Find three consecutive integers whose sum is 216. (Hint: if n represents the smallest of the three integers, n + 1 and n + 2 represent the other two numbers.)

71, 72, 73

Step-by-step explanation:

So we want to find three consecutive integers that equal 216.

Let the first integer be n.

Then the second integer is n+1, and the third integer is n+2.

Thus:

Combine like terms:

Subtract 3 from both sides:

Divide 3 from both sides:

So, the first term is 71.

And the other two is 72 and 73.

And we are done :)

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)