3. [tex]\dfrac{9x+1}4>2x-1[/tex]
Multiply both sides by 4:
[tex]4\times\dfrac{9x+1}4>4\times(2x-1)\implies9x+1>8x-4[/tex]
Subtract [tex]8x[/tex] and 1 from both sides:
[tex]9x+1-8x-1>8x-4-8x-1\implies x>-5[/tex]
7. [tex]\dfrac x3-1<\dfrac x2+3[/tex]
Multiply both sides by 6; this choice is motivated by the fact that [tex]\dfrac63=2[/tex] and [tex]\dfrac62=3[/tex]. In other words, 6 is the least common multiple of 3 and 2, which allows us to eliminate the denominators of the fractions in this inequality.
[tex]6\left(\dfrac x3-1\right)<6\left(\dfrac x2+3\right)\implies2x-6<3x+18[/tex]
Subtract [tex]2x[/tex] and 18 from both sides:
[tex]2x-6-2x-18<3x+18-2x-18\implies -24<x[/tex]
8. Since these are all linear inequalities, I think this is supposed to read
[tex]2x-1>4-\dfrac12x\iff2x-1>4-\dfrac x2[/tex]
Multiply both sides by 2:
[tex]2(2x-1)>2\left(4-\dfrac x2\right)\implies4x-2>8-x[/tex]
Subtract [tex]-x[/tex] and -2 from both sides:
[tex]4x-2-(-x)-(-2)>8-x-(-x)-(-2)\implies 5x>10[/tex]
Divide both sides by 5:
[tex]\dfrac{5x}5>\dfrac{10}5\implies x>2[/tex]
12. [tex]12\left(\dfrac14+\dfracx3\right)>15[/tex]
12 is the least common multiple of 4 and 3; distributing it to both terms on the left gives
[tex]\dfrac{12}4+\dfrac{12x}3=3+4x[/tex]
so we have
[tex]3+4x>15[/tex]
Subtract 3 from both sides:
[tex]3+4x-3>15-3\implies4x>12[/tex]
Divide both sides by 4:
[tex]\dfrac{4x}4>\dfrac{12}4\implies x>3[/tex]
