Solve simultaneously: { <mtable columnalign="right center left" rowspacing="4pt" col

Wotjake3jev8

Wotjake3jev8

Answered question

2022-06-01

Solve simultaneously:
{ 6 ( x + y ) = 5 x y 21 ( y + z ) = 10 y z 14 ( z + x ) = 9 z x
Obviously one of the solutions is ( 0 , 0 , 0 ) but I'm more interested in the other .
I actually solved this question , but using hard way.
I expanded each equation then got one variable in terms of other. For example after expanding first equation I get ,
6 x + 6 y = 5 x y
and then
5 x y 6 x = 6 y
So,
x ( 5 y 6 ) = 6 y
then,
x = 6 y / ( 5 y 6 )
Similarly I found the value of x from the third equation ( in terms of z), equated both and got z in terms of y, substituted that in equation two and finally got the values. The solution is ( 2 , 3 , 7 ). You might have guessed that this is a very tedious method.

Answer & Explanation

ran1suel23

ran1suel23

Beginner2022-06-02Added 3 answers

You can divide the equations, to get
1 x + 1 y = 5 6 = 35 42 1 y + 1 z = 10 21 = 20 42 1 x + 1 z = 9 14 = 27 42
Subtracting the second equation from the first yields
1 x 1 z = 15 42
and adding that to the last
2 x = 42 42 = 1 x = 2.
Subtracting from the last yields
2 z = 12 42 z = 7.
Inserting x = 2 for example into the first equation yields y = 3.
Lesly Weiss

Lesly Weiss

Beginner2022-06-03Added 2 answers

If all you want to do is get a quick, quick answer and you're confident that there is only one answer, with integers, try this. 6 | 5 x y, so 6 | x y. Similarly 21 | y z and 14 | x z.
There may be many ways to distribute 2 , 3 , 7 among x , y , z, but the most "natural" to make these three properties all true is: 2 | x, 3 | y, 7 | z. As it happens, that's already a solution, but if it weren't I'd try x = 2 x , y = 3 y , z = 7 z ; this will at least allow you to cancel the 6 , 21 , 14 on the LHS's.

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