No, because it is not an equation (or inequality), but simply two expressions separated by a space.
n = x2 + 1 n = y3 - 1 x2 + 1 = y3 - 1 x2 = y3 - 2 x = (y3 - 2).5 This means that you need to find all cubes where subtracting 2 yields a square: Cubes: 1, 8, 27, 64, 125,... (-2): -1, 6, 25, 62, 123,... (27,25) works. n = 26
10xy3 * 8x5y3 = (10*8)*(x*x5)*(y3*y3) = 80x6y6
In algebra, you cannot solve 5x25 since there is no equation nor an inequality, only an expression.
4
125 goes into 1000 exactly 8 times. You can solve this by simply diving 1000 by 125, which would give you a result of 8.
Lets start by putting the known number on the other side. Y3 -7500y=125000. There is your start. Take it from here.
16y3 / 20y4 = 16/20 * y3/y4 = 4/5 * y3-4 = 4/5 * y-1 or 4/5y
6y-y3 = 3
x3-y3
-y3 + 7y3
You would need to know the value of either x or y, so that you could solve for the one independent variable.
In the absence of any equality signs and proper grammar with great difficulty if not impossible.
n = x2 + 1 n = y3 - 1 x2 + 1 = y3 - 1 x2 = y3 - 2 x = (y3 - 2).5 This means that you need to find all cubes where subtracting 2 yields a square: Cubes: 1, 8, 27, 64, 125,... (-2): -1, 6, 25, 62, 123,... (27,25) works. n = 26
y3 + 9y + 10y - 5/y + 2 can not be solved, as it is not an equation. If you would like to simplify it, you can reduce it a bit: = y3+ 19y - 5/y + 2
72
27-y3 factored completely = 24
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