# System of equalities 3

#### solakis

##### Active member
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) $$\displaystyle \frac{1}{z}+\frac{1}{w}=\frac{5}{18}$$

#### Monoxdifly

##### Well-known member
Starting at the third equation, the only couple of Egyptian fraction which sum is $$\displaystyle \frac7{12}$$ are $$\displaystyle \frac1{3}$$ and $$\displaystyle \frac14$$, so x = 3 and y = 4 (interchangeable).
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)}.

#### Country Boy

##### Well-known member
MHB Math Helper
Since the last two equation have only two unknowns each, I would start with them.

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

#### solakis

##### Active member
Starting at the third equation, the only couple of Egyptian fraction which sum is $$\displaystyle \frac7{12}$$ are $$\displaystyle \frac1{3}$$ and $$\displaystyle \frac14$$, so x = 3 and y = 4 (interchangeable).
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)}.
is this the only solution to the problem?

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#### Monoxdifly

##### Well-known member
I don't know. That's as far as I can get.

#### Opalg

##### MHB Oldtimer
Staff member
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) $$\displaystyle \frac{1}{z}+\frac{1}{w}=\frac{5}{18}$$
From 3) and 4), $12(x+y) = 7xy$ and $18(z+w) = 5zw$. Multiply those two equations and use 2), to get $18*12(x+y)(z+w) = 35*648$, which reduces to $(x+y)(z+x) = 105$. From 1), $(x+y) + (z+w) = 22$. It follows that $x+y$ and $z+w$ are the solutions of the quadratic equation $s^2 - 22s + 105=0$, namely $7$ and $15$. So there are two possible cases.

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.

#### Monoxdifly

##### Well-known member
From 3) and 4), $12(x+y) = 7xy$ and $18(z+w) = 5zw$. Multiply those two equations and use 2), to get $18*12(x+y)(z+w) = 35*648$, which reduces to $(x+y)(z+x) = 105$. From 1), $(x+y) + (z+w) = 22$. It follows that $x+y$ and $z+w$ are the solutions of the quadratic equation $s^2 - 22s + 105=0$, namely $7$ and $15$. So there are two possible cases.

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