Question

Pls helpp with que 4a and 4b

Answer 1

4a.
`\bold{(-x+7)/(6\left(2x-1\right))}`

4b.
`(-19x-4)/(\left(3x+1\right)\left(3x-1\right))`

Explanation:

The expression for problem (4a) is

`(x-1)/(4x-2)-(5-2x)/(3-6x)`

Least Common Multiplier of the denominators. The easiest way to do this is to simply multiply the denominators together.

LCM of
`4x-2,\:and\;3-6x`

But first, let's simplify the two expressions by factoring

`4x-2 = 2(2x-1)` and

`3-6x = -3(1-2x) = 3(2x - 1)`

`2(2x-1) -3(2x-1)`

Multiply the coefficients 2 and -3 to get -6 and find an expression that appears in both.

This gives LCM as
`6(2x-1)`

`\mathrm{Multiply\:each\:numerator\:by\:the\:same\:amount\:needed\:to\:multiply\:its}`
`\mathrm{corresponding\:denominator\:to\:turn\:it\:into\:the\:LCM}\:6\left(2x-1\right)`

For
`\:(x-1)/(4x-2)`
`\mathrm{multiply\:the\:denominator\:and\:numerator\:by\:}\:3` since
`3(2x-1) = 6x -1`

`(x-1)/(4x-2)=(\left(x-1\right)\cdot \:3)/(\left(4x-2\right)\cdot \:3)=(\left(x-1\right)\cdot \:3)/(6\left(2x-1\right))`

`\mathrm{For}\:(5-2x)/(3-6x):\:\mathrm{multiply\:the\:denominator\:and\:numerator\:by\:}\:-2` since

`-2(3-6x) = (-6 +12x) = -6(-1+12x) = -6 + 12x = 6(2x-1) = 12x - 6 = 6(2x-1)`

We get

`(5-2x)/(3-6x)=(\left(5-2x\right)\left(-2\right))/(\left(3-6x\right)\left(-2\right))=(-2\left(5-2x\right))/(6\left(2x-1\right))`

Since the denominators are the same, we can apply the fraction rule:
`(a)/(c)-(b)/(c)=(a-b)/(c)`

`=(\left(x-1\right)\cdot \:3-\left(-2\left(5-2x\right)\right))/(6\left(2x-1\right))`

Numerator
`\left(x-1\right)\cdot \:3-\left(-2\left(5-2x\right)\right)=\left(x-1\right)\cdot \:3+2\left(5-2x\right)` becomes

`3\left(x-1\right)+2\left(5-2x\right)` =
`3x - 3 + 10x - 4x = -x + 7`

Therefore the expression result is

`\bold{(-x+7)/(6\left(2x-1\right))}`

Part (b)

Note that
`9x^2-1 = (3x+1)(3x-1) \textrm{using the property} (a+b)(a-b) = a^2-b^2`

So the LCM is
`=\left(3x+1\right)\left(3x-1\right)` =
`(3x+1)(3x-1)`

`\mathrm{Multiply\:each\:numerator\:by\:the\:same\:amount\:needed\:to\:multiply\:its}`
`\mathrm{corresponding\:denominator\:to\:turn\:it\:into\:the\:LCM}\; (3x+1)(3x-1)`

Multiply the second term in the expression by
`3x+1` and subtract from the first term

==>
`(2x+3)/((3x+1)(3x-1)) - (7(3x+1))/((3x-1)(3x+1)`

==>
`(2x+3)/(9x^2-1) - (7(3x+1))/(9x^2-1)`

==>
`(2x+3-7\left(3x+1\right))/(\left(3x+1\right)\left(3x-1\right))`

==>
`(2x+3-7\left(3x+1\right))/(\left(3x+1\right)\left(3x-1\right))`

==>
`(2x+3-21x-7)/(\left(3x+1\right)\left(3x-1\right)) \\`

==>
`(-19x-4)/(\left(3x+1\right)\left(3x-1\right))`