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I've stumbled upon a very good simultaneous equation with 4 variables and 4 equations, the are as follows

\begin{array}{r l} bc+3d+15a-db-15c-3b=60 & (1) \\ d-15c=0 & (2) \\ 3a-b= -6 & (3) \\ b-d-c+a=0 & (4) \end{array}

Just need the values and how to to them.. not case sensitive just in case Thanks

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  • $\begingroup$ Where did you “stumble” upon this system of equations? $\endgroup$ – amd Jan 15 at 20:44
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From $(2)$, $$d=15c\ \ \ \ldots(5)$$ Also from $(3)$, $b=3a+6$. Putting this and $(5)$ in $(4)$, $$3a+6-15c-c+a=0$$ $$4a-16c+6=0$$ $$a = 4c-\frac32\ \ \ \ldots(6)$$ So$$\begin{align} b&=3\left(4c-\frac32\right)+6\\ &=12c+\frac32\ \ \ \ldots(7) \end{align}$$ Taking values of $a$, $b$ and $d$ from $(5)$, $(6)$ and $(7)$ and putting in $(1)$, we get a quadratic equation in $c$. I hope you can take it from here.

Note: You can do this by solving any three variables with respect to the fourth one. I just solved $a$, $b$ and $d$ w.r.t. $c$ because it seemed easier.

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Hint: use the last three equations to solve for three variables in terms of the fourth. Substitute in to the first equation.

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