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The right answer is: The new drug appears to reduce the number of eczema breakouts since the rerandomized mean differences are considerably less than the mean difference found between the two groups studied.

sqrt{2x+1} + sqrt{x-3} = 3\sqrt x +7

\(\sqrt{2x+1} + \sqrt{x-3} = 3\sqrt x +7 \)

Since we are in the real number plane,  x>=3

\(LHS\\ =\sqrt{2x+1} + \sqrt{x-3} \\ <\sqrt{3x}+\sqrt x \\ <(\sqrt3 +1)\sqrt x\\ < 2.8\sqrt x \\~\\ RHS\\ =3\sqrt x +7\\ >3\sqrt x \\ \)

So the LHS is always smaller than \(2.8\sqrt x\)

And the RHS is always bigger than \(3 \sqrt x \)

So they are never equal.  there are no solutions.

I will admit that I tried conventional methods to find the intersestions to start with.

When I got nowhere I plotted them using Desmos.

I could see that they were not going to intersect.  The LHS was always  less than the RHS.

Then I set out to prove it algebraically.  (Which was not difficult.)

y=3(x+7/2)2−131/4
(3,7/2,-131/4)

This can be 11 choose 3, or 11*10*9.

\(990\)

If we graph this equation on Desmos, we will see the range is [-2.5,8]

We have the system of equations: 

\(2x + y = 14\)

\(2L + z = 20\)

\(x +y + z = 14\)

\(x + z + L = 18\)

Subtracting \(y\) from both the first and third equations gives us: \(x + z = 14 - y\) and \(2x = 14-y\)

Because both equations are equal to each other, we can set them equal. This gives us: \(x + z = 2x\), meaning \(x = z \)

Substituting this in, we have a new system in terms of \(x\):

 \(2x + y = 14\)

\(2L + x = 20\)

\(2x + L = 18\)

Now, we are looking to find the value of \(2x\). 

We can do this by disregarding the first equation, which gives us this system: 

\(2L + x = 20\)

\(2x + L = 18\)

Multiplying the second equation by 2 gives us: \(4x + 2L = 36\)

Now, subtracting the first equation from this shows: \(3x = 16\).

Multiplying both sides by \(2 \over 3\) shows \(x +z = \color{brown}\boxed{32 \over 3}\), just as Guest found 

We can simplify the expression to: \(x^{2}+12x+5=0\).

By Vieta's formula, the sum of the roots (\(r + s\)), equals \({-12 \over 1} = -12 \), and the product of the roots (\(r \times s \)) equals \({5 \over 1 } = 5 \).

Note that \(r^2 + s^2 = (r+s)^2 - 2rs\). 

Can you take it from here?

Let \(x = 0.3 \overline 2 \).

This means \(10x = 3. \overline 2\) and \(100x = 32. \overline 2\)

Subtracting \(10x \) from \(100x\) gives us: \(90x=29\), because the repeating part cancels each other out. 

Can you find the value of x from here?

Note that for this problem, you don't need to find the value of each variable individually, you just need to know the value of \(a + b + c \).

Adding all 3 equations together gives us: \(2(a+b+c) = 18\). Can you find the value of \(a + b + c \) from here?

\(0.3 \overline{2} = \dfrac{32}{99}\)

Solve the following system:
{y + 2 x = 14 | (equation 1)
z + 2 L = 20 | (equation 2)
x + y + z = 14 | (equation 3)
L + x + z = 18 | (equation 4)

Subtract 1/2 × (equation 1) from equation 3:
{2 x + y + 0 z + 0 L = 14 | (equation 1)
0 x + 0 y + z + 2 L = 20 | (equation 2)
0 x + y/2 + z + 0 L = 7 | (equation 3)
x + 0 y + z + L = 18 | (equation 4)

Multiply equation 3 by 2:
{2 x + y + 0 z + 0 L = 14 | (equation 1)
0 x + 0 y + z + 2 L = 20 | (equation 2)
0 x + y + 2 z + 0 L = 14 | (equation 3)
x + 0 y + z + L = 18 | (equation 4)

Subtract 1/2 × (equation 1) from equation 4:
{2 x + y + 0 z + 0 L = 14 | (equation 1)
0 x + 0 y + z + 2 L = 20 | (equation 2)
0 x + y + 2 z + 0 L = 14 | (equation 3)
0 x - y/2 + z + L = 11 | (equation 4)

Multiply equation 4 by 2:
{2 x + y + 0 z + 0 L = 14 | (equation 1)
0 x + 0 y + z + 2 L = 20 | (equation 2)
0 x + y + 2 z + 0 L = 14 | (equation 3)
0 x - y + 2 z + 2 L = 22 | (equation 4)

Swap equation 2 with equation 3:
{2 x + y + 0 z + 0 L = 14 | (equation 1)
0 x + y + 2 z + 0 L = 14 | (equation 2)
0 x + 0 y + z + 2 L = 20 | (equation 3)
0 x - y + 2 z + 2 L = 22 | (equation 4)

Add equation 2 to equation 4:
{2 x + y + 0 z + 0 L = 14 | (equation 1)
0 x + y + 2 z + 0 L = 14 | (equation 2)
0 x + 0 y + z + 2 L = 20 | (equation 3)
0 x + 0 y + 4 z + 2 L = 36 | (equation 4)

Divide equation 4 by 2:
{2 x + y + 0 z + 0 L = 14 | (equation 1)
0 x + y + 2 z + 0 L = 14 | (equation 2)
0 x + 0 y + z + 2 L = 20 | (equation 3)
0 x + 0 y + 2 z + L = 18 | (equation 4)

