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.
That equals 2xy + 3y which factors to y(2x + 3)
3xy^(3)2xy(6) = 6x^(2)y^(9).
The greatest common factor of 4xy and -6xy is 2xy. To find the greatest common factor, we first factor out the numbers 4 and -6 to get 2(2xy) and -2(3xy). The greatest common factor is the highest common factor that divides both terms evenly, which in this case is 2xy.
√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.
(2x2+4xy-3)-(x2-2xy-4)Answer: x2+6x+1
18x3y5 + 9xy3 - 3xy = 3xy(6x2y4 + 3y2 - 1)