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Equations Iis it possible to solve for x?

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Equations Iis it possible to solve for x?

by Menzorra » Tue Mar 19, 2013 3:51 am
Hello,
I am having trouble with basic equations.
Most of you will probably think "this is to easy" but for me its not.
I am having trouble with math insight, my score on the gmat was 390.
I am retaking it and want to get at least 500 now.

The issue:
As the first image explains, is it possible to solve for x? Yes or No question.
3a+2b+x=8 and 12a+8b+2x=4

When I look at the explanation I get confused. its the second image.

Thank you in advance for helping me people :)

Explanation/My question:
[spoiler]why does it put 12+8b+4x=32 and then subtract it with 12a+8b+2x=4
the 4x comes out of nowhere to me, and I get the 32 Sum 1 + sum 2 (8*4=32)
I am lost here. [/spoiler]
There is another sum that has this type of math that makes me confused, but i am not sure if I can post 2 questions in 1 topic.

Question:
Image

Answer:

Image
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by Anju@Gurome » Tue Mar 19, 2013 4:01 am
3a + 2b + x = 8 ................... Equation (1)
12a + 8b + 2x = 4 ................... Equation (2)

Now, we need to solve for x.
In other words, we need to get rid of a and b and express x in some definite number.

How can we get rid of a and b?
Look at the equations :
  • Equation (2) has 12a which is exactly four times of what is there in equation (1), i.e. 3a
    Also, equation (2) has 8b which is again exactly four times of what is there in equation (1), i.e. 2b
Hence, if we multiply equation (1) with 4, we will get 4*(3a + 2b + x) = 4*8 ---> 12a + 8b + 4x = 32

So, now the equations are...
  • 12a + 8b + 4x = 32 ................... Equation (1A)
    12a + 8b + 2x = 4 ................... Equation (2)
We can now simply subtract one equation from the other and get rid of a and b.

Subtract equation (2) from equation (1), and we get...
--> (12a + 8b + 4x) - (12a + 8b + 2x) = 32 - 4
--> (4x - 2x) = 28
--> 2x = 28
--> x = 14

Hope that helps.
Anju Agarwal
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by Menzorra » Tue Mar 19, 2013 5:35 am
This made it clear, finally!
Thank you a lot.

I have another question I will type it when I get home.
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by Menzorra » Tue Mar 19, 2013 6:42 am
Can you explain me why if you have x^2-x-12=0. the step to solve it is (x-4)(x+3)=0?
It makes scense to do this, but the gmat seems to just randomize this with (x+4)(x-3)=0
You get two different answers in the end.

So how do I know which version of 12 i have to take? the (x-4)(x+3) or the (x+4)(x-3).

Thank you in advance.
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by Anju@Gurome » Tue Mar 19, 2013 7:26 am
Menzorra wrote:So how do I know which version of 12 i have to take? the (x-4)(x+3) or the (x+4)(x-3).
(x - 4)(x + 3) = x² - 4x + 3x - 12 = x² - x - 12 ---> This is the equation you started with.
(x + 4)(x - 3) = x² + 4x - 3x - 12 = x² + x - 12

Hence, the two products are actually different.

Now, as I understand, you are not sure how to express (x² - x - 12) as the product of two factors.

To understand that let us take a general expression of such equations (quadratic equations) : (x² + px + q) and the factorized form of the equation : (x - a)(x - b) = 0

Now, (x - a)(x - b) = x² - ax - bx + ab = x² - (a + b)x + ab
So, comparing the terms, p = -(a + b) and q = ab

Hence, the trick is to exploit this fact, i.e. to express the co-efficient (often called middle-term) as the negative of the sum and the constant as the product.

Here, sum = 1 and product = -12
Now, in how many ways -12 can be expressed as the product of two integers such that sum of the two integers = 1? Only one way : -12 = 4*(-3)

Therefore, x² - x - 12 = (x - 4)(x - (-3)) = (x - 4)(x + 3)

This is the logic behind the factorization.

Often you'll see we do the factorization as follows...
--> x² - x - 12 = 0
--> x² - 4x + 3x - 12 = 0 .......... [We are expressing -x as -4x + 3x as -4*3 = -12]
--> x(x - 4) + 3(x - 4) = 0
--> (x - 4)(x + 3) = 0

I hope I didn't confuse you much. :)
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by Menzorra » Tue Mar 19, 2013 8:05 am
It is a bit confusing but let me evaluate this.
If I am correct I can understand why I should choose the (x+4)(x-3) If you have the sum=1 and the product=-12 then 4*-3= -12 and 4-3=1 Is this correct? My question then, how do you figure out that the sum is 1?

If my thought with the sum is not correct then I still dont know why I should choose 4*-3 over -4*3.
And the answer was {-4,3} So this is not correct with the sum, the sum should be -1 then (-4+3=-1) :(
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by Anju@Gurome » Tue Mar 19, 2013 8:34 am
Menzorra wrote:If I am correct I can understand why I should choose the (x+4)(x-3) If you have the sum=1 and the product=-12 then 4*-3= -12 and 4-3=1 Is this correct? My question then, how do you figure out that the sum is 1?
Yes, that is correct.
Now, "how do you figure out that the sum is 1?"

Look at my previous post again.
the trick is to exploit this fact, i.e. to express the co-efficient (often called middle-term) as the negative of the sum and the constant as the product.
Hence, sum = negative of the co-efficient of x

Here, coefficient of x = -1 (as we have -x in the equation : x² - x - 12)
Hence, sum = -(-1) = 1

Let us take a couple of examples...
## x² - x - 6 = 0
  • Sum = -(-1) = 1 and product = -6
    Now, -6 = -1*6 = -2*3
    We can get a sum of 1 only if we take 2*(-3)
    Hence, x² - x - 6 = (x - 2)(x - (-3)) = (x - 2)(x + 3)
## x² + 5x + 6 = 0
  • Sum = -(5) = -5 and product = 6
    Now, 6 = 1*6 = 2*3
    We can get a sum of -5 only if we take (-2)*(-3)
    Hence, x² + 5x + 6 = (x - (-2))(x - (-3)) = (x + 2)(x + 3)
Hope that helps.
Anju Agarwal
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Backup Methods : General guide on plugging, estimation etc.
Wavy Curve Method : Solving complex inequalities in a matter of seconds.

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by Menzorra » Tue Mar 19, 2013 12:33 pm
I see, So I have to look closely to the " x" if this is -x, then the sum is positive and if its +x, the sum is negative.
I am sure the GMAT will have tricky questions that will confuse me or maybe with fractures.

Thank you for helping me with this problem!
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