Assuming the two equations are:
3xy - 3 = 0 ............................................. 1
2xy = 0 ............................................. 2
Then
from eq 1 xy = 1
from eq 2 xy = 0
That is not possible.
Or, assuming the equation is 3xy - 3 + 2xy = 0
Then
5xy = 3
or
y = 3/5x
Substitute for any value of x to get the value of y
√18xy = √(9*2xy) = √9*√2xy = 3*√(2xy)
(2x2+4xy-3)-(x2-2xy-4)Answer: x2+6x+1
2xy*9 if x=3 and y = 2 2(3)(2)*9 12*9 108 You substitute x and y for their equivalent (3 and 2) then simply multiply 2*x*y, then multiply the answer of that by 9.
(3m2+4n)3
2x^2+3xy-4y2(4)+3(2)(-4)-(4)(-4)8-24+16=0
3xy and 2xy. You can see that 3xy and 2xy have something in common with each other. They both have xy and the end of them. Hope this helped.
-1
That equals 2xy + 3y which factors to y(2x + 3)
√18xy = √(9*2xy) = √9*√2xy = 3*√(2xy)
3xy
3xy - 3xy = 0
2xy substitute 3 for y and 5 for x 2(5)(3) = 30
the difference between 2x2 +4xy-3 and x2-2xy-4 is?
(x^2 + 2y)(x^3 - 2xy + y^3) = x^2(x^3 - 2xy + y^3) + 2y(x^3 - 2xy + y^3) Now, let's distribute each term: = x^2 * x^3 - x^2 * 2xy + x^2 * y^3 + 2y * x^3 - 2y * 2xy + 2y * y^3 Now, simplify each term: = x^5 - 2x^3y + x^2y^3 + 2x^3y - 4xy^2 + 2y^4 Now, combine like terms: = x^5 + x^2y^3 - 4xy^2 + 2y^4 So, the expanded form of (x^2 + 2y)(x^3 - 2xy + y^3) is x^5 + x^2y^3 - 4xy^2 + 2y^4.
2xy
(2xy)x=2x2y
2xy - 10x + 3y -15 = 2x(y - 5) + 3(y - 5) =(y - 5)(2x + 3)