## Elementary Algebra

Hana must bowl at least 130 in the third game. The solution set is {x|x $\geq$ 130}.
Let x represent the score in the third game. Average for the three games = $\frac{144+176+x}{3}$ Because the average must be at least 150, we solve the following inequality. $\frac{144+176+x}{3}$ $\geq$ 150 Multiply both sides by 3. 144 + 176 + x $\geq$ 450 320 + x $\geq$ 450 x $\geq$ 130 Hana must bowl at least 130 in the third game. The solution set is {x|x $\geq$ 130}.