0
What else can I help you with?
How do you factor x-y plus xy-1?
(x-y) + (xy - 1) = (x - 1)(y + 1)
Factor xy plus 2x plus 4y plus 8?
xy plus 2x plus 4y plus 8 or (xy+2x) + (4y+8) or x(y+2) + 4(y+2) or (x+4)(y+2)
The sum of two number is 10 and their product is 30 the sum of their reciprocals is?
Suppose the numbers are x and y. The sum of their reciprocals = 1/x + 1/y = y/xy + x/xy = (y+x)/xy = (x+y)/xy = 10/30 = 1/3
If the sum of two numbers is 12 and the product is 24 What is the sum of the reciprocals?
1/x+1/y y/xy+x/xy x+y/xy=12/24 12/24=1/2 1/2 is the answer.
The sum of two numbers is 10 and their product is 20 find the sum of the reciprocal of the number.?
1/x + 1/y y/xy + x/xy (y+x)/xy= 10/20 10/20=1/2 1/2 is the answer.
How do you factor xy x squared y?
If that's xy + x^2y, it factors to xy(x + 1) If it's something else, please re-submit your question with any plus signs written out.
How do you factor x-y plus xy-1?
(x-y) + (xy - 1) = (x - 1)(y + 1)
Factor 2 plus 2y plus x plus xy?
2+2y+x+xy=(x+2)(y+1)
How xy plus x plus y plus 1 will solve?
xy + x + y + 1 is an expression, not an equation. An expression cannot be solved.
Factor 2x squared plus 2x plus xy plus y?
4
X plus XY equals 8 solve for Y?
x+xy=8 xy=-x+8 y=-1+8/x
What is the answer to xy-x plus 2y-2?
0
What is 2 over X plus 1 over Y simplified they are fractions?
I understand you want [(2/x) + (1/y)] as a single fraction. The common denominator is (xy): 2/x = (2y)/(xy); and 1/y = x/(xy), so the answer is (2y + x) / (xy)
What is the inverse of y(2-x plus 3) plus 1?
The multiplicative inverse of 5y -xy + 1 is 1/5y -xy + 1 The additive inverse of 5y - xy + 1 is -5y + xy - 1
Are the factors of x2 are x and x plus 1?
The factors of x2 are x and x x times x plus 1 = x2 + x
How do you solve for x when y equals quantity x plus 2 divided by quantity x plus 10?
y=(x+2)/(x+10) xy+10y=x+2 xy-x=2-10y x(y-1)=2-10y x=(2-10y)/(y-1)
Factoring this expression Check by multiplying factors xy plus ay plus ab plus bx?
xy + ay + ab + bx = y(x + a) + b(a + x) = y(x + a) + b(x + a) = y(x + a) + b(x + a) = (y + b)(x + a) To check, multiply out the two brackets making sure that each pair is evaluated.
