QUESTION 16
The given inequality is [tex]n-3\geq \frac{n+1}{2}[/tex].
We first multiply through by the least common multiple, which is [tex]2[/tex] to get,
[tex]2(n-3)\geq \frac{n+1}{2} \times 2[/tex]
This simplifies to
[tex]2(n-3)\geq n+1[/tex]
We expand brackets to get,
[tex]2n-6\geq n+1[/tex]
We group like terms to get,
[tex]2n-n\geq 1+6[/tex]
[tex]\Rightarrow n\geq 7[/tex].
We now choose the values from the replacement set
{[tex]-10,-9,-8,...,8,9,10[/tex]}.
The solution set is therefore
{[tex]7,8,9,10[/tex]}
QUESTION 17
We want to solve [tex]\frac{2(x+2)}{3}\:<\:4[/tex]
We multiply through by 3 to get,
[tex]3\times \frac{2(x+2)}{3}\:<\:4\times 3[/tex]
This simplifies to, [tex]2(x+2)\:<\:12[/tex]
We now expand brackets to get,
[tex]2x+4\:<\:12[/tex]
We group like terms to get,
[tex]2x\:<\:12-4[/tex]
[tex]\Rightarrow 2x\:<\:8[/tex]
[tex]\Rightarrow x\:<\:4[/tex]
We now choose the values from the replacement set
{[tex]-10,-9,-8,...,8,9,10[/tex]}.
The solution set is therefore
{[tex]-10,-9,-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3[/tex]}
QUESTION 18
We want to solve [tex]1.3y-12\:<\:0.9y+4[/tex].
We group like terms to get,
[tex]1.3y-0.9y\:<\:4+12[/tex].
We simplify to get,
[tex]0.4y\:<\:16[/tex].
We divide through by 0.4 to get,
[tex]y\:<\:40[/tex].
This time the complete replacement is the solution set of the inequality because all the elements satisfies [tex]y\:<\:40[/tex].
The solution set is
{[tex]-10,-9,-8,...,8,9,10[/tex]}
QUESTION 19
We want to solve [tex]-20\geq 8+7k[/tex].
We group like terms to get,
[tex]-20-8\geq 7k[/tex]
This implies that,
[tex]-28\geq 7k[/tex]
We divide through by 7 to get,
[tex]-4\geq k[/tex]
We can rewrite this as [tex]k\leq -4[/tex].
We now choose the values from the replacement set {[tex]-10,-9,-8,...,8,9,10[/tex]}. The solution set is therefore
{[tex]-10,-9,-8,-7,-6,-5,-4[/tex]}
