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Solve Compound Inequalities

Solve compound inequalities with “and”

Solve Compound Inequalities with “and”

Now that we know how to solve linear inequalities, the next step is to look at compound inequalities. A compound inequality is made up of two inequalities connected by the word “and” or the word “or.” For example, the following are compound inequalities.

\[x+3>-4\ \text{and}\ 4x-5\le 3\]\[2(y+1)<0\ \text{or}\ y-5\ge -2\]

To solve a compound inequality means to find all values of the variable that make the compound inequality a true statement. We solve compound inequalities using the same techniques we used to solve linear inequalities. We solve each inequality separately and then consider the two solutions.

To solve a compound inequality with the word “and,” we look for all numbers that make both inequalities true. To solve a compound inequality with the word “or,” we look for all numbers that make either inequality true.

Let’s start with the compound inequalities with “and.” Our solution will be the numbers that are solutions to both inequalities known as the intersection of the two inequalities. Consider the intersection of two streets—the part where the streets overlap—belongs to both streets.

To find the solution of the compound inequality, we look at the graphs of each inequality and then find the numbers that belong to both graphs—where the graphs overlap.

For the compound inequality $x>-3$ and $x\le 2,$ we graph each inequality. We then look for where the graphs “overlap”. The numbers that are shaded on both graphs, will be shaded on the graph of the solution of the compound inequality. See .

We can see that the numbers between $-3$ and $2$ are shaded on both of the first two graphs. They will then be shaded on the solution graph.

Example

Try it.

Solve $6x-3<9$ and $2x+9\ge 3.$ Graph the solution and write the solution in interval notation.

Solution

	$6x-3<9\$	and	$2x+9\ge 3\$
Step 1. Solve each inequality.	$6x-3<9\$		$2x+9\ge 3\$
	$6x<12$		$2x\ge -6$
	$x<2\$	and	$x\ge -3$
Step 2. Graph each solution. Then graph the numbers that make both inequalities true. The final graph will show all the numbers that make both inequalities true—the numbers shaded on both of the first two graphs.
Step 3. Write the solution in interval notation.	$[-3,2)$
All the numbers that make both inequalities true are the solution to the compound inequality.

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Solve Compound Inequalities with “or”

To solve a compound inequality with “or”, we start out just as we did with the compound inequalities with “and”—we solve the two inequalities. Then we find all the numbers that make either inequality true.

Just as the United States is the union of all of the 50 states, the solution will be the union of all the numbers that make either inequality true. To find the solution of the compound inequality, we look at the graphs of each inequality, find the numbers that belong to either graph and put all those numbers together.

To write the solution in interval notation, we will often use the union symbol, $\cup$ to show the union of the solutions shown in the graphs.

Example

Try it.

Solve $5-3x\le -1$ or $8+2x\le 5.$ Graph the solution and write the solution in interval notation.

Solution

	$5-3x\le -1$	or	$8+2x\le 5\$
Solve each inequality.	$5-3x\le -1$		$8+2x\le 5\$
	$-3x\le -6$		$\ 2x\le -3\$
	$x\ge 2\$	or	$\ x\le -\frac{3}{2}$
Graph each solution.
Graph numbers that make either inequality true.
	$(\text{-}\infty ,-\frac{3}{2}]\cup [2,\infty )$

Example

Try it.

Solve $\frac{2}{3}x-4\le 3$ or $\frac{1}{4}(x+8)\ge -1.$ Graph the solution and write the solution in interval notation.

Solution

	$\frac{2}{3}x-4\le 3\$	or	$\frac{1}{4}(x+8)\ge -1\$
Solve each inequality.	$\ 3(\frac{2}{3}x-4)\le 3(3)$		$4\cdot \frac{1}{4}(x+8)\ge 4\cdot (-1)$
	$2x-12\le 9\$		$x+8\ge -4\$
	$2x\le 21\$		$x\ge -12\$
	$x\le \frac{21}{2}\$
	$x\le \frac{21}{2}\$	or	$x\ge -12\$
Graph each solution.
Graph numbers that make either inequality true.
	The solution covers all real numbers.
	$(\text{-}\infty ,\infty )$

Solve Applications with Compound Inequalities

Situations in the real world also involve compound inequalities. We will use the same problem solving strategy that we used to solve linear equation and inequality applications.

