Well, you can factor this.
(10X^2 + 5X)/5x
5X(2X + 1)/5X
cancel 5X; top and bottom
2X + 1
set to 0 and see
2X+ 1 = 0
2X = - 1
X = -1/2
-----------check to see what we get
[10(- 1/2)^2 + 5(- 1/2)/5(- 1/2)
10/4 - 5/2/- 5/2
5/2 - 5/2/- 5/2
= 0
----------- I think so because my TI-84 agrees with this answer
(2x - 1)(5x + 8)
5x + 10x doesn't need factorization as you can just add them up (5x+10x=15x). However, if you really want to, 5x+10x=5x(1+2)=5x(3)=15x
0.5
5
6-3x = 5x-10x+10 6-3x = -5x+10 -3x+5x = 10-6 2x = 4 x = 2
10x - 5x + 5x = 10x
-10x + 5x = -5x
10x ÷ 5x = 2
(2x - 1)(5x + 8)
10x+7 = 5x+32 10x-5x = 32-7 5x = 25 x = 5
5x(x - 2)
If: 10x+7 = 5x+22 Then: 10x-5x = 22-7 And: 5x = 15 So: x = 3
5x/x-2
5
5x + 10x doesn't need factorization as you can just add them up (5x+10x=15x). However, if you really want to, 5x+10x=5x(1+2)=5x(3)=15x
5x + 4 = 10x - 5 10x - 5x = 4 + 5 5x = 9 x = 1.8
The question contains an expression. An expression does not have an answer - it cannot be solved.