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Pls helpp with que 4a and 4b

Pls helpp with que 4a and 4b-example-1
User Iulia
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1 Answer

1 vote

Answer:

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))

User Andrei Tita
by
4.8k points