Recall the problem solving strategies are to first read the problem and make sure all the words are understood. Then, identify what we are looking for and assign a variable to represent it. Next, restate the problem in one sentence to make it easy to translate into a compound inequality. Last, we will solve the compound inequality.

Example

Try it.

Due to the drought in California, many communities have tiered water rates. There are different rates for Conservation Usage, Normal Usage and Excessive Usage. The usage is measured in the number of hundred cubic feet (hcf) the property owner uses.

During the summer, a property owner will pay $24.72 plus $1.54 per hcf for Normal Usage. The bill for Normal Usage would be between or equal to $57.06 and $171.02. How many hcf can the owner use if he wants his usage to stay in the normal range?

Solution

Identify what we are looking for.	The number of hcf he can use and stay in the “normal usage” billing range.
Name what we are looking for.	Let $x=$ the number of hcf he can use.
Translate to an inequality.	Bill is $24.72 plus $1.54 times the number of hcf he uses or $24.72+1.54x.$

Solve the inequality.
Answer the question.	The property owner can use 21–95 hcf and still fall within the “normal usage” billing range.

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Key Concepts

How to solve a compound inequality with “and”
1. Solve each inequality.
2. Graph each solution. Then graph the numbers that make both inequalities true. This graph shows the solution to the compound inequality.
3. Write the solution in interval notation.
Double Inequality
- A double inequality is a compound inequality such as $a\(\begin{array}{llllllllllllllllllllllllllllllllllllllllllllllllllll}\text{Other forms:} & & & \begin{array}{lllllllllllll}ax>b & & & \text{is equivalent to} & & & a>x & & & \text{and} & & & x>b \\ a\ge x\ge b & & & \text{is equivalent to} & & & a\ge x & & & \text{and} & & & x\ge b\end{array}\end{array}$
How to solve a compound inequality with “or”
1. Solve each inequality.
2. Graph each solution. Then graph the numbers that make either inequality true.
3. Write the solution in interval notation.

Solve Compound Inequalities

Solve Compound Inequalities with “and”

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

Try it.

$x<3$ and $x\ge 1$

Try it.

$x\le 4$ and $x>-2$

Solution

Try it.

$x\ge -4$ and $x\le -1$

Try it.

$x>-6$ and $x<-3$

Solution

Try it.

$5x-2<8$ and $6x+9\ge 3$

Try it.

$4x-1<7$ and $2x+8\ge 4$

Solution

Try it.

$4x+6\le 2$ and
$2x+1\ge -5$

Try it.

$4x-2\le 4$ and
$7x-1>-8$

Solution

Try it.

$2x-11<5$ and
$3x-8>-5$

Try it.

$7x-8<6$ and
$5x+7>-3$

Solution

Try it.

$4(2x-1)\le 12$ and
$2(x+1)<4$

Try it.

$5(3x-2)\le 5$ and
$3(x+3)<3$

Solution

Try it.

$3(2x-3)>3$ and
$4(x+5)\ge 4$

Try it.

$-3(x+4)<0$ and
$-1(3x-1)\le 7$

Solution

Try it.

$\frac{1}{2}(3x-4)\le 1$ and
$\frac{1}{3}(x+6)\le 4$

Try it.

$\frac{3}{4}(x-8)\le 3$ and
$\frac{1}{5}(x-5)\le 3$

Solution

Try it.

$5x-2\le 3x+4$ and
$3x-4\ge 2x+1$

Try it.

$\frac{3}{4}x-5\ge -2$ and
$-3(x+1)\ge 6$

Solution

Try it.

$\frac{2}{3}x-6\ge -4$ and
$-4(x+2)\ge 0$

Try it.

$\frac{1}{2}(x-6)+2<-5$ and
$4-\frac{2}{3}x<6$

Solution

Try it.

$-5\le 4x-1<7$

Try it.

$-3<2x-5\le 1$

Solution

Try it.

$5<4x+1<9$

Try it.

