Help me with this algebraic expressions.?

Question

Please help to to factorise these expressions. Show all working and steps clearly.

a) px -py + qx - qy

b) 6ab + 6b^ -a - b

The sign ^ is square or power to 2

Thanks a lot in advance.

Answer 1

Factor by grouping:
a) px -py + qx - qy =
p(x - y) + q(x - y) =
(p + q)(x - y)

b) 6ab + 6b^2 -a - b =
6b(a + b) - (a + b) =
(6b - 1)(a + b)

Answer 2

a) px - py + qx - qy

a lot of ways to answer this one, you could try:
( px - py ) + ( qx - qy )
= p ( x - y ) + q ( x - y )
= ( x - y ) ( p + q )
OR
px + qx - py - qy
= ( px + qx ) - ( py + qy )
= x ( p + q ) - y ( p + q )
= ( p + q ) ( x - y )
= ( x - y ) ( p + q )

b) 6ab + 6b^2 - a - b

similarly, for this one, you could try:

( 6ab + 6b^2 ) - ( a + b )
= 6b ( a + b ) - ( a + b )
= ( a + b ) ( 6b - 1 )

OR

( 6ab - a ) + ( 6b^2 - b )
= a ( 6b - 1 ) + b ( 6b - 1 )
= ( 6b - 1 ) ( a + b )
= ( a + b ) ( 6b - 1 )

*cheers*

Answer 3

a) px - py + qx - qy <--assumed there are two parts
= p(x-y) + q(x-y) <-- take the common factors of the two
= (x-y)(p+q) <-- x-y is common for both

b) 6ab + 6b^ - a - b
= 6ab - a + 6b^ - b <-- rearrange
= a(6b-1) + b(6b-1) <-- (6b-1) is common
=(6b-1)(a+b)

It is easier if you do it this way because the other method you have to change the sign and you might do it wrongly..

And which grade are you in?

Answer 4

to factor the first one:

factor out a p from the frist two expressions to get:

p(x-y) + qx-qy

then factor out a q from the second half to get:

p(x-y) + q(x-y)

Now, notice that both p and q have x-y as a factor, so now we can simplyfy this to:

(p+q)(x-y)


and for the second one, the same techicque is used:
place alike terms together to get:

6ab-a+6b^-b

now, factor out an a from the first part and a b from the second part to get:

a(6b-1)+b(6b-1)

Again, notice that both terms have a 6b-1 as a factor, so simplfy this to:

(a+b)(6b-1)

Hope this helps...:)

Answer 5

a)

px - py + qx -qy
px + qx - py - qy
x(p+q) -y(p+q)

take out the common factor (p+q)
(x-y)(p+q)

b)
6ab + 6b^ -a - b
a (6b - 1) + b (6b - 1)
take out the common factor (6b - 1)
(6b - 1)(a+b)

Answer 6

I think its:

a) px-py+qx-qy
x(p+q)-y(p+q)
(x-y)(p+q)

b) 6ab-6b^2-a-b
6b(a+b)-(a+b)
(a+b)(6b-1)

Answer 7

a) px-py + qx-qy
p(x-y) + q(x-y)
(p+q)(x-y)

b) 6ab + 6b^ -a -b
6ab - a + 6b^ - b
a(6b-1) + b(6b-1)
(a + b)(6b-1)

Answer 8

interior the simplest words, an algebraic expression is the 0.5 without the equivalent verify in an algebraic equation. expression: 2x+3 equation: 2x+3 = 7 Expressions are no longer solvable, and the variable purely symbolize some unknown quantity. Equations on the different hand, provide you the particularly unknown value.

Answer 9

a) p (x - y) + q (x - y)

b) a (6b - 1) + b (6b - 1)