2y+3k=5u
2u+4y=3k
3k=ux
x = ?
Thanks Best regards
Hello, blackjack21!
$\displaystyle \begin{array}{cc}2y+3k\:=\:5u & [1] \\ 2u+4y\:=\:3k & [2] \\ ux\:=\:3k & [3] \end{array}$
$\displaystyle \text{Solve for }x.$
From [1]: .$\displaystyle y \:=\:\frac{5u-3k}{2}\;\;[4]$
From [2]: .$\displaystyle y \:=\:\frac{3k-2u}{4}\;\;[5]$
Equate [4] and [5]: .$\displaystyle \frac{5u-3k}{2} \:=\:\frac{3k-2u}{4} \quad\Rightarrow\quad 20y - 12k \:=\:6k-4u$
. . . $\displaystyle 24u \:=\:18k \quad\Rightarrow\quad k \:=\:\tfrac{4}{3}u\;\;[6]$
Substitute into [3]: .$\displaystyle ux \:=\:3\left(\tfrac{4}{3}u\right) \quad\Rightarrow\quad ux \:=\:4u $
Divide by $\displaystyle u\!:\;\;x \:=\:4$
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After dividing by $\displaystyle u$, I had to consider: what if $\displaystyle u = 0\,?$
Then equation [6] says: $\displaystyle k = 0.$
And equation [3] becomes: .$\displaystyle 0\cdot x \:=\:3\cdot0 \quad\Rightarrow\quad 0 \:=\:0$
. . which is true for any value of $\displaystyle x.$
I suppose that counts as another solution.