$-1<3x+2<8$

Solution

Try it.

$-8<5x+2\le -3$

Try it.

$-6\le 4x-2<-2$

Solution

Solve Compound Inequalities with “or”

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

Try it.

$x\le -2$ or $x>3$

Try it.

$x\le -4$ or $x>-3$

Solution

Try it.

$x<2$ or $x\ge 5$

Try it.

$x<0$ or $x\ge 4$

Solution

Try it.

$2+3x\le 4$ or
$5-2x\le -1$

Try it.

$4-3x\le -2$ or
$2x-1\le -5$

Solution

Try it.

$2(3x-1)<4$ or
$3x-5>1$

Try it.

$3(2x-3)<-5$ or
$4x-1>3$

Solution

Try it.

$\frac{3}{4}x-2>4$ or $4(2-x)>0$

Try it.

$\frac{2}{3}x-3>5$ or $3(5-x)>6$

Solution

Try it.

$3x-2>4$ or $5x-3\le 7$

Try it.

$2(x+3)\ge 0$ or
$3(x+4)\le 6$

Solution

Try it.

$\frac{1}{2}x-3\le 4$ or
$\frac{1}{3}(x-6)\ge -2$

Try it.

$\frac{3}{4}x+2\le -1$ or
$\frac{1}{2}(x+8)\ge -3$

Solution

Mixed practice

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

Try it.

$3x+7\le 1$ and
$2x+3\ge -5$

Try it.

$6(2x-1)>6$ and
$5(x+2)\ge 0$

Solution

Try it.

$4-7x\ge -3$ or
$5(x-3)+8>3$

Try it.

$\frac{1}{2}x-5\le 3$ or
$\frac{1}{4}(x-8)\ge -3$

Solution

Try it.

$-5\le 2x-1<7$

Try it.

$\frac{1}{5}(x-5)+6<4$ and
$3-\frac{2}{3}x<5$

Solution

Try it.

$4x-2>6$ or
$3x-1\le -2$

Try it.

$6x-3\le 1$ and
$5x-1>-6$

Solution

Try it.

$-2(3x-4)\le 2$ and
$-4(x-1)<2$

Try it.

$-5\le 3x-2\le 4$

Solution

Solve Applications with Compound Inequalities

In the following exercises, solve.

Try it.

Penelope is playing a number game with her sister June. Penelope is thinking of a number and wants June to guess it. Five more than three times her number is between 2 and 32. Write a compound inequality that shows the range of numbers that Penelope might be thinking of.

Try it.

Gregory is thinking of a number and he wants his sister Lauren to guess the number. His first clue is that six less than twice his number is between four and forty-two. Write a compound inequality that shows the range of numbers that Gregory might be thinking of.

Solution

$5\le n\le 24$

Try it.

Andrew is creating a rectangular dog run in his back yard. The length of the dog run is 18 feet. The perimeter of the dog run must be at least 42 feet and no more than 72 feet. Use a compound inequality to find the range of values for the width of the dog run.

Try it.

Elouise is creating a rectangular garden in her back yard. The length of the garden is 12 feet. The perimeter of the garden must be at least 36 feet and no more than 48 feet. Use a compound inequality to find the range of values for the width of the garden.

Solution

$6\le w\le 12$

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Simplify: $\frac{2}{5}\ (x+10).$
If you missed this problem, review .

Jawaabta muuji

$\frac{2}{5}x+4$
Simplify: $\text{-}(x-4).$
If you missed this problem, review .

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$-x+4$

Solve $6x-3<9$ and $2x+9\ge 3.$ Graph the solution and write the solution in interval notation.

Jawaabta muuji

	$6x-3<9\$	and	$2x+9\ge 3\$
Step 1. Solve each inequality.	$6x-3<9\$		$2x+9\ge 3\$
	$6x<12$		$2x\ge -6$
	$x<2\$	and	$x\ge -3$
Step 2. Graph each solution. Then graph the numbers that make both inequalities true. The final graph will show all the numbers that make both inequalities true—the numbers shaded on both of the first two graphs.
Step 3. Write the solution in interval notation.	$[-3,2)$
All the numbers that make both inequalities true are the solution to the compound inequality.

