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- #1

- Thread starter solakis
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- Thread starter
- #1

- Aug 6, 2015

- 271

Using the same method for the fourth equation, we get\(\displaystyle \frac16+\frac19=\frac5{18}\), so z = 6 and w = 9 (also interchangable).

Check them on the first equation:

x + y + z + w = 22

3 + 4 + 6 + 9 = 22

22 = 22 (TRUE)

Also check them on the second equation:

xyzw = 648

3(4)(6)(9) = 648

648 = 648 (TRUE)

So, if the solution is denoted as (w, x, y, z), the solution set is {(6, 3, 4, 9), (6, 4, 3, 9), (9, 3, 4, 6), (9, 4, 3, 6)}.

- Jan 30, 2018

- 371

$\frac{1}{x}+ \frac{1}{y}= \frac{7}{12}$

Multiply both sides by 12xy:

$12y+ 12x= 7xy$

$12y= 7xy- 12x= 7x(y- \frac{12}{7})$

$\frac{12y}{y- \frac{12}{7}}= \frac{7(12)y}{7y- 12}= 7x$.

$x= \frac{12y}{7y- 12}$.

$\frac{1}{z}+ \frac{1}{w}= \frac{5}{18}$

Multiply both sides by 18wz.

$18w+ 18z= 5wz$

$18w= 5wz- 18z= 5z(w- \frac{18}{5})$

$\frac{18w}{w}- \frac{18}{5}= \frac{5(18)w}{5w- 18}= 5z$

$z= \frac{18w}{5w- 18}$.

Now replace x and z in the first two equations so we have two equations in y and w.

$x+ y+ z+ w= \frac{12y}{7y- 12}+ y+ \frac{18w}{5w- 18}+ w= \frac{7y^2}{7y- 12}+ \frac{5w^2}{5w- 18}= 22$.

$xyzw= \frac{12y}{7y- 12}(y)\left(\frac{18w}{5w- 18}\right)w= \frac{216yw}{(7y-12)(5w-18)}= 648$

I will leave solving those last two equations to you!

- Thread starter
- #4

is this the only solution to the problem?

Using the same method for the fourth equation, we get\(\displaystyle \frac16+\frac19=\frac5{18}\), so z = 6 and w = 9 (also interchangable).

Check them on the first equation:

x + y + z + w = 22

3 + 4 + 6 + 9 = 22

22 = 22 (TRUE)

Also check them on the second equation:

xyzw = 648

3(4)(6)(9) = 648

648 = 648 (TRUE)

So, if the solution is denoted as (w, x, y, z), the solution set is {(6, 3, 4, 9), (6, 4, 3, 9), (9, 3, 4, 6), (9, 4, 3, 6)}.

Last edited:

- Aug 6, 2015

- 271

I don't know. That's as far as I can get.

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- #6

- Feb 7, 2012

- 2,679

Solve the following sustem

1) \(\displaystyle x+y+z+w=22\)

2) \(\displaystyle xyzw=648\)

3)\(\displaystyle \frac{1}{x}+\frac{1}{y}=\frac{7}{12}\)

4) \(\displaystyle \frac{1}{z}+\frac{1}{w}=\frac{5}{18}\)

- Aug 6, 2015

- 271

Ah, so I needn't go further. Thanks for your confirmation.

Case 1: $x+y = 7$ and $z+w=15$. Then $xy=12$, $zw=54$ and we get the solutions $\{x,y\} = \{3,4\}$, $\{z,w\} = \{6,9\}$, as found by Monoxdifly .

Case 2: $x+y = 15$ and $z+w = 7$. Then $xy = \frac{180}7$ and $zw = \frac{126}5$. But then $z$ and $w$ would have to be the solutions of the equation $t^2 - 7t + \frac{126}5 = 0$. Since that equation has no real solutions, Case 2 cannot arise. So the solutions found by Monoxdifly are the only ones.