Which Compound Inequality is Represented by the Graph
Which Compound Inequality is Represented by the Graph
2.7: Solve Compound Inequalities
- Folio ID
- 17390
By the end of this section, you will be able to:
- Solve compound inequalities with “and”
- Solve compound inequalities with “or”
- Solve applications with compound inequalities
Before you become started, take this readiness quiz.
- Simplify: \(\frac{ii}{five}(x+10)\).
If you missed this trouble, review
[link]. - Simplify: \(−(x−4)\).
If you missed this trouble, review
[link].
Solve Compound Inequalities with “and”
Now that we know how to solve linear inequalities, the next step is to await at chemical compound inequalities. A
compound inequality
is made upwards of 2 inequalities connected past the give-and-take “and” or the word “or.” For case, the following are compound inequalities.
\[\begin{array} {lll} {x+3>−4} &{\text{and}} &{4x−v\leq 3} \\ {2(y+1)<0} &{\text{or}} &{y−5\geq −2} \\ \stop{array} \nonumber\]
A
compound inequality
is made up of two inequalities connected past the word “and” or the word “or.”
To solve a compound inequality means to find all values of the variable that make the compound inequality a true argument. We solve compound inequalities using the aforementioned techniques we used to solve linear inequalities. We solve each inequality separately and then consider the two solutions.
To solve a chemical compound inequality with the word “and,” we wait for all numbers that brand
both
inequalities true. To solve a compound inequality with the word “or,” we look for all numbers that make
either
inequality true.
Let’due south commencement with the compound inequalities with “and.” Our solution will be the numbers that are solutions to
both
inequalities known as the intersection of the ii inequalities. Consider the intersection of 2 streets—the part where the streets overlap—belongs to both streets.
To observe the solution of an “and” compound inequality, we look at the graphs of each inequality and and so find the numbers that belong to both graphs—where the graphs overlap.
For the compound inequality \(ten>−three\) and \(x\leq 2\), we graph each inequality. We so 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 Effigy \(\PageIndex{ane}\).
We can come across that the numbers between \(−3\) and \(2\) are shaded on both of the first two graphs. They volition then be shaded on the solution graph.
The number \(−3\) is not shaded on the first graph and then since it is not shaded on both graphs, it is not included on the solution graph.
The number two is shaded on both the first and 2nd graphs. Therefore, it is be shaded on the solution graph.
This is how we will show our solution in the next examples.
Solve \(6x−three<9\) and \(2x+7\geq 3\). Graph the solution and write the solution in interval note.
- Reply
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\(6x−three<9\) and \(2x+nine\geq iii\) Step 1.
Solve each
inequality.\(6x−three<nine\) \(2x+9\geq 3\) \(6x<12\) \(2x\geq −6\) \(x<2\) and \(x\geq −3\) Stride 2.
Graph each solution. Then graph the numbers that make both inequalities true. The final graph will bear witness all the numbers that brand both inequalities truthful—the numbers shaded on
both
of the first two graphs.
Step 3.
Write the solution in interval notation.\([−three,2)\) All the numbers that make both inequalities true are the solution to the chemical compound inequality.
Solve the compound inequality. Graph the solution and write the solution in interval notation: \(4x−7<9\) and \(5x+viii\geq 3\).
- Answer
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Solve the compound inequality. Graph the solution and write the solution in interval note: \(3x−iv<5\) and \(4x+9\geq i\).
- Answer
-
- Solve each inequality.
- Graph each solution. And then graph the numbers that make
both
inequalities true.
This graph shows the solution to the chemical compound inequality. - Write the solution in interval notation.
Solve \(iii(2x+5)\leq 18\) and \(2(x−vii)<−6\). Graph the solution and write the solution in interval notation.
- Answer
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\(iii(2x+5)\leq eighteen\) and \(2(x−7)<−half dozen\) Solve each
inequality.\(6x+15\leq 18\) \(2x−fourteen<−half-dozen\) \(6x\leq three\) \(2x<8\) \(x\leq \frac{1}{2}\) and \(x<iv\) Graph each
solution.
Graph the numbers
that make both
inequalities true.
Write the solution
in interval note.\((−\infty, \frac{ane}{2}]\)
Solve the chemical compound inequality. Graph the solution and write the solution in interval notation: \(2(3x+1)\leq xx\) and \(4(x−1)<2\).
- Answer
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Solve the chemical compound inequality. Graph the solution and write the solution in interval notation: \(5(3x−1)\leq 10\) and \(4(ten+3)<8\).
- Answer
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Solve \(\frac{1}{3}x−4\geq −ii\) and \(−2(x−three)\geq iv\). Graph the solution and write the solution in interval notation.
- Answer
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\(\frac{one}{iii}ten−4\geq −two\) and \(−2(ten−3)\geq 4\) Solve each inequality. \(\frac{1}{3}x−4\geq −two\) \(−2x+6\geq 4\) \(\frac{i}{3}ten\geq ii\) \(−2x\geq −2\) \(ten\geq half dozen\) and \(10\leq ane\) Graph each solution. 
