# Thread: An Equation with 4 variable

1. ## An Equation with 4 variable

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

x = ?

Thanks Best regards

2. ## Re: An Equation with 4 variable

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.