# How to solve these simultaneous equation?

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

• Where did you “stumble” upon this system of equations? – amd Jan 15 at 20:44

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.