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Given

$$\frac{a}{6x-1}-\frac1{3x-1}\equiv\frac{b}{(6x-1)(3x+1)}$$

Where $a$ and $b$ are both constants, find the values of $a$ and $b$

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How far could you get? Also, shouldn't be one of $3x+1$ and $3x-1$ be the same as the other one? –  Berci Feb 3 '13 at 21:15
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Are you sure that you have the correct denominators? I’d expect either $\frac1{3x+1}$ on the left or $\frac{b}{(6x-1)(3x-1)}$ on the right. –  Brian M. Scott Feb 3 '13 at 21:16
    
yes, the denominators are correct, thanks for your help! –  Fede_Felds Feb 3 '13 at 21:22
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Then there’s probably a typo in the problem because there are no values of $a$ and $b$ that make that an identity. –  Brian M. Scott Feb 3 '13 at 21:23
    
well maybe there is, I checked and double checked my textbook and many friends are also having problems with this particular exercise, maybe there is a mistake in the textbook. Thanks! –  Fede_Felds Feb 3 '13 at 21:29

2 Answers 2

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Assuming you mean that the denominator for the right side is to be $3x - 1$:

$$\frac{a}{6x-1}-\frac1{3x-1} = \frac{b}{(6x-1)(3x-1)}$$

$$\frac{a(3x-1)}{(6x-1)(3x-1)}-\frac{6x - 1}{(6x - 1)(3x-1)} = \frac{b}{(6x-1)(3x-1)}$$

$$ a(3x - 1) - 6x + 1 = b $$

Note that this is an identity, so it must hold for any value of $x$. Choose $x = 1/3$, which makes the $a$ term vanish:

$$ a(3(1/3) - 1) - 6(1/3) + 1 = b $$ $$ -2 + 1 = b $$ $$ -1 = b $$

Now returning to the equation:

$$ a(3x - 1) - 6x + 1 = b $$ $$ a(3x - 1) - 6x + 1 = -1 $$

Let $x = 1$. Then we have: $$ a(3 \cdot 1 - 1) - 6 + 1 = -1 $$ $$ 2a - 5 = -1 $$ $$ 2a = 4 $$ $$ a = 2 $$

So: $$ a = 2 $$ $$ b = -1 $$

Assuming you mean that the denominator for the left side is to be $3x + 1$:

$$\frac{a}{6x-1}-\frac{1}{3x+1} = \frac{b}{(6x-1)(3x+1)}$$ $$\frac{a(3x+1)}{(6x-1)(3x + 1)}-\frac{6x - 1}{(6x - 1)(3x+1)} = \frac{b}{(6x-1)(3x+1)}$$ $$a(3x + 1) - (6x - 1) = b$$ $$a(3x + 1) - 6x + 1 = b $$

Since this is an identity, it holds for any value of $x$. Choose $x = -1/3$, because it makes the $a$ term vanish. Then we have: $$a(3 \cdot -1/3 + 1) - 6 \cdot -1/3 + 1 = b $$ $$2 + 1 = b$$ $$3 = b$$

Returning to: $$a(3x + 1) - 6x + 1 = 3 $$

Let $x=1$. Now we have: $$4a - 6 + 1 = 3$$ $$4a - 5 = 3$$ $$4a = 8$$ $$a = 2$$

So: $$ a = 2 $$ $$ b = 3 $$

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great!! but just for curiosity, what if instead of using using -1/3 for the value of x, you use a different number such as 3? would it still be possible to find that a=2 and that b=3? –  Fede_Felds Feb 3 '13 at 23:47
    
@Fede_Felds, yes. You could choose two values of $x$, (ex. $x=1$ and $x=2$), and then solve the system of two equations for $a$ and $b$ (viz. $4a - 5 = b, 7a - 11 = b$). There's no good reason to do that as opposed to just choosing a value of $x$ which makes the coefficient of $a$ vanish, and this method is quicker. –  George V. Williams Feb 4 '13 at 1:20

HINT

Find common denominators for both sides, and set the coefficients to be equal to each other.

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