my parents and i couldnt figure out part (b), but we figured out part (a), so i wont post that.

In this multi-part problem, we will consider this system of simultaneous equations:

$$\begin{array}{r@{~}c@{~}l l}

3x+5y-6z &=&2, & \textrm{(i)} \\

5xy-10yz-6xz &=& -41, & \textrm{(ii)} \\

xyz&=&6. & \textrm{(iii)}

\end{array}$$

Let $a=3x$,$b=5y$ , and $c=-6z$ .

Given that $(x,y,z)$ is a solution to the original system of equations, determine all distinct possible values of $x+y$. (Suggestion: Using the substitutions in part (a), first determine all possible values of the ordered triple $(a,b,c)$, then determine the possible solutions $(x,y,z)$.)

i figured out that (a,b,c) is (-12,5,9). i dont know for sure if this is correct could someone crosscheck this?

and also please dont copy paste the same answer from the other post, it didnt explain much. thanks!

Guest Jul 26, 2021

#1**+1 **

IF this is true

i figured out that (a,b,c) is (-12,5,9) then x = -4 y =1 z = -1.5 sub those into the equations to see if it makes all three of them true...

ElectricPavlov Jul 26, 2021

#3**+1 **

SInce we know what x y and z supposedly are, we just sub them into the equations to see if true:

Here is the second equation

5 (-4)(1) -10 ( 1)(-1.5) -6 (-4)(-1.5) does this equal -41 ???? do the same for the other two equations to check your answer....

ElectricPavlov Jul 26, 2021

#5**+1 **

5xy -10yz - 6xz since x = -4 y = 1 z = -1.5 then

5 (-4)(1) -10 ( 1)(-1.5) -6 (-4)(-1.5)

ElectricPavlov
Jul 26, 2021

#6**0 **

oh wait i think i explained this wrong ill link you the oringinal post on this.

https://web2.0calc.com/questions/some-of-my-questions-got-deleted-question

if you read this i think you'll understand, its part b.

Guest Jul 26, 2021