Welcome to our community

Be a part of something great, join today!

System of equalities 3

solakis

Active member
Dec 9, 2012
304
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
Aug 6, 2015
271
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
Jan 30, 2018
371
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
Dec 9, 2012
304
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?
 
Last edited:

Monoxdifly

Well-known member
Aug 6, 2015
271
I don't know. That's as far as I can get.
 

Opalg

MHB Oldtimer
Staff member
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) \(\displaystyle \frac{1}{z}+\frac{1}{w}=\frac{5}{18}\)
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
Aug 6, 2015
271
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.