Graph the numbers that
make both inequalities
true.
At that place are no numbers that brand both inequalities true. This is a contradiction so at that place is no solution.There are no numbers that make both inequalities true.
This is a contradiction so at that place is no solution.At that place are no numbers that brand both inequalities truthful.
This is a contradiction and then in that location is no solution.
Solve the compound inequality. Graph the solution and write the solution in interval notation: \(\frac{one}{4}x−iii\geq −1\) and \(−3(x−ii)\geq two\).
- Respond
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Solve the compound inequality. Graph the solution and write the solution in interval notation: \(\frac{i}{5}x−5\geq −iii\) and \(−iv(ten−1)\geq −two\).
- Respond
-
Sometimes nosotros take a compound inequality that can exist written more concisely. For example, \(a<ten\) and \(x<b\) can be written but as \(a<x<b\) and then we call it a
double inequality. The ii forms are equivalent.
A double inequality is a compound inequality such as \(a<ten<b\). Information technology is equivalent to \(a<10\) and \(x<b\).
\[\text{Other forms:} \quad \begin{array} {lllll} {a<10<b} &{\text{is equivalent to }} &{a<x} &{\text{and}} &{x<b} \\ {a\leq x\leq b} &{\text{is equivalent to }} &{a\leq x} &{\text{and}} &{x\leq b} \\ {a>x>b} &{\text{is equivalent to }} &{a>x} &{\text{and}} &{ten>b} \\ {a\geq x\geq b} &{\text{is equivalent to }} &{a\geq x} &{\text{and}} &{10\geq b} \\ \end{array} \nonumber\]
To solve a double inequality we perform the same performance on all three “parts” of the double inequality with the goal of isolating the variable in the eye.
Solve \(−4\leq 3x−7<8\). Graph the solution and write the solution in interval notation.
- Respond
-
\(-4 \leq 3x – 7 < eight\) Add together 7 to all three parts. \( -4 \,{\colour{red}{+\, seven}} \leq 3x – 7 \,{\color{cerise}{+ \,7}} < 8 \,{\color{red}{+ \,7}}\) Simplify. \( 3 \le 3x < xv \) Dissever each part by iii. \( \dfrac{3}{\color{blood-red}{3}} \leq \dfrac{3x}{\color{red}{3}} < \dfrac{15}{\colour{cerise}{3}} \) Simplify. \( one \leq x < 5 \) Graph the solution. 
Write the solution in interval annotation. \( [1, five) \)
When written as a double inequality, \(i\leq ten<5\), it is like shooting fish in a barrel to encounter that the solutions are the numbers caught between one and five, including one, simply non v. We can then graph the solution immediately as we did above.
Another fashion to graph the solution of \(one\leq x<5\) is to graph both the solution of \(x\geq 1\) and the solution of \(10<5\). We would then observe the numbers that brand both inequalities true as we did in previous examples.
Solve the compound inequality. Graph the solution and write the solution in interval notation: \(−v\leq 4x−1<7\).
- Answer
-
Solve the compound inequality. Graph the solution and write the solution in interval note: \(−3<2x−5\leq 1\).
- Reply
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Solve Chemical compound Inequalities with “or”
To solve a
compound inequality
with “or”, we start out merely as nosotros did with the compound inequalities with “and”—we solve the ii inequalities. Then we find all the numbers that make
either
inequality true.
Merely equally the Us is the union of all of the fifty states, the solution volition be the marriage of all the numbers that make either inequality truthful. To find the solution of the compound inequality, nosotros expect 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, nosotros will oftentimes use the
union symbol, \(\cup\), to show the wedlock of the solutions shown in the graphs.
- Solve each inequality.
- Graph each solution. Then graph the numbers that brand either inequality truthful.
- Write the solution in interval notation.
Solve \(5−3x\leq −1\) or \(viii+2x\leq 5\). Graph the solution and write the solution in interval annotation.
- Answer
-
\(5−3x\leq −1\) or \(8+2x\leq v\) Solve each inequality. \(5−3x\leq −ane\) \(viii+2x\leq v\) \(−3x\leq −6\) \(2x\leq −3\) \(x\geq 2\) or \(x\leq −\frac{3}{2}\) Graph each solution. 
Graph numbers that
make either inequality
true.
\((−\infty,−32]\cup[two,\infty)\)
Solve the compound inequality. Graph the solution and write the solution in interval notation: \(1−2x\leq −3\) or \(seven+3x\leq 4\).
- Answer
-
Solve the chemical compound inequality. Graph the solution and write the solution in interval notation: \(ii−5x\leq −3\) or \(5+2x\leq three\).
- Reply
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Solve \(\frac{2}{3}ten−4\leq three\) or \(\frac{1}{4}(x+8)\geq −i\). Graph the solution and write the solution in interval notation.
