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

Question

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 $b$

Answer 1

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

Answer 2

HINT

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