1. ## Simplifying expressions

Ok this is probably REALLY easy, but i have this as part of a few tests I have to do for college work, i dont take maths and havent done since school (about 5 years ago) so im really rusty on this sort of stuff so if anyone could help me out with the answers and if possible how they did it, it would be of great help!

Ok here it goes!

(------ means divide and (sq) means squared!)

3x - 6
--------
4x - 8

Question 2

2a - b
----------
4a(sq) - b(sq)

Question 3

4a - 2 x 5a(squared) - 25
---------- ------------------
3a(sq) - 15 8a - 4

Ok! any help with these is much appreciated!!!!!!!

sorry one more!

x - 3 - x-5
-------- ------------
4 3

Thank you!

2. Originally Posted by jimmybob
Ok this is probably REALLY easy, but i have this as part of a few tests I have to do for college work, i dont take maths and havent done since school (about 5 years ago) so im really rusty on this sort of stuff so if anyone could help me out with the answers and if possible how they did it, it would be of great help!

Ok here it goes!

(------ means divide and (sq) means squared!)

3x - 6
--------
4x - 8

Question 2

2a - b
----------
4a(sq) - b(sq)

Question 3

4a - 2 x 5a(squared) - 25
---------- ------------------
3a(sq) - 15 8a - 4

Ok! any help with these is much appreciated!!!!!!!

sorry one more!

x - 3 - x-5
-------- ------------
4 3

Thank you!

1) $\frac{{3x - 6}}{{4x - 8}} = \frac{{3(x - 2)}}{{4(x - 2)}} = \frac{3}{4}$

2) $\frac{{2a - b}}{{4a^2 - b^2 }} = \frac{{2a - b}}{{(2a - b)(2a + b)}} = \frac{1}{{2a + b}}$

3) $\frac{{4a - 2}}{{3a^2 - 15}} \cdot \frac{{5a^2 - 25}}{{8a - 4}} = \frac{{2(2a - 1)}}{{3(a^2 - 5)}} \cdot \frac{{5(a^2 - 5)}}{{4(2a - 1)}} = \frac{2}{3} \cdot \frac{5}{4} = \frac{{10}}{{12}} = \frac{5}{6}$

4) $\frac{{x - 3}}{4} - \frac{{x - 5}}{3} = \frac{{3(x - 3) - 4(x - 5)}}{{12}} = \frac{{3x - 9 - 4x + 20}}{{12}} = \frac{{11 - x}}{{12}}$

3. Wow thats great, thank you!

Just a few more i need to complete i have

Q1. 8p - 3q +5p - 2q + 4p

Q2. 5x + 5y divide 4x + 4y
---------- --------------
2x(sq) +6x 6x

Q3. x + x + x
-- -- --
4 + 5 + 6

Q4. 5x - 2 - 2x - 5
-------- ---------
10 15

Sorry for asking so many questions

Really appreciate the help!

4. Originally Posted by jimmybob
...
Just a few more...

Q1. 8p - 3q +5p - 2q + 4p

Q2. 5x + 5y divide 4x + 4y
---------- --------------
2x(sq) +6x 6x

Q3. x + x + x
-- -- --
4 + 5 + 6

Q4. 5x - 2 - 2x - 5
-------- ---------
10 15
Hello, Jimmybob,

to Q1: you can only add the same variables:

8p - 3q +5p - 2q + 4p = 8p + 5p + 4p - 3q - 2q = 17p - 5q

to Q2: I assume that you know that division by a fraction could be rewritten as a product by the inverse fraction:
$\frac{5x+5y}{2x^2+6x} \div \frac{4x+4y}{6x} = \frac{5x+5y}{2x^2+6x} \cdot \frac{6x}{4x+4y}=$.
Now factorize and cancel out equal factors:

$\frac{5(x+y)}{2x(x+3)} \cdot \frac{6x}{4(x+y)} =\frac{15}{4(x+3)}$

to Q3: The LCM of 4, 5 and 6 is 60. That's the new denominator:
$\frac{x}{4}+\frac{x}{5}+\frac{x}{6}=x\left(\frac{1 5}{60}+\frac{12}{60}+\frac{10}{60}\right) = \frac{37}{60}x$

to Q4: The LCM of 10 and 15 is 30. This is the new denominator:

$\frac{5x-2}{10}-\frac{2x-5}{15}=\frac{3(5x-2)}{3\cdot 10}-\frac{2(2x-5)}{2 \cdot 15}= \frac{11x+4}{30}$

EB

5. Hello, jimmybob!

$Q1)\;\;8p - 3q +5p - 2q + 4p$

We can combine only 'like' terms ("apples and oranges").

We have: . $(8p + 5p + 4p) + (-3q - 2q) \;= \;17p - 5q$

$Q2)\;\;\frac{5x + 5y}{2x^2+6x} \div \frac{4x + 4y}{6x}$

To divide fractions: invert (the second fraction) and multiply.

We have: . $\frac{5x + 5y}{2x^2 + 6x}\cdot \frac{6x}{4x+4y}$

Factor: . $\frac{5(x+y)}{2x(x+3)} \cdot \frac{6x}{4(x+y)}$

Reduce: . $\frac{5(x+y)}{\not{2}\not{x}(x+3)} \cdot \frac{\not{6}^3\not{x}}{4(x+y)}$ .
and the (x+y) also cancels out

And we have: . $\frac{15}{4(x+3)}$

$Q3)\;\;\frac{x}{4} + \frac{x}{5} + \frac{x}{6}$

To add or subtract fractions, they must have a common denominator.
The LCD for this problem is $60.$
We must convert all the fractions to have $60$ in their denominators.

We have: . $\frac{15}{15}\!\cdot\!\frac{x}{4} + \frac{12}{12}\!\cdot\!\frac{x}{5} + \frac{10}{10}\!\cdot\!\frac{x}{6} \:=\:\frac{15x}{60} + \frac{12x}{60} + \frac{10x}{60}$

Now they can be added: combine the numerators and "keep" the denominator.

. . $\frac{15x + 12x + 10x}{60} \;=\;\frac{37x}{60}$

$Q4)\;\;\frac{5x-2}{10} - \frac{2x-5}{15}$

The LCD is $30$.

We have: . $\frac{3}{3}\!\cdot\!\frac{5x-2}{10} - \frac{2}{2}\!\cdot\!\frac{2x-5}{15} \;= \;\frac{15x - 6}{30} - \frac{4x - 10}{30}$

Combine: . $\frac{(15x - 6) - (4x - 10)}{30} \;= \;\frac{15x - 6 - 4x + 10}{30} \;= \;\frac{11x + 4}{30}$