Solve the compound inequality. Graph the solution and write the solution in interval notation: $4x-7<9$ and $5x+8\ge 3.$

Jawaabta muuji
Solve the compound inequality. Graph the solution and write the solution in interval notation: $3x-4<5$ and $4x+9\ge 1.$

Jawaabta muuji

Solve $3(2x+5)\le 18$ and $2(x-7)<-6.$ Graph the solution and write the solution in interval notation.

Jawaabta muuji

	$3(2x+5)\le 18$	and	$2(x-7)<-6$
Solve each inequality.	$6x+15\le 18$		$2x-14<-6$
	$6x\le 3\$		$2x<8\$
	$x\le \frac{1}{2}\$	and	$x<4\$
Graph each solution.
Graph the numbers that make both inequalities true.
Write the solution in interval notation.	$(\text{-}\infty ,\frac{1}{2}]$

Solve the compound inequality. Graph the solution and write the solution in interval notation: $2(3x+1)\le 20$ and $4(x-1)<2.$

Jawaabta muuji
Solve the compound inequality. Graph the solution and write the solution in interval notation: $5(3x-1)\le 10$ and $4(x+3)<8.$

Jawaabta muuji

Solve $\frac{1}{3}x-4\ge -2$ and $-2(x-3)\ge 4.$ Graph the solution and write the solution in interval notation.

Jawaabta muuji

	$\frac{1}{3}x-4\ge -2$	and	$-2(x-3)\ge 4\$
Solve each inequality.	$\frac{1}{3}x-4\ge -2$		$-2x+6\ge 4\$
	$\frac{1}{3}x\ge 2\$		$-2x\ge -2$
	$x\ge 6\$	and	$x\le 1\$
Graph each solution.
Graph the numbers that make both inequalities true.
	There are no numbers that make both inequalities true. This is a contradiction so there is no solution.

Solve the compound inequality. Graph the solution and write the solution in interval notation: $\frac{1}{4}x-3\ge -1$ and $-3(x-2)\ge 2.$

Jawaabta muuji
Solve the compound inequality. Graph the solution and write the solution in interval notation: $\frac{1}{5}x-5\ge -3$ and $-4(x-1)\ge -2.$

Jawaabta muuji
Solve $-4\le 3x-7<8.$ Graph the solution and write the solution in interval notation.

Jawaabta muuji

Add 7 to all three parts.
Simplify.
Divide each part by three.
Simplify.
Graph the solution.
Write the solution in interval notation.
Solve the compound inequality. Graph the solution and write the solution in interval notation: $-5\le 4x-1<7.$

Jawaabta muuji
Solve the compound inequality. Graph the solution and write the solution in interval notation: $-3<2x-5\le 1.$

Jawaabta muuji

Solve $5-3x\le -1$ or $8+2x\le 5.$ Graph the solution and write the solution in interval notation.

Jawaabta muuji

	$5-3x\le -1$	or	$8+2x\le 5\$
Solve each inequality.	$5-3x\le -1$		$8+2x\le 5\$
	$-3x\le -6$		$\ 2x\le -3\$
	$x\ge 2\$	or	$\ x\le -\frac{3}{2}$
Graph each solution.
Graph numbers that make either inequality true.
	$(\text{-}\infty ,-\frac{3}{2}]\cup [2,\infty )$

Solve the compound inequality. Graph the solution and write the solution in interval notation: $1-2x\le -3$ or $7+3x\le 4.$

Jawaabta muuji
Solve the compound inequality. Graph the solution and write the solution in interval notation: $2-5x\le -3$ or $5+2x\le 3.$

Jawaabta muuji

Solve $\frac{2}{3}x-4\le 3$ or $\frac{1}{4}(x+8)\ge -1.$ Graph the solution and write the solution in interval notation.

