# An Equation with 4 variable

• Oct 15th 2012, 04:40 PM
blackjack21
An Equation with 4 variable
2y+3k=5u
2u+4y=3k
3k=ux

x = ?

Thanks Best regards
• Oct 15th 2012, 05:37 PM
Soroban
Re: An Equation with 4 variable
Hello, blackjack21!

Quote:

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

$\text{Solve for }x.$

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

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

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.
• Oct 15th 2012, 05:49 PM
blackjack21
Re: An Equation with 4 variable