Question

Find many solutions to the system of inequalities:
​

Answer 1

1) Solve each inequality individually.

`\frac{2x-3}5 - \frac{4x-9}6 > 1 \\\\ 6(2x-3) - 5(4x-9) > 30 \\\\ (12x-18) - (20x+45) > 30 \\\\ -8x > 3 \\\\ x < -\frac38`

and

`5(x-1) + 7(x+2) > 3 \\\\ (5x-5) + (7x+14) > 3 \\\\ 12x > -6 \\\\ x > -\frac12`

Then the solution set is

`-\frac12 < x < -\frac38`

since -1/2 = -4/8 is smaller than -3/8.

2)

`\frac{x+1}2 - \frac{x+2}3 < \frac{x+12}6 \\\\ 3(x+1) - 2(x+2) < x+12 \\\\ (3x+3) - (2x+4) < x + 12 \\\\ x - 1 < x + 12 \\\\ -1 < 12`

This is true for all values of
`x`.

For the second inequality, (I use periods in place of commas because it renders better as TeX)

`0.3x - 19 \le 1.7x - 5 \\\\ -1.4x \le 14 \\\\ x \ge -10`

Taken together, the solution set is

`\boxed{x\ge-10}`

3)

`(x-6)^2 < (x-2)^2 - 8 \\\\ x^2 - 12x + 36 < (x^2 - 4x + 4) - 8 \\\\ -8x < -40 \\\\ x > 5`

and

`3(2x-1) - 8 < 34 - 3(5x-9) \\\\ (6x - 3) - 8 < 34 - (15x - 27) \\\\ 21x < 72 \\\\ x < (72)/(21) = \frac{24}7`

But
`x` cannot be both larger than 5 = 35/7 and smaller than 24/7, so there are no solutions.

4)

`\frac{3x-2}3 - \frac{4x+1}4 \le 1 \\\\ 4(3x-2) - 3(4x+1) \le 12 \\\\ (12x-8) - (12x + 3) \le 12 \\\\ -11 \le 12`

This is true for all
`x`.

`(x-1)(x-2) > (x+4)(x-7) \\\\ x^2 - 3x + 2 > x^2 - 3x - 28 \\\\ 2 > -28`

This is also true for all
`x`.

Any value of
`x` is a solution.