- Answer
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\(\frac{2}{three}x−4\leq 3\) or \(\frac{1}{iv}(x+viii)\geq −1\) Solve each
inequality.\(three(\frac{two}{three}x−four)\leq 3(3)\) \(4⋅\frac{1}{4}(ten+8)\geq iv⋅(−i)\) \(2x−12\leq 9\) \(ten+eight\geq −4\) \(2x\leq 21\) \(x\geq −12\) \(ten\leq \frac{21}{2}\) \(ten\leq \frac{21}{ii}\) or \(x\geq −12\) Graph each
solution.
Graph numbers
that brand either
inequality true.
The solution covers all real numbers. \((−\infty ,\infty )\)
Solve the compound inequality. Graph the solution and write the solution in interval note: \(\frac{3}{5}x−seven\leq −1\) or \(\frac{i}{iii}(x+six)\geq −2\).
- Reply
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Solve the compound inequality. Graph the solution and write the solution in interval annotation: \(\frac{3}{4}ten−3\leq 3\) or \(\frac{2}{5}(10+10)\geq 0\).
- Answer
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Solve Applications with Compound Inequalities
Situations in the existent world also involve chemical compound inequalities. We will use the aforementioned trouble 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, place what we are looking for and assign a variable to correspond it. Adjacent, restate the problem in one sentence to make it like shooting fish in a barrel to translate into a
compound inequality. Final, we will solve the compound inequality.
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 holding possessor 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 tin the possessor use if he wants his usage to stay in the normal range?
- Answer
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Place what we are looking for. The number of hcf he tin can employ and stay in the “normal usage” billing range. Name what we are looking for. Let ten=x= the number of hcf he can use. Translate to an inequality. Bill is $24.72 plus $ane.54 times the number of hcf he uses or \(24.72+1.54x\). \(\color{Cerulean}{\underbrace{\color{black}{\text{His bill will be between or equal to }$57.06\text{ and }$171.02.}}}\)
\(57.06 \leq 24.74 + 1.54x \leq 171.02 \)
Solve the inequality. \(57.06 \leq 24.74 + ane.54x \leq 171.02\)
\(57.06 \,{\color{red}{- \,24.72}}\leq 24.74 \,{\color{red}{- \,24.72}} + one.54x \leq 171.02 \,{\color{reddish}{- \,24.72}}\)
\( 32.34 \leq 1.54x \leq 146.three\)
\( \dfrac{32.34}{\color{blood-red}{1.54}} \leq \dfrac{1.54x}{\colour{carmine}{1.54}} \leq \dfrac{146.three}{\colour{red}{ane.54}}\)
\( 21 \leq x \leq 95 \)
Answer the question. The property possessor 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 h2o 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 belongings owner volition pay $24.72 plus $i.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 possessor use if she wants her usage to stay in the conservation range?
- Answer
-
The homeowner tin use \(v–20\) hcf and still fall within the “conservation usage” billing range.
Due to the drought in California, many communities have tiered h2o rates. At that place are dissimilar 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 wintertime, a holding owner will pay $24.72 plus $1.54 per hcf for Normal Usage. The pecker for Normal Usage would be betwixt or equal to $49.36 and $86.32. How many hcf will he be allowed to utilize if he wants his usage to stay in the normal range?
- Reply
-
The homeowner can use \(16–twoscore\) hcf and withal autumn within the “normal usage” billing range.
Access this online resources for additional instruction and exercise with solving compound inequalities.
- Compound inequalities
Key Concepts
-
How to solve a compound inequality with “and”
- Solve each inequality.
- Graph each solution. Then graph the numbers that make
both
inequalities truthful. This graph shows the solution to the compound inequality. - Write the solution in interval note.
-
Double Inequality
- A
double inequality
is a compound inequality such as \(a<10<b\). It is equivalent to \(a<10\) and \(x<b.\)Other forms: \[\begin{align*} a<x<b & & \text{is equivalent to} & & a<x\;\text{and}\;x<b \\
a≤10≤b & & \text{is equivalent to} & & a≤x\;\text{and}\;x≤b \\
a>x>b & & \text{is equivalent to} & & a>x\;\text{and}\;x>b \\
a≥ten≥b & & \text{is equivalent to} & & a≥x\;\text{and}\;x≥b \end{marshal*}\]
- A
-
How to solve a compound inequality with “or”
- Solve each inequality.
- Graph each solution. Then graph the numbers that make either inequality true.
- Write the solution in interval notation.
Glossary
- compound inequality
- A compound inequality is made upwards of two inequalities connected by the word “and” or the word “or.”
Which Compound Inequality is Represented by the Graph
Source: https://math.libretexts.org/Bookshelves/Algebra/Book%3A_Intermediate_Algebra_(OpenStax)/02%3A_Solving_Linear_Equations/2.07%3A_Solve_Compound_Inequalities