Swap equation 3 with equation 4:
{2 x + y + 0 z + 0 L = 14 | (equation 1)
0 x + y + 2 z + 0 L = 14 | (equation 2)
0 x + 0 y + 2 z + L = 18 | (equation 3)
0 x + 0 y + z + 2 L = 20 | (equation 4)

Subtract 1/2 × (equation 3) from equation 4:
{2 x + y + 0 z + 0 L = 14 | (equation 1)
0 x + y + 2 z + 0 L = 14 | (equation 2)
0 x + 0 y + 2 z + L = 18 | (equation 3)
0 x + 0 y + 0 z + (3 L)/2 = 11 | (equation 4)

Multiply equation 4 by 2:
{2 x + y + 0 z + 0 L = 14 | (equation 1)
0 x + y + 2 z + 0 L = 14 | (equation 2)
0 x + 0 y + 2 z + L = 18 | (equation 3)
0 x + 0 y + 0 z + 3 L = 22 | (equation 4)

Divide equation 4 by 3:
{2 x + y + 0 z + 0 L = 14 | (equation 1)
0 x + y + 2 z + 0 L = 14 | (equation 2)
0 x + 0 y + 2 z + L = 18 | (equation 3)
0 x + 0 y + 0 z + L = 22/3 | (equation 4)

Subtract equation 4 from equation 3:
{2 x + y + 0 z + 0 L = 14 | (equation 1)
0 x + y + 2 z + 0 L = 14 | (equation 2)
0 x + 0 y + 2 z + 0 L = 32/3 | (equation 3)
0 x + 0 y + 0 z + L = 22/3 | (equation 4)

Divide equation 3 by 2:
{2 x + y + 0 z + 0 L = 14 | (equation 1)
0 x + y + 2 z + 0 L = 14 | (equation 2)
0 x + 0 y + z + 0 L = 16/3 | (equation 3)
0 x + 0 y + 0 z + L = 22/3 | (equation 4)

Subtract 2 × (equation 3) from equation 2:
{2 x + y + 0 z + 0 L = 14 | (equation 1)
0 x + y + 0 z + 0 L = 10/3 | (equation 2)
0 x + 0 y + z + 0 L = 16/3 | (equation 3)
0 x + 0 y + 0 z + L = 22/3 | (equation 4)

Subtract equation 2 from equation 1:
{2 x + 0 y + 0 z + 0 L = 32/3 | (equation 1)
0 x + y + 0 z + 0 L = 10/3 | (equation 2)
0 x + 0 y + z + 0 L = 16/3 | (equation 3)
0 x + 0 y + 0 z + L = 22/3 | (equation 4)

Divide equation 1 by 2:
{x + 0 y + 0 z + 0 L = 16/3 | (equation 1)
0 x + y + 0 z + 0 L = 10/3 | (equation 2)
0 x + 0 y + z + 0 L = 16/3 | (equation 3)
0 x + 0 y + 0 z + L = 22/3 | (equation 4)

Collect results:

 x = 16/3
y = 10/3
z = 16/3
L = 22/3                            x  +  z ==16/3  +  16/3==32/3

Here's a hint: Let z = a + bi, where a and b are real numbers.

Thanks for posting the "solution" without an explanation! But it would be nice if you showed me how you solved it :)

to take the reciprocal of a number you raise it to the power of negative one, or divide 1 by the number (same thing)

so the reciprocal of x would be x^(-1), or just 1/x. we know that a number is 3 larger than its reciprocal, lets call this number x.

so 1/x <-- the reciprocal, plus 3 would be equal to x. we can write this as 1/x + 3 = x

lets add 1/x and 3, and we get (3x+1)/x = x

multiply both sides by x, and we get 3x+1 = x^2

now lets move this into a quadratic that equals zero

x^2-3x-1 = 0

we can use the quadratic formula to find x, and we end up with

x= (3 plusorminus sqrt(13))/2

now plug these values into (x-(1/x))^4

Let the amount of money that Annie brought with her==A

Then, the amount of money that Bennie brought with him ==86 - A

A - 4/9A ==5/9A - amount Annie has left

86  -  A  - 16 ==70  - A - amount that Bennie has left

5/9A ==70  - A, solve for A

A + 5/9A ==70

14/9A ==70

14A ==70 * 9 ==630

A ==630 / 14 ==$45 - amount of money that Annie brought

86  -  45 ==$41 - amount of money that Bennie brought.

284 ==(1, 2, 4, 71, 142, 284)>>Total = 6 divisors

ADD them up on your calculator except 284 itself.

sumfor(n, 51, 250, 2*n - 1) ==60,000 sum of all odd numbers between 100 and 500

1 / 0.714285 ==~ 1.4

5^100 * 2^100 ==10^100 - which means 1 followed by 100 zeros.

So: 0 + 0 + 0==0

3 * Sqrt(2)^N ==768 * sqrt(2), solve for N

N ==17

17 + 1 ==18 terms of this GP

Levans writes a positive fraction in which the numerator and denominator are integers, and the numerator is 2 greater than the denominator. He then writes several more fractions. To make each new fraction, he increases both the numerator and the denominator of the previous fraction by 1. He then multiplies all his fractions together. He has 3 fractions, and their product equals 10. What is the value of the first fraction he wrote?

  3         4          5  

 ––   x   ––   x    ––  

  1         2           3

The 3's cancel, leaving (4/2) x 5 = 10  

                                    3  

The first fraction was   ––  

                                     1  

This is the third time I've answered this same problem. 

_.