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
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(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
6-3x = 5x-10x+10 6-3x = -5x+10 -3x+5x = 10-6 2x = 4 x = 2
No.