Question

Simplify the expression completely.

Answer 1

\[\frac{ 7x }{ 8x+3 }-\frac{ 6 }{ 8x ^{2}+3x }+\frac{ 2 }{ x }\]

Answer 2

You are again adding and subtracting fractions.
That means you need the LCD.

Answer 3

To find the LCD, take each denominator and factor it.

Answer 4

\(8x + 3 = 8x + 3\) This is already factored.
\(8x^2 + 3x = x(8x + 3)\) This is how this denominator factors.
\(x = x\) This is already factored.

Answer 5

Now that we have all denominators factored, we need to find the LCD of all denominators.

Answer 6

x and 8x+3

Answer 7

Correct.
That means the first fraction needs to be multiplied by x/x
The middle fraction is good.
The last fraction needs to by multiplied by (x + 3)/(x + 3).

Ok?

Answer 8

yes

Answer 9

\(\large \dfrac{x}{x} \cdot \dfrac{7x}{8x+3}-\dfrac{6}{x(8x+3)}+\dfrac{8x + 3}{8x + 3} \cdot \dfrac{2}{x} \)

Answer 10

yes

Answer 11

\(\large \dfrac{7x^2}{x(8x+3)}-\dfrac{6}{x(8x+3)}+\dfrac{16x + 6}{x(8x + 3)} \)

Answer 12

yes

Answer 13

Now we add and subtract the fractions:

\(\large \dfrac{7x^2 - 6 + 16x + 6}{x(8x + 3)} \)

\(\large \dfrac{7x^2 + 16x}{x(8x + 3)} \)

Answer 14

Now we factor the numerator to try to reduce the fraction.

Answer 15

\[\frac{ 7x+16 }{ 8x+3 }\]

Answer 16

\(\large \dfrac{x(7x + 16)}{x(8x + 3)}\)

\(\large \dfrac{\cancel{x}(7x + 16)}{\cancel{x}(8x + 3)}\)

\(\large \dfrac{7x + 16}{8x + 3}\)

Answer 17

Yes, you are correct.