Jawaabta muuji

	$\frac{2}{3}x-4\le 3\$	or	$\frac{1}{4}(x+8)\ge -1\$
Solve each inequality.	$\ 3(\frac{2}{3}x-4)\le 3(3)$		$4\cdot \frac{1}{4}(x+8)\ge 4\cdot (-1)$
	$2x-12\le 9\$		$x+8\ge -4\$
	$2x\le 21\$		$x\ge -12\$
	$x\le \frac{21}{2}\$
	$x\le \frac{21}{2}\$	or	$x\ge -12\$
Graph each solution.
Graph numbers that make either inequality true.
	The solution covers all real numbers.
	$(\text{-}\infty ,\infty )$

Solve the compound inequality. Graph the solution and write the solution in interval notation: $\frac{3}{5}x-7\le -1$ or $\frac{1}{3}(x+6)\ge -2.$

Jawaabta muuji
Solve the compound inequality. Graph the solution and write the solution in interval notation: $\frac{3}{4}x-3\le 3$ or $\frac{2}{5}(x+10)\ge 0.$

Jawaabta muuji

Jawaabta muuji

Identify what we are looking for.	The number of hcf he can use and stay in the “normal usage” billing range.
Name what we are looking for.	Let $x=$ the number of hcf he can use.
Translate to an inequality.	Bill is $24.72 plus $1.54 times the number of hcf he uses or $24.72+1.54x.$

Solve the inequality.
Answer the question.	The property owner can use 21–95 hcf and still fall within the “normal usage” billing range.

Due to the drought in California, many communities now have tiered water rates. There are different rates for Conservation Usage, Normal Usage and Excessive Usage. The usage is measured in the number of hundred cubic feet (hcf) the property owner uses.
During the summer, a property owner will pay $24.72 plus $1.32 per hcf for Conservation Usage. The bill for Conservation Usage would be between or equal to $31.32 and $52.12. How many hcf can the owner use if she wants her usage to stay in the conservation range?

Jawaabta muuji

The homeowner can use 5–20 hcf and still fall within the “conservation usage” billing range.
Due to the drought in California, many communities have tiered water rates. There are different rates for Conservation Usage, Normal Usage and Excessive Usage. The usage is measured in the number of hundred cubic feet (hcf) the property owner uses.
During the winter, a property owner will pay $24.72 plus $1.54 per hcf for Normal Usage. The bill for Normal Usage would be between or equal to $49.36 and $86.32. How many hcf will he be allowed to use if he wants his usage to stay in the normal range?

Jawaabta muuji

The homeowner can use 16–40 hcf and still fall within the “normal usage” billing range.
$x\le 4$ and $x>-2$

Jawaabta muuji
$x\ge -4$ and $x\le -1$
$x>-6$ and $x<-3$

Jawaabta muuji
$4x+6\le 2$ and
$2x+1\ge -5$
$4x-2\le 4$ and
$7x-1>-8$

Jawaabta muuji
$2x-11<5$ and
$3x-8>-5$
$7x-8<6$ and
$5x+7>-3$

Jawaabta muuji
$4(2x-1)\le 12$ and
$2(x+1)<4$
$5(3x-2)\le 5$ and
$3(x+3)<3$

Jawaabta muuji
$3(2x-3)>3$ and
$4(x+5)\ge 4$
$-3(x+4)<0$ and
$-1(3x-1)\le 7$

Jawaabta muuji
$\frac{1}{2}(3x-4)\le 1$ and
$\frac{1}{3}(x+6)\le 4$
$\frac{3}{4}(x-8)\le 3$ and
$\frac{1}{5}(x-5)\le 3$

Jawaabta muuji
$5x-2\le 3x+4$ and
$3x-4\ge 2x+1$
$\frac{3}{4}x-5\ge -2$ and
$-3(x+1)\ge 6$

Jawaabta muuji
$\frac{2}{3}x-6\ge -4$ and
$-4(x+2)\ge 0$
$\frac{1}{2}(x-6)+2<-5$ and
$4-\frac{2}{3}x<6$

Jawaabta muuji

Symbols used here

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

A \cup B,\ A \cap B,\ A \setminus B

union, intersection, difference
In either; in both; in A but not B.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\sup,\ \inf

supremum, infimum
Least upper bound, greatest lower bound.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\neq

not equal
The two sides are different.

