2 equations, 3 variables!

$\begin{cases} a - 7b + 8c = 4 & ... (1) \\ 8a + 4b - c = 7 & ...(2) \end{cases}$

$\begin{aligned} (1)+8(2): \quad \quad \quad \ 65a + 25b & = 60 \\ \implies b & = \frac{12-13a}{5} \\ (2): \quad 8a + 4(2.4-2.6a) - c & = 7 \\ \implies c & = \frac{13-12a}{5} \end{aligned}$

Now, we have:

$\begin{aligned} 60(a^2-b^2+c^2) & = 60\left(a^2 - \left(\frac{12-13a}{5}\right)^2 + \left(\frac{13-12a}{5}\right)^2\right) \\ & = \frac{60}{25}\left(25a^2 - \left(12-13a\right)^2 + \left(13-12a\right)^2\right) \\ & = \frac{60}{25}\left(25a^2 - \left(144-312a+169a^2 \right) + \left(169-312a +144a^2 \right)\right) \\ & = \frac{60}{25}\left(25a^2 + 25 - 25a^2 \right) \\ & = \boxed{60} \end{aligned}$

$a-7b+8c=4$ implies $a+8c=4+7b$

Squaring on both sides,

$a^{2}+64c^{2}+16ac=16+49b^{2}+56b$ $----------$ $(1)$

$8a+4b-c=7$ implies $8a-c=7-4b$

Squaring on both sides,

$64a^{2}+c^{2}-16ac=49+16b^{2}-56b$ $----------$ $(2)$

$(1)$ $+$ $(2)$ yields $65a^{2}+65c^{2}=65b^{2}+65$

Dividing by 65 and rearranging, we get $a^{2}-b^{2}+c^{2}=1$

$60(a^{2}-b^{2}+c^{2}) =60$

don't know the explanation and I was hoping someone could explain why what I did works. I recall someone once saying in problems like these u can set one variable to 0 and then solve. I set a=0 and I solved for b and c and got final answer as 60. Thanks!

I don't know the explanation and I was hoping someone could explain why what I did works. I recall someone once saying in problems like these u can set one variable to 0 and then solve. I set a=0 and I solved for b and c and got final answer as 60. Thanks!