1.5E: Exercises

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Sep 9, 2022
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- 1.5: Solve Linear Inequalities
- 1.6: Exponents and Radicals

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Practice Makes Perfect

Graph Inequalities

In the following exercises, graph each inequality on the number line and write in interval notation.

$x>3$
$x\leq −0.5$
$x\geq \frac{1}{3}$
$x\leq 5$
$x\geq −1.5$
$x<−\frac{7}{3}$
$−2<x<0$
$−5\leq x<−3$
$0\leq x\leq 3.5$
$−4<x<2$
$−5<x\leq −2$
$−3.75\leq x\leq 0$

Answer

$The solution for x is greater than 3 on a number line has a left bracket 3 with shading to the right. The solution in interval notation is 3 to infinity within parentheses.$
$The solution for x is less than or equal to negative 0.5 on a number line has a right bracket at negative 0.5 with shading to the left. The solution in interval notation is negative infinity to negative 0.5 within a parenthesis and a bracket.$
$The solution for x is greater than or equal to one-third on a number line has a left bracket at one-third with shading to the right. The solution in interval notation is one-third to infinity within a bracket and a parenthesis.$
$The solution for x is less than or equal to 5 on a number line has a right bracket with shading to the left. The solution in interval notation is negative infinity to 5 within a parenthesis and a bracket.$
$The solution for x is greater than or equal to negative 1.5 on a number line has a left bracket with shading to the right. The solution in interval notation is negative 1.5 to infinity within a bracket and a parenthesis.$
$The solution for x is less than negative seven-thirds on a number line has a right parenthesis with shading to the left. The solution in interval notation is negative infinity to negative seven-thirds within parentheses.$
$Negative 2 is less than x which is less than 0. There is an open circle at negative 2 and an open circle at 0 and shading between negative 2 and 0 on the number line. The interval notation is negative 2 and 0 within parentheses.$
$Negative 5 is less than or equal to x which is less than negative 3. There is a closed circle at negative 5 and an open circle at negative 3 and shading between negative 5 and negative 3 on the number line. The interval notation is negative 5 and negative 3 within a bracket and a parenthesis.$
$0 is less than or equal to x which is less than or equal to 3.5. There is a closed circle at 0 and a closed circle at 3.5 and shading between 0 and 3.5 on the number line. The interval notation is 0 and 3.5 within brackets.$
$Negative 4 is less than x which is less than 2. There is a open circle at negative 4 and an open circle at 2 and shading between negative 4 and 2 on the number line. The interval notation is negative 4 and 2 within parentheses.$
$Negative 5 is less than x which is less than or equal to 2. There is a open circle at negative 5 and a closed circle at negative 2 and shading between negative 5 and negative 2 on the number line. The interval notation is negative 5 and negative 2 within a parenthesis and a bracket.$
$Negative 3.75 is less than or equal to x which is less than or equal to 0. There is a closed circle at negative 3.75 and a closed circle at 0 and shading between negative 3.75 and 0 on the number line. The interval notation is negative 3.75 and 0 within brackets.$

Solve Inequalities

In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval notation.

$b+\frac{7}{8}\geq \frac{1}{6}$
$6y<48$
$40<\frac{5}{8}k$
$g−\frac{11}{12}<−\frac{5}{18}$
$7s<−28$
$\frac{9}{4}g\leq 36$
$−8v\leq 96$
$\frac{b}{−10}\geq 30$
$−7d>105$
$−18>\frac{q}{−6}$

Answer

$The solution is b is greater than or equal to negative seventeen twenty-fourths. The solution on a number line has a left bracket negative seventeen twenty-fourths with shading to the right. The solution in interval notation is negative seventeen twenty-fourths to infinity within a bracket and a parenthesis.$
$The solution is y is less than 8. The solution on a number line has a right parenthesis at 8 with shading to the left. The solution in interval notation is negative infinity to 8 within parentheses.$
$The solution is k is greater than 64. The solution on a number line has a left parenthesis at 64 with shading to the right. The solution in interval notation is 64 to infinity within parentheses.$
$The solution is g is less than twenty-three thirty-sixths. The solution on a number line has a right parenthesis at twenty-three thirty-sixths with shading to the left. The solution in interval notation is negative infinity to twenty-three thirty-sixths within parentheses.$
$The solution is s is less than negative 4. The solution on a number line has a right parenthesis at negative 4 with shading to the left. The solution in interval notation is negative infinity to negative 4 within parentheses.$
$The solution is g is less than or equal to 16. The solution on a number line has a right bracket at 16 with shading to the left. The solution in interval notation is negative infinity to 16 within parenthesis and a bracket.$
$The solution is v is greater than or equal to negative 12. The solution on a number line has a left bracket with shading to the right. The solution in interval notation is negative 12 to infinity within a bracket and a parenthesis.$
$The solution is b is less than or equal to negative 300. The solution on a number line has a right bracket at negative 300 with shading to the left. The solution in interval notation is negative infinity to negative 300 within a parenthesis and a bracket.$
$The solution is d is less than negative 15. The solution on a number line has a right parentheses with shading to the left. The solution in interval notation is negative infinity to negative 15 within parentheses.$
$The solution is q is greater than 108. The solution on a number line has a left parentheses at 108 with shading to the right. The solution in interval notation is 108 to infinity within parentheses.$

Now with Interval Notation

In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval notation.

$5u\leq 8u−21$
$9p>14p+18$
$9y+5(y+3)<4y−35$
$4k−(k−2)\geq 7k−26$
$6n−12(3−n)\leq 9(n−4)+9n$
$9u+5(2u−5)\geq 12(u−1)+7u$
$12v+3(4v−1)\leq 19(v−2)+5v$
$35k\geq −77$
$18q−4(10−3q)<5(6q−8)$
$−\frac{21}{8}y\leq −\frac{15}{28}$

Answer

$The solution is u is greater than or equal to 7. The solution on a number line has a left bracket at 7 with shading to the right. The solution in interval notation is 7 to infinity within a a bracket and a parenthesis.$
$The solution is p is less than eighteen fifths. The solution on a number line has a right parenthesis at eighteen fifths with shading to the left. The solution in interval notation negative infinity to eighteen fifths within parentheses.$
$The solution is y is less than negative 5. The solution on a number line has a right parenthesis at negative 5 with shading to the left. The solution in interval notation is negative infinity to negative 5 within parentheses.$
$The solution is k is less than or equal to 7. The solution on a number line has a right bracket at 7with shading to the left. The solution in interval notation is negative infinity to 7 within a parenthesis and a bracket.$
$The inequality is an identity. Its solution on the number line is shaded for all values. The solution in interval notation is negative infinity to infinity within parentheses.$
$The inequality is a contradiction. So, there is no solution. As a result, there is no graph on the number line or interval notation.$
$The inequality is a contradiction. So, there is no solution. As a result, there is no graph on the number line or interval notation.$
$The solution is k is greater than or equal to negative eleven fifthss. The solution on a number line has a left bracket at negative eleven fifths with shading to the right. The solution in interval notation is negative eleven fifths to negative infinity within a bracket and a parenthesis.$
$The inequality is a contradiction. So, there is no solution. As a result, there is no graph on the number line or interval notation.$
$The solution is y is greater than or equal to ten twenty-ninths. The solution on a number line has a left bracket at ten twenty-ninths with shading to the right. The solution in interval notation is ten twenty-ninths to infinity within a bracket and a parenthesis.$