Question 1205480
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Note that the problem does not ask us to find the values of a, b, and c.<br>
In fact there are an infinite number of triples of numbers a, b, and c which all give the same value for {{{2a^2 - 3b^2 + c^2}}}.  This is easy to see empirically by choosing arbitrary values for a and finding the corresponding values of b and c using the given equations; in every case the value of {{{2a^2 - 3b^2 + c^2}}} is the same.<br>
ANSWER: {{{2a^2 - 3b^2 + c^2 = 54}}}.<br>
Let's use algebra to show that 54 is always the answer.<br>
One common way to solve problems like this is to square the given equations.  But that introduces "ab" and "bc" terms, which we really don't want.<br>
So another way to solve the problem is to look for examples of expressions of the form {{{x^2-y^2=(x+y)(x-y)}}} in the given expression {{{2a^2 - 3b^2 + c^2}}}.<br>
{{{2a^2 - 3b^2 + c^2=2a^2-2b^2-b^2+c^2=2(a^2-b^2)-(b^2-c^2)=2(a-b)(a+b)-(b+c)(b-c)=2(a-b)(-3)-(b+c)(6)=-6a+6b-6b-6c=-6(a+c)}}}.<br>
But eliminating b from the original two equations gives us {{{a+c=-9}}}.  And so<br>
{{{2a^2-3b^2+c^2=(-6)(-9)=54}}}<br>
ANSWER: 54<br>