An Equation with 4 variable

**blackjack21** · October 15th 2012, 03:40 PM

2y+3k=5u
2u+4y=3k
3k=ux

x = ?

Thanks Best regards

**Soroban** · October 15th 2012, 04:37 PM

Hello, blackjack21!

$\begin{array}{cc}2y+3k\:=\:5u & [1] \\ 2u+4y\:=\:3k & [2] \\ ux\:=\:3k & [3] \end{array}$

$\text{Solve for }x.$

From [1]: . $y \:=\:\frac{5u-3k}{2}\;\;[4]$

From [2]: . $y \:=\:\frac{3k-2u}{4}\;\;[5]$

Equate [4] and [5]: . $\frac{5u-3k}{2} \:=\:\frac{3k-2u}{4} \quad\Rightarrow\quad 20y - 12k \:=\:6k-4u$

. . . $24u \:=\:18k \quad\Rightarrow\quad k \:=\:\tfrac{4}{3}u\;\;[6]$

Substitute into [3]: . $ux \:=\:3\left(\tfrac{4}{3}u\right) \quad\Rightarrow\quad ux \:=\:4u$

Divide by $u\!:\;\;x \:=\:4$

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After dividing by $u$ , I had to consider: what if $u = 0\,?$

Then equation [6] says: $k = 0.$

And equation [3] becomes: . $0\cdot x \:=\:3\cdot0 \quad\Rightarrow\quad 0 \:=\:0$

. . which is true for any value of $x.$

I suppose that counts as another solution.

**blackjack21** · October 15th 2012, 04:49 PM

thank you so much for your explanatory answer and Best regards

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