Let the unknown number be represented by the variable '$x$'. The problem states that one-sixth of the number exceeds its one-ninth by 100.
We can translate the problem into an algebraic equation:
$ \frac{x}{6} = \frac{x}{9} + 100 $
To find the value of '$x$', we need to solve the equation:
Therefore, the number is 1800.
Check if the solution is correct:
The number is 1800.
Simplify: $3((\frac{5}{3})x^2 - 28x + 15) - 5(x^2 + 6x - 15)$
If m + n = 24, then (m - 16)³ + (n - 8)³ is ____.
For the following equations, what are the values of a and b to have infinitely many solutions?
ax + by = 2
3x - (5 - 2ay) = 6
If m + n = 24, then (m - 16)³ + (n - 8)³ is ____.
For the following equations, what are the values of a and b to have infinitely many solutions?
ax + by = 2
3x - (5 - 2ay) = 6
Simplify the following expression:
\(\frac{(x - y)^3 + (y - z)^3 + (z - x)^3}{(x - y)(y - z)(z - x)}\)