What are the number of integral solutions of the equation 2x+2y+z=20 such that x>=0 , y>=0 , z>=0?
1 answer:
Now, for z to be an integer, the sum x+y must be a divisor of 10.
It has to be a positive divisor since z≥0. Also x≠y.
x≥0 and y≥0 so y can be equal to either 0 or 1. There are 2 solution in this case.
In this case, y can be equal to 0,1, or 2, but for y=1 ⇒ x=1, so there are two solutions.
y can be 0,1,2,3,4 or 5 - 6 solutions
y can be 0,1,2,3,4,5,6,7,8,9,10, but for y=5 ⇒ x=5, so 10 solutions.
2+2+6+10=20 solutions in total.
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We can do like we did before, distribute the denominator,
Answer:
Step-by-step explanation:
Given equation can be re written as
............(i)
Now it is given that y(π/2) = 2e
Applying value in (i) we get
ln(2e) = sin(π/2) + c
=> ln(2) + ln(e) = 1+c
=> ln(2) + 1 = 1 + c
=> c = ln(2)
Thus equation (i) becomes
ln(y) = sin(x) + ln(2)
ln(y) - ln(2) = sin(x)
ln(y/2) = sin(x)