# Factorisation/Algebra

• Feb 7th 2008, 12:58 AM
andrew2322
Factorisation/Algebra
Gah, i'm beat i give up... i lose, maths wins :(

Can someone help me solve these?

1.) (X-1)/2 - (x-2)/3 = 1

2.) 2x/3x+1 = 0.85

3.) (2x - 1)/x = 1

4.) (x-1)(x+3) = 5

5.) -10x^2 +15

6.) 1/4 (x+1/2)^2 - 1/4

Could someone please leave working out and explanations particularly with the factorisation ones. Thanks, appreciate it.
• Feb 7th 2008, 01:48 AM
earboth
Quote:

Originally Posted by andrew2322
Can someone help me solve these?

1.) (X-1)/2 - (x-2)/3 = 1

2.) 2x/3x+1 = 0.85

3.) (2x - 1)/x = 1

4.) (x-1)(x+3) = 5

5.) -10x^2 +15

6.) 1/4 (x+1/2)^2 - 1/4

to #1)

multiply the equation by 6:
$3(x-1) - 2(x-2) = 6$ ...... Expand the brackets, collect like terms, separate x at the LHS, constants to the RHS, I've got x = 5

to #2)

I assume that you mean
$\frac{2x}{3x+1}=0.85$ ...... Muliply by (3x+1), expand the bracket at the RHS, I've got $x = -\frac{17}{11}$

to #3)
multiply by x, collect the x at the RHS, divide by the coefficient of x (by the way, it's (-1)), I've got x = 1

to #4)

expand the brackets, collect like terms, solve the quadratic for x:

$x^2+2x-8=0~\iff~(x-2)(x+4)=0$...... A product equals zero if one factor equals zero. Thus:
$x-2=0~\vee~x+4=0~\iff~x=2~\vee~x=-4$

to #5

there is nothing to solve because this is only a term. (I assume that there is a typo)

to #6)

$\frac14 \cdot \left(x+\frac12\right)^2-\frac14 = \frac14 \cdot \left(\left(x+\frac12\right)^2-1\right) = \frac14 \cdot \left(\left(x+\frac12+1\right)\left(x+\frac12-1\right)\right)$ $= \frac14 \cdot \left(\left(x+\frac32\right)\left(x-\frac12\right)\right) =$
• Feb 7th 2008, 01:51 AM
andrew2322
Hey thanks for helping out
Question 5 is not a typo, Could you please explain on question 6, im still lost
• Feb 7th 2008, 02:04 AM
andrew2322
Sorry i have another quick question
hey sorry to bother you, but i really want to learn.

consider: 0.01(2y+0.6)^2 - 0.04

I took out the common factor here as 0.01 hence

0.01 (2y+0.6) ^ 2 - 4
0.01 (2y+0.6)^2 -(2)^2
0.01 (2y + 0.6-2)(2y+0.6+2)
0.01(2y-1.4)(2y+2.6)

i checked the answer and it appeared as 0.04(y-0.7)(y+1.3)

I realised it could possibly due to the simplification of the two 2y's but on other problems when i removed 1 '2y' the other side got the 2 chopped off as well which leads me to another answer 0.02 (y-0.07)(y+1.3) could you please explain how it works? thanks.
• Feb 7th 2008, 02:40 AM
earboth
Quote:

Originally Posted by andrew2322
...
0.01(2y-1.4)(2y+2.6)

i checked the answer and it appeared as 0.04(y-0.7)(y+1.3)

....

You've done a great job. Now continue:

$0.01(2y-1.4)(2y+2.6)= 0.01 \cdot \left(2(x-0.7)\right) \left(2(x+1.3) \right) = 4 \cdot 0.01 (x-0.7)(x+1.3)$

which will give you the result you found obviously at the back of your book.
• Feb 7th 2008, 06:29 PM
andrew2322
Sorry to bother again
Consider: (2x+4)(x+6)

If i were to remove the 2 as the common factor.

2(x+2)(x+3)

Why is the second term affected by me removing two as the common factor for the first term?