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3 years ago

Help .. 3(x+2)=5x+1-2x+5

Mathematics

1 answer:

Hunter-Best [27]3 years ago

8 0

Answer:

All real numbers are solutions

Step-by-step explanation:

Let's solve your equation step-by-step

3(x+2)=5x+1−2x+5

Step 1: Simplify both sides of the equation

3(x+2)=5x+1−2x+5

(3)(x)+(3)(2)=5x+1+−2x+5(Distribute)

3x+6=5x+1+−2x+5

3x+6=(5x+−2x)+(1+5)(Combine Like Terms)

3x+6=3x+6

Step 2: Subtract 3x from both sides

3x+6−3x=3x+6−3x

6=6

Step 3: Subtract 6 from both sides

6−6=6−6

0=0

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$\bold{\huge{\pink{\underline{ Solution }}}}$

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We have given two linear equations that is 2x - 3y = -6 and x + 3y = 12 .

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$\sf{ 2x - 3y = -6 ...eq(1)}$

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Multiply eq( 2 ) by 2 :-

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Subtract eq(1) from eq(2) :-

$\sf{ 2x + 6y -( 2x - 3y) = 24 -(-6)}$

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Read 2 more answers

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Zigmanuir [339]

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For <img src="https://tex.z-dn.net/?f=e%5E%7B-x%5E2%2F2%7D" id="TexFormula1" title="e^{-x^2/2}" alt="e^{-x^2/2}" align="absmiddl

nevsk [136]

I'm assuming you're talking about the indefinite integral

$\displaystyle\int e^{-x^2/2}\,\mathrm dx$

and that your question is whether the substitution $u=\dfrac x{\sqrt2}$ would work. Well, let's check it out:

$u=\dfrac x{\sqrt2}\implies\mathrm du=\dfrac{\mathrm dx}{\sqrt2}$
$\implies\displaystyle\int e^{-x^2/2}\,\mathrm dx=\sqrt2\int e^{-(\sqrt2\,u)^2/2}\,\mathrm du$
$=\displaystyle\sqrt2\int e^{-u^2}\,\mathrm du$

which essentially brings us to back to where we started. (The substitution only served to remove the scale factor in the exponent.)

What if we tried $u=\sqrt t$ next? Then $\mathrm du=\dfrac{\mathrm dt}{2\sqrt t}$ , giving

$=\displaystyle\frac1{\sqrt2}\int \frac{e^{-(\sqrt t)^2}}{\sqrt t}\,\mathrm dt=\frac1{\sqrt2}\int\frac{e^{-t}}{\sqrt t}\,\mathrm dt$

Next you may be tempted to try to integrate this by parts, but that will get you nowhere.

So how to deal with this integral? The answer lies in what's called the "error function" defined as

$\mathrm{erf}(x)=\displaystyle\frac2{\sqrt\pi}\int_0^xe^{-t^2}\,\mathrm dt$

By the fundamental theorem of calculus, taking the derivative of both sides yields

$\dfrac{\mathrm d}{\mathrm dx}\mathrm{erf}(x)=\dfrac2{\sqrt\pi}e^{-x^2}$

and so the antiderivative would be

$\displaystyle\int e^{-x^2/2}\,\mathrm dx=\sqrt{\frac\pi2}\mathrm{erf}\left(\frac x{\sqrt2}\right)$

The takeaway here is that a new function (i.e. not some combination of simpler functions like regular exponential, logarithmic, periodic, or polynomial functions) is needed to capture the antiderivative.

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