SOLUTION: Please help me solve the equation below. I seem to get the variables mixed up. This equation is not from a textbook. x2y – 3x – y when x = -1 and y = 1 Thank you for yo

Algebra ->  Equations -> SOLUTION: Please help me solve the equation below. I seem to get the variables mixed up. This equation is not from a textbook. x2y – 3x – y when x = -1 and y = 1 Thank you for yo      Log On


   

Question 73779: Please help me solve the equation below. I seem to get the variables mixed up. This equation is not from a textbook.

x2y – 3x – y when x = -1 and y = 1

Thank you for your help.
LadyK
Found 3 solutions by don vito, checkley75, bucky:
Answer by don vito(18) About Me  (Show Source):
You can put this solution on YOUR website!
x=-1 y=1
x2y-3x-y=(-1)*(2*1)-(3*-1)-1=0
there's your answer, po-po

Answer by checkley75(3666) About Me  (Show Source):
You can put this solution on YOUR website!
x^2y-3x-y
(-1)^2-3(-1)-1
1+3-1
4-1
3

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
As I understand your problem, you are to evaluate the expression:
.
%28x%5E%282%29%2Ay%29-%283%2Ax%29-y
.
when you are given that x = -1 and y = 1
.
When you substitute -1 for x and +1 for y the expression becomes:
.
%28-1%29%5E2%2A%281%29+-+3%2A%28-1%29+-+%281%29
.
Since %28-1%29%5E2 = (-1)*(-1) = 1 you can replace %28-1%29%5E2 by +1 to get:
.
%281%29%2A%281%29+-+3%2A%28-1%29+-+%281%29
.
The first product of (1)*(1) is just +1 so substitute +1 for it to get:
.
1+-+3%2A%28-1%29+-+%281%29
.
Next work on the middle term of -3*(-1). Multiplying the -1 times -3 results in +3 because
the rule is that multiplying a negative by a negative gives you a positive. So the -3*(-1)
can be replaced by +3 and the expression then becomes:
.
1+%2B+3+-+%281%29
.
The final term is simply -1 (or if you prefer a -(+1) equals -1). Substitute this
to get:
.
1+%2B+3+-+1 which reduces to 4+-1 and this further becomes just 3.
.
I hope this works for you. If I misunderstood your problem, just let me know in your response
by providing a correction to the problem statement and I'll correct the error.