can you find it?

Question

find the set values of x for which:

a) 3(2x+1)>5-2x
b) 2x^2-7x + 3> 0
c) both 3(2x+1) > 5-2x and 2x^-7x +3 >0

thanks for your help in advance.

Answer 1

a) 3(2x+1)>5-2x
6x+3>5-2x
6x+2x>5-3
8x>2
x>1/4
b) 2x^2-7x + 3> 0
(2x-1)(x-3)>0
x>1/2 and x>3
c) both 3(2x+1) > 5-2x and 2x^-7x +3 >0
3(2x+1) > 5-2x
6x+3>5-2x
6x+2x>5-3
8x>2
x>1/4
2x^-7x +3 >0
(2x-1)(x-3)>0
x>1/2 and x>3

Answer 2

for a) what you should do is distribute the 3 into the () and combine all like terms, get x on one side (in other words isolate x), remember when you devide by a negative number, the inequality sign will have to flip but in this case this is not necessary.
x >= 1/4

for b) factor the polynomial into (2x-1)(x-3) and solve for x by setting both factors greater than zero, you should come up with x>1/2 and x>3

for c it will be the union

Answer 3

a) 3(2x+1) > 5 - 2x
=> 6x +3 > 5-2x
=> 6x + 2x > 5-3
=> 8x>2
=> x>1/4

b) 2x^2 - 7x + 3>0
2x^2 - 6x - x + 3>0
2x(x-3) -1 (x-3)>0
(x-3)(2x-1)>0
x>3, x>1/2

Answer 4

3(2x+1)>5-2x then
6x+3 > 5-2x
6x + 2x + 3 > 5
8x > 5-3
8x > 2
x > 1/4
You do the rest the same way

Doug

Answer 5

a) 6x+1>5-2x = 8x>4 = x>0.5