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

imaginary unit
i² = −1.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Solve Compound Inequalities

Solve compound inequalities with “and”
Solve compound inequalities with “or”
Solve applications with compound inequalities
Solve each inequality.
Graph each solution. Then graph the numbers that make
Write the solution in interval notation.
Solve each inequality.
Graph each solution. Then graph the numbers that make either inequality true.

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

Ku day inaad ku

Parts of this page are adapted from OpenStax Intermediate Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

In ka badan Algebra

Linear equations Quadratic equations Systems of equations Inequalities Factoring Expanding Simplifying expressions Functions and graphs Exponential and logarithmic equations Polynomial equations Absolute value

	\(6x-3<9\\)	and	\(2x+9\ge 3\\)
Step 1. Solve each inequality.	\(6x-3<9\\)		\(2x+9\ge 3\\)
	\(6x<12\)		\(2x\ge -6\)
	\(x<2\\)	and	\(x\ge -3\)
Step 2. Graph each solution. Then graph the numbers that make both inequalities true. The final graph will show all the numbers that make both inequalities true—the numbers shaded on both of the first two graphs.
Step 3. Write the solution in interval notation.	\([-3,2)\)
All the numbers that make both inequalities true are the solution to the compound inequality.

	\(5-3x\le -1\)	or	\(8+2x\le 5\\)
Solve each inequality.	\(5-3x\le -1\)		\(8+2x\le 5\\)
	\(-3x\le -6\)		\(\ 2x\le -3\\)
	\(x\ge 2\\)	or	\(\ x\le -\frac{3}{2}\)
Graph each solution.
Graph numbers that make either inequality true.
	\((\text{-}\infty ,-\frac{3}{2}]\cup [2,\infty )\)

	\(\frac{2}{3}x-4\le 3\\)	or	\(\frac{1}{4}(x+8)\ge -1\\)
Solve each inequality.	\(\ 3(\frac{2}{3}x-4)\le 3(3)\)		\(4\cdot \frac{1}{4}(x+8)\ge 4\cdot (-1)\)
	\(2x-12\le 9\\)		\(x+8\ge -4\\)
	\(2x\le 21\\)		\(x\ge -12\\)
	\(x\le \frac{21}{2}\\)
	\(x\le \frac{21}{2}\\)	or	\(x\ge -12\\)
Graph each solution.
Graph numbers that make either inequality true.
	The solution covers all real numbers.
	\((\text{-}\infty ,\infty )\)

	\(6x-3<9\\)	and	\(2x+9\ge 3\\)
Step 1. Solve each inequality.	\(6x-3<9\\)		\(2x+9\ge 3\\)
	\(6x<12\)		\(2x\ge -6\)
	\(x<2\\)	and	\(x\ge -3\)
Step 2. Graph each solution. Then graph the numbers that make both inequalities true. The final graph will show all the numbers that make both inequalities true—the numbers shaded on both of the first two graphs.
Step 3. Write the solution in interval notation.	\([-3,2)\)
All the numbers that make both inequalities true are the solution to the compound inequality.

	\(3(2x+5)\le 18\)	and	\(2(x-7)<-6\)
Solve each inequality.	\(6x+15\le 18\)		\(2x-14<-6\)
	\(6x\le 3\\)		\(2x<8\\)
	\(x\le \frac{1}{2}\\)	and	\(x<4\\)
Graph each solution.
Graph the numbers that make both inequalities true.
Write the solution in interval notation.	\((\text{-}\infty ,\frac{1}{2}]\)

	\(\frac{1}{3}x-4\ge -2\)	and	\(-2(x-3)\ge 4\\)
Solve each inequality.	\(\frac{1}{3}x-4\ge -2\)		\(-2x+6\ge 4\\)
	\(\frac{1}{3}x\ge 2\\)		\(-2x\ge -2\)
	\(x\ge 6\\)	and	\(x\le 1\\)
Graph each solution.
Graph the numbers that make both inequalities true.
	There are no numbers that make both inequalities true. This is a contradiction so there is no solution.


Add 7 to all three parts.
Simplify.
Divide each part by three.
Simplify.
Graph the solution.
Write the solution in interval notation.