SOLUTION: if a + b = -3 and b - c = 6, find the value of {{{2a^2

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Question 1205480: if a + b = -3 and b - c = 6, find the value of

Found 3 solutions by greenestamps, math_tutor2020, ikleyn:
Answer by greenestamps(13215)   (Show Source):
You can put this solution on YOUR website!

Note that the problem does not ask us to find the values of a, b, and c.

In fact there are an infinite number of triples of numbers a, b, and c which all give the same value for . 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 is the same.

ANSWER: .

Let's use algebra to show that 54 is always the answer.

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.

So another way to solve the problem is to look for examples of expressions of the form in the given expression .

.

But eliminating b from the original two equations gives us . And so

ANSWER: 54

Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!

This is a system of 2 equations with 3 unknowns.
The fact we have more unknowns than equations leads to "infinitely many solutions" for this system.
It turns out that each solution is of the form (a,b,c) = (-9-c,6+c,c) which I explain in a later section below.

Let's say c = 0
b-c = 6
b-0 = 6
b = 6
Then,
a+b = -3
a+6 = -3
a = -3-6
a = -9
Or you could say
(a,b,c) = (-9-c,6+c,c)
(a,b,c) = (-9-0,6+0,0)
(a,b,c) = (-9,6,0)

Therefore,

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Another example

Let c = 1
b-c = 6
b-1 = 6
b = 6+1
b = 7
Then,
a+b = -3
a+7 = -3
a = -3-7
a = -10
Or
(a,b,c) = (-9-c,6+c,c)
(a,b,c) = (-9-1,6+1,1)
(a,b,c) = (-10,7,1)

Therefore,

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One more example

Let's say c = 2
b-c = 6
b-2 = 6
b = 6+2
b = 8
Then,
a+b = -3
a+8 = -3
a = -3-8
a = -11
Or
(a,b,c) = (-9-c,6+c,c)
(a,b,c) = (-9-2,6+2,2)
(a,b,c) = (-11,8,2)

Therefore,

It appears we keep landing on 54.
Is this a coincidence? Or is this always going to happen?
The next section will shed light on that.

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A more generalized approach.

b-c = 6
b = 6+c
a+b = -3
a+(6+c) = -3
a = -3-6-c
a = -9-c

We have
a = -9-c
b = 6+c
c = c
in which we can say
(a,b,c) = (-9-c,6+c,c)
This confirms that the system a+b = -3 and b-c = 6 has infinitely many solutions.

So,

This proves that if a+b = -3 and b-c = 6, then will always land on 54.

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Answer: 54

Answer by ikleyn(52903)   (Show Source):
You can put this solution on YOUR website!
.
if a + b = -3 and b - c = 6, find the value of .
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        Here is another solution.

if a + b = -3 and b - c = 6, then a = -3 - b (1) and c = b - 6 (2) Substitute these expressions (1) and (2) into . You will get then 2a^2 - 3b^2 + c^2 = 2*(-3-b)^2 - 3b^2 + (b-6)^2 = 2*(9 + 6b + b^2) - 3b^2 + b^2 - 12b + 36 = = 18 + 12b + 2b^2 - 3b^2 + b^2 - 12b + 36 = 54 + (12b-12b) + (3b^2-3b^2) = 54. ANSWER. Under the given constrains, = 54.

Solved.

SOLUTION: if a + b = -3 and b - c = 6, find the value of {{{2a^2 - 3b^2 + c^2}}}