Can someone please help with solving this linear equation

Question

Can you please help me with this high school math problem, thanks ! :)

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

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

$$a^2 - b^2 + c^2 = ?$$

Answer 1

Yes, it is true that the value is 1.

First an algebraic point of view. Let us consider the system as a linear parametric system with parameter $a$ and unknowns $b$ and $c$:

$$\begin{cases} -7b+8c&=&4-a \\ \ \ 4b-c&=&7-8a \end{cases}$$

Its solutions are:

$$\tag{1}\begin{cases}b&=&(12-13a)/5\\c&=&(13-12a)/5\end{cases}$$

Now, a geometrical view. Let us add to (1) the trivial equation $a=a$, we obtain

$$\tag{2}\begin{cases}a&=&a\\b&=&(12-13a)/5\\c&=&(13-12a)/5\end{cases}$$

which is a parametric representation of a line $(L)$ in $\mathbb{R}^3$. This is not surprising because the constraints brought by the 2 initial equations are the intersection of 2 planes, thus a line (see also the Remark at the bottom). Using (2): $$a^2-b^2+c^2=\dfrac{25 a^2-(12-13a)^2+(13-12a)^2}{25}=\dfrac{25}{25}=1.$$

This, as well, is not geometrically surprising. In fact, equation

$$a^2-b^2+c^2=1$$

is that of a hyperboloid of one sheet, which is a ruled surface (http://mathworld.wolfram.com/One-SheetedHyperboloid.html), thus able to contain a whole straight line.

Remark: A supplementary geometrical interpretation:

Let us write (2) under the form:

$$\pmatrix{a\\b\\c}=\pmatrix{0\\12/5\\13/5}+(a/5)\pmatrix{5\\13\\12}$$

This vectorial representation shows in particular that line $(L)$ is directed by vector

$$\pmatrix{5\\13\\12} \ \ \text{which is proportional to} \ \ \pmatrix{ \ \ 1\\-7\\ \ \ 8}\times \pmatrix{8\\4\\-1}$$

i.e., the cross product of the two normal vectors to the initial planes.

Answer 2

Try putting $b$ and $c$ in terms of $a$ as follows: $$\begin{cases} a-7b+8c=4 \\ 8a+4b-c=7 \end{cases}$$ Eliminating $c$: $$\begin{cases} a-7b+8c=4 \\ 64a+32b-8c=56 \end{cases}$$

Which implies that: $b=\frac{12-13a}{5}$.

Now, try the same by eliminating $b$:

$$\begin{cases} 4a-28b+32c=16 \\ 56a+28b-7c=49\end{cases}$$

Which implies that $c=\frac{13-12a}{5}$

From here, I will let you do the rest.

Substitute the equations you have for $b$ and $c$ into $a^2-b^2+c^2$. The $a$ terms should cancel out very nicely and you should be left with a constant value for $a^2-b^2+c^2$ (The answer you have been given).

Feel free to comment on your progress on solving the problem, and ask if you have any doubts on my answer.

Answer 3

$$a-7b+8c=4 \tag{1}$$

$$8a+4b-c= 7 \tag{2}$$

Tricks

Considering $(\lambda+8)^2=(-7\lambda+4)^2 \implies \lambda=2 \; $ or $\; -\dfrac{1}{2}$

$2\times (1)+(2)$,

$$10a-10b+15c=15$$

$$a-b=\frac{3(1-c)}{2} \tag{3}$$

$2\times (2)-(1)$,

$$15a+15b-10c=10$$

$$a+b=\frac{2(1+c)}{3} \tag{4}$$

$(3) \times (4)$,

$$a^2-b^2=1-c^2$$

$$\fbox{$a^2-b^2+c^2=1$}$$