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

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

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

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

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    This is not my homework , I am practicing for SAT. – 2017-02-13
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    The answer to last equation is 1. But I cant solve it, tried it but still can find even 1 variable. – 2017-02-13
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    No that is the answer to the equation (answer on the back of the book ). all I know is : a−7b+8c=4 8a+4b−c=7 and from there I have to find a^2−b^2+c^2= – 2017-02-13
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    @JMoravitz Although your remarks about underdetermined systems are generally correct, if you check out this particular example, you will find that the first two equations do indeed imply that $a^2-b^2+c^2=1$. – 2017-02-13
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    Is it possible to find values for a,b and c ? And if so, how ? – 2017-02-13
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    @David I should have tried the problem before immediately dismissing it as having a typo or no solution before commenting. Yes, indeed after going through the effort every point along the line created by the intersection of those two planes do indeed satisfy the equation $a^2-b^2+c^2=1$. @ Zoran, "is it possible to find values for $a,b,c$" no, it is not possible to find a single specific value for each as the first two equations each represent a plane in three dimensional space and their intersection is a line. E.g. $(4,-8,-7)$ and $(9,-21,-19)$ are both solutions among infinite others. – 2017-02-13
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    @JMoravitz see the geometrical interpretation I give in my solution. – 2017-02-13
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    @ZoranMladenovski: You should probably put any accompanying information (e.g., that this is a practice problem for the SAT, that you know the answer is $1$, but you don't know how or why) in the actual question. It will help you avoid downvotes and attract more people to answer the question. – 2017-02-13

3 Answers 3

5

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.

2

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

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

Tricks

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

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

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

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

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

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

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

$(3) \times (4)$,

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

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

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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.

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    I agree with your computations. I will try in my coming answer to give a geometrical interpretation. – 2017-02-13
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    @JeanMarie This works because $a^2-b^2-c^2=\hbox{constant}$ is a [ruled surface](https://en.wikipedia.org/wiki/Ruled_surface) - although it is curved, it contains (infinitely many) straight lines. – 2017-02-13
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    @David I had not seen your comment, but I arrived at the same geometrical explanation... (see my solution) – 2017-02-13