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!
+37159 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...
+37159 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....
+37159 5xy -10yz - 6xz since x = -4 y = 1 z = -1.5 then
5 (-4)(1) -10 ( 1)(-1.5) -6 (-4)(-1.5)
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.
