Bath filling problem

[cohen27]

Dear Colleagues,
I have been reading David Acheson - 'The Spirit of Mathematics'.
How long would it take for C to fill the bath alone?.
I understand the reasoning of the three equations he sets out, but stuck on the process circled in red in the problem attached.
My question is: How did he get rid of A in the second equation ( a+c=1/5)?. I understand where he rewrote the equation to get a by itself (a=1/5-c),
but I don't understand how he got from a=1/5-c to 1/5-c+2c=1/4 ?.
The question is attached.
Thank you.

 

[fresh_42]

[math]\begin{array}{lll} a+b&=\dfrac{1}{4}\\[8pt] a+c&=\dfrac{1}{5}\\[8pt] b&=2c \end{array}[/math]
The first step was to plug in [imath] b [/imath] from (3) into (1), i.e., [imath] a+2c=1/4, [/imath] call it (4). Then we rewrite (2) as [imath] a=1/5 - c, [/imath] call it (5). Finally, we take the first equation and plug in [imath] b [/imath] from equation (3) and [imath] a [/imath] from equation (5), i.e., [math] \dfrac{1}{4}=a+b= \left(\dfrac{1}{5}-c\right)+2c.[/math]
Note that such a way to solve a linear equation system is a bit arbitrary, depending on which equations you choose to do what. There is a risk of ending up with an expression like [imath] 1=1 [/imath] at some point of the calculation. My suggestion would be to use (1) and (2) for eliminating [imath] a [/imath] and this result together with (3) for eliminating [imath] b. [/imath] This would be more systematic. If you want to look up how it is done in general, then we need more notations such as matrices and vectors. Your search key should be "Gaussian Elimination".