Answer:
4a. [tex]\bold{\frac{-x+7}{6\left(2x-1\right)}}[/tex]
4b. [tex]\frac{-19x-4}{\left(3x+1\right)\left(3x-1\right)}[/tex]
Step-by-step explanation:
The expression for problem (4a) is
[tex]\frac{x-1}{4x-2}-\frac{5-2x}{3-6x}[/tex]
Least Common Multiplier of the denominators. The easiest way to do this is to simply multiply the denominators together.
LCM of [tex]4x-2,\:and\;3-6x[/tex]
But first, let's simplify the two expressions by factoring
[tex]4x-2 = 2(2x-1)[/tex] and
[tex]3-6x = -3(1-2x) = 3(2x - 1)[/tex]
[tex]2(2x-1) -3(2x-1)[/tex]
Multiply the coefficients 2 and -3 to get -6 and find an expression that appears in both.
This gives LCM as [tex]6(2x-1)[/tex]
[tex]\mathrm{Multiply\:each\:numerator\:by\:the\:same\:amount\:needed\:to\:multiply\:its}[/tex] [tex]\mathrm{corresponding\:denominator\:to\:turn\:it\:into\:the\:LCM}\:6\left(2x-1\right)[/tex]
For [tex]\:\frac{x-1}{4x-2}[/tex] [tex]\mathrm{multiply\:the\:denominator\:and\:numerator\:by\:}\:3[/tex] since [tex]3(2x-1) = 6x -1[/tex]
[tex]\frac{x-1}{4x-2}=\frac{\left(x-1\right)\cdot \:3}{\left(4x-2\right)\cdot \:3}=\frac{\left(x-1\right)\cdot \:3}{6\left(2x-1\right)}[/tex]
[tex]\mathrm{For}\:\frac{5-2x}{3-6x}:\:\mathrm{multiply\:the\:denominator\:and\:numerator\:by\:}\:-2[/tex] since
[tex]-2(3-6x) = (-6 +12x) = -6(-1+12x) = -6 + 12x = 6(2x-1) = 12x - 6 = 6(2x-1)[/tex]
We get
[tex]\frac{5-2x}{3-6x}=\frac{\left(5-2x\right)\left(-2\right)}{\left(3-6x\right)\left(-2\right)}=\frac{-2\left(5-2x\right)}{6\left(2x-1\right)}[/tex]
Since the denominators are the same, we can apply the fraction rule: [tex]\frac{a}{c}-\frac{b}{c}=\frac{a-b}{c}[/tex]
[tex]=\frac{\left(x-1\right)\cdot \:3-\left(-2\left(5-2x\right)\right)}{6\left(2x-1\right)}[/tex]
Numerator [tex]\left(x-1\right)\cdot \:3-\left(-2\left(5-2x\right)\right)=\left(x-1\right)\cdot \:3+2\left(5-2x\right)[/tex] becomes
[tex]3\left(x-1\right)+2\left(5-2x\right)[/tex] = [tex]3x - 3 + 10x - 4x = -x + 7[/tex]
Therefore the expression result is
[tex]\bold{\frac{-x+7}{6\left(2x-1\right)}}[/tex]
Part (b)
Note that [tex]9x^2-1 = (3x+1)(3x-1) \textrm{using the property} (a+b)(a-b) = a^2-b^2[/tex]
So the LCM is [tex]=\left(3x+1\right)\left(3x-1\right)[/tex] = [tex](3x+1)(3x-1)[/tex]
[tex]\mathrm{Multiply\:each\:numerator\:by\:the\:same\:amount\:needed\:to\:multiply\:its}[/tex] [tex]\mathrm{corresponding\:denominator\:to\:turn\:it\:into\:the\:LCM}\; (3x+1)(3x-1)[/tex]
Multiply the second term in the expression by [tex]3x+1[/tex] and subtract from the first term
==> [tex]\frac{2x+3}{(3x+1)(3x-1)} - \frac{7(3x+1)}{(3x-1)(3x+1}[/tex]
==> [tex]\frac{2x+3}{9x^2-1} - \frac{7(3x+1)}{9x^2-1}[/tex]
==> [tex]\frac{2x+3-7\left(3x+1\right)}{\left(3x+1\right)\left(3x-1\right)}[/tex]
==> [tex]\frac{2x+3-7\left(3x+1\right)}{\left(3x+1\right)\left(3x-1\right)}[/tex]
==> [tex]\frac{2x+3-21x-7}{\left(3x+1\right)\left(3x-1\right)} \\[/tex]
==> [tex]\frac{-19x-4}{\left(3x+1\right)\left(3x-1\right)}[/tex]
