divide both sides by 2
X = -6
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2y^2\,\mathrm dx-(x+y)^2\,\mathrm dy=0

Divide both sides by x^2\,\mathrm dx to get

2\left(\dfrac yx\right)^2-\left(1+\dfrac yx\right)^2\dfrac{\mathrm dy}{\mathrm dx}=0

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2\left(\frac yx\right)^2}{\left(1+\frac yx\right)^2}

Substitute v(x)=\dfrac{y(x)}x, so that \dfrac{\mathrm dv(x)}{\mathrm dx}=\dfrac{x\frac{\mathrm dy(x)}{\mathrm dx}-y(x)}{x^2}. Then

x\dfrac{\mathrm dv}{\mathrm dx}+v=\dfrac{2v^2}{(1+v)^2}

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The remaining ODE is separable. Separating the variables gives

\dfrac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=-\dfrac{\mathrm dx}x

Integrate both sides. On the left, split up the integrand into partial fractions.

\dfrac{(1+v)^2}{v(1+v^2)}=\dfrac{v^2+2v+1}{v(v^2+1)}=\dfrac av+\dfrac{bv+c}{v^2+1}

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Then

\displaystyle\int\frac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=\int\left(\frac1v+\frac2{v^2+1}\right)\,\mathrm dv=\ln|v|+2\tan^{-1}v

On the right, we have

\displaystyle-\int\frac{\mathrm dx}x=-\ln|x|+C

Solving for v(x) explicitly is unlikely to succeed, so we leave the solution in implicit form,

\ln|v(x)|+2\tan^{-1}v(x)=-\ln|x|+C

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\ln\left|\frac{y(x)}x\right|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C

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3 years ago
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3 years ago
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