Math, asked by jatish2, 1 year ago

plz help ......
(x+y) (2x+3y)-(x+y) (x+1)

Answers

Answered by Anonymous
2
Here is ur ans:-
Simplifying
(x + y)(2x + 3y) + -1(x + y)(x + 1) = 0

Multiply (x + y) * (2x + 3y)
(x(2x + 3y) + y(2x + 3y)) + -1(x + y)(x + 1) = 0
((2x * x + 3y * x) + y(2x + 3y)) + -1(x + y)(x + 1) = 0

Reorder the terms:
((3xy + 2x2) + y(2x + 3y)) + -1(x + y)(x + 1) = 0
((3xy + 2x2) + y(2x + 3y)) + -1(x + y)(x + 1) = 0
(3xy + 2x2 + (2x * y + 3y * y)) + -1(x + y)(x + 1) = 0
(3xy + 2x2 + (2xy + 3y2)) + -1(x + y)(x + 1) = 0

Reorder the terms:
(3xy + 2xy + 2x2 + 3y2) + -1(x + y)(x + 1) = 0

Combine like terms: 3xy + 2xy = 5xy
(5xy + 2x2 + 3y2) + -1(x + y)(x + 1) = 0

Reorder the terms:
5xy + 2x2 + 3y2 + -1(x + y)(1 + x) = 0

Multiply (x + y) * (1 + x)
5xy + 2x2 + 3y2 + -1(x(1 + x) + y(1 + x)) = 0
5xy + 2x2 + 3y2 + -1((1 * x + x * x) + y(1 + x)) = 0
5xy + 2x2 + 3y2 + -1((1x + x2) + y(1 + x)) = 0
5xy + 2x2 + 3y2 + -1(1x + x2 + (1 * y + x * y)) = 0

Reorder the terms:
5xy + 2x2 + 3y2 + -1(1x + x2 + (xy + 1y)) = 0
5xy + 2x2 + 3y2 + -1(1x + x2 + (xy + 1y)) = 0

Reorder the terms:
5xy + 2x2 + 3y2 + -1(1x + xy + x2 + 1y) = 0
5xy + 2x2 + 3y2 + -1(1x + xy + x2 + 1y) = 0
5xy + 2x2 + 3y2 + (1x * -1 + xy * -1 + x2 * -1 + 1y * -1) = 0
5xy + 2x2 + 3y2 + (-1x + -1xy + -1x2 + -1y) = 0

Reorder the terms:
-1x + 5xy + -1xy + 2x2 + -1x2 + -1y + 3y2 = 0

Combine like terms: 5xy + -1xy = 4xy
-1x + 4xy + 2x2 + -1x2 + -1y + 3y2 = 0

Combine like terms: 2x2 + -1x2 = 1x2
-1x + 4xy + 1x2 + -1y + 3y2 = 0

Solving
-1x + 4xy + 1x2 + -1y + 3y2 = 0

Solving for variable 'x'.
This equation cannot be determined..

jatish2: sorry factorise this
jatish2: i forgot to teold that
shanu85: hlo
Answered by Pavi789
0
=x+y[2x+3y-(x+1)]
=x+y[2x+3y-x-1]
=x+y[2x-x+3y-1]
=x+y[x+3y-1]
=x^2+3xy-x+xy+3y^2-y
=x^2+3xy+xy-x+3y^2-y
=x^2+4xy-x+3y^2-y
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