How to common factor (a - b)x - (b - a)y?

(a- b)x - (b - a)y
= (a-b)x-(-1)(-b+a)y
= (a-b)x-(-1)(a-b)y
= (a-b)x+(a-b)y
= (a-b)(x+y)

2x^2(3a- 2b) + 5y(3a - 2b) + (-2b + 3a)
= 2x^2(3a- 2b) + 5y(3a - 2b) + (3a-2b)
= (3a-2b)(2x^2+5y+1)

Hope this helps.

(a- b)x - (b - a)y
=x(a-b)-y(-a+b)
=x(a-b)+y(a-b)
=(a-b)(x+y) answer//


2x^2(3a-2b)+5y(3a-2b)+(-2b+3a)
=2x^2(3a-2b)+5y(3a-2b)+(3a-2b)
=(3a-2b)(2x^2+5y+1) answer//

hi,
given( a-b)x -(b-a)y,
take the portion-(b-a)y alone.multiplying and dividing it with(-) same time,net value remains same.so [-(-)(b-a)y/(-)],(-)(-)=+,+b/(-)=-b,-a/(-)=+a
then[-(-)(b-a)y/(-)]=+(a-b)y and
(a-b)x-(b-a)y=(a-b)x+(a-b)y.Dividing with(a-b) both sides we get (x+y) reminder.
So
(a- b)x - (b - a)y=(a-b)+(x+y)

given.
2x²(3a- 2b) + 5y(3a - 2b) + (-2b + 3a)
=2x²(3a- 2b) + 5y(3a - 2b) +(1) (3a-2b)
taking out the common (3a-2b), result is
(3a-2b)(2x²+5y+1)

Howdy Deer Eyes, & I sure hope all is well. :)

As for your question,

(a -- b)X -- (b -- a)Y

Since X is multiplied by (a -- b), whereas Y is multiplied by the reverse, & since the X & Y terms are being subtracted from each other, multiply the (b -- a) by the -- 1.

When you write X -- Y, then what you're really saying is X + (-- 1)*Y. That's where the -- 1 comes from.

So, expanding yields:

(a -- b)X + (a -- b)Y =

(a -- b)(X + Y)

Question 2)

(2X^2)(3a -- 2b) + (5Y)(3a -- 2b) + (-- 2b + 3a)

Well, this is similar to the first question. Assuming you didn't copy anything wrong, that the last term really is (-- 2b + 3a), not (2b -- 3a), then notice that:

(3a -- 2b) is the same as (-- 2b + 3a). They're just re-arranged.

In this case, the common factor is (3a -- 2b). So we have:

(3a -- 2b)(2X^2 + 5Y + 1)

However, if you meant to write (2b -- 3a) for the last term, then notice that:

(-- 1)*(3a -- 2b) = (2b -- 3a)

Once again, we factor out (3a -- 2b). We get:

(3a -- 2b)(2X^2 + 5Y -- 1)

I sure hope that was as clear & as helpful as you needed it to be. :) Good luck, take care, & have a great day & a great weekend. :)

Cheers! :)

(a- b)x - (b - a)y
ax - bx - by + ay
ax + ay -bx - by
a (x + y) -b (x + y)
(x + y ) ( a -b )

(3a-2b)(2x^2+5y+1)

1)
(a - b)x - (b - a)y
= (a - b)x - [-(-b + a)y]
= (a - b)x + (a - b)y
= (a - b)(x + y)

2)
2x^2(3a - 2b) + 5y(3a - 2b) + (-2b + 3a)
= 2x^2(3a - 2b) + 5y(3a - 2b) + 1(3a - 2b)
= (3a - 2b)(2x^2 + 5y + 1)

Q:(a-b)x-(b-a)y
A:multiplying with '-' to (b-a)y we get (-b+a)-y which can be written as, (a-b)-y now in the equation
(a-b)x-(a-b)-y
take a-b common
a-b(x+y) //i know u r wndring how did -y bcm + this is cozz - * -y=y//

there you have it (a-b)(x+y)
now use the same method on the next sum of urs there multiply '-' with (-2a+3b) you will get it all the best

first-
(a-b)x-(b-a)y=ax-bx-by+ay=ax+ay-(bx+by)=a(x+y)-b(x+y)=(a-b)(x+y)
hope its right
second question is not clear to me

(a- b)x - (b - a)y
= ax- bx-by+ay
= (a-b)(x+y)