Which Compound Inequality is Represented by the Graph

Which Compound Inequality is Represented by the Graph

2.7: Solve Compound Inequalities

  • Folio ID
    17390
  • Learning Objectives

    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.

    1. Simplify: \(\frac{ii}{five}(x+10)\).
      If you missed this trouble, review
      [link].
    2. 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.

    \[\begin{array} {lll} {x+3>−4} &{\text{and}} &{4x−v\leq 3} \\ {2(y+1)<0} &{\text{or}} &{y−5\geq −2} \\ \stop{array} \nonumber\]

    COMPOUND INEQUALITY

    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 \(\PageIndex{ane}\).

    The figure shows the graph of x is greater than negative 3 with a left parenthesis at negative 3 and shading to its right, the graph of x is less than or equal to 2 with a bracket at 2 and shading to its left, and the graph of x is greater than negative 3 and x is less than or equal to 2 with a left parenthesis at negative 3 and a right parenthesis at 2 and shading between negative 3 and 2. Negative 3 and 2 are marked by lines on each number line.
    Figure \(\PageIndex{one}\)

    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.

    Instance \(\PageIndex{1}\)

    Solve \(6x−three<9\) and \(2x+7\geq 3\). Graph the solution and write the solution in interval note.

    Reply
    \(6x−three<9\) 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.
    Example \(\PageIndex{2}\)

    Solve the compound inequality. Graph the solution and write the solution in interval notation: \(4x−7<9\) and \(5x+viii\geq 3\).

    Answer

    The solution is negative 1 is less than or equal to x which is less than 4. On a number line it is shown with a closed circle at negative 1 and an open circle at 4 with shading in between the closed and open circles. Its interval notation is negative 1 to 4 within a bracket and a parenthesis.

    SOLVE A COMPOUND INEQUALITY WITH “AND.”
    1. Solve each inequality.
    2. Graph each solution. And then graph the numbers that make
      both
      inequalities true.
      This graph shows the solution to the chemical compound inequality.
    3. Write the solution in interval notation.
    Example \(\PageIndex{4}\)

    Solve \(iii(2x+5)\leq 18\) and \(2(x−vii)<−6\). Graph the solution and write the solution in interval notation.

    Answer
    \(iii(2x+5)\leq eighteen\) 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.
    \((−\infty, \frac{ane}{2}]\)
    Example \(\PageIndex{5}\)

    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

    The solution is x is less than three-halves. On a number line it is shown with an open circle at three-halves with shading to its left. Its interval notation is negative infinity to three-halves within a parentheses.

    Example \(\PageIndex{6}\)

    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

    The solution is x is less than negative 1. On a number line it is shown with an open circle at 1 with shading to its left. Its interval notation is negative infinity to negative 1 within parentheses.

    Example \(\PageIndex{7}\)

    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
    \(\frac{one}{iii}ten−4\geq −two\) 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.

    Instance \(\PageIndex{viii}\)

    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

    The inequality is a contradiction. So, there is no solution. As a result, there is no graph of the number line or interval notation.

    Instance \(\PageIndex{9}\)

    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

    The inequality is a contradiction. So, there is no solution. As a result, there is no graph or the number line or interval notation.

    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.

    DOUBLE INEQUALITY

    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.

    Case \(\PageIndex{10}\)

    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.

    Example \(\PageIndex{12}\)

    Solve the compound inequality. Graph the solution and write the solution in interval note: \(−3<2x−5\leq 1\).

    Reply

    The solution is 1 is less than x which is less than or equal to 3. Its graph has an open circle at 1 and a closed circle at 3 with shading between the closed and open circles. Its interval notation is negative 1 to 3 within a parenthesis and a bracket.

    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 A Chemical compound INEQUALITY WITH “OR.”
    1. Solve each inequality.
    2. Graph each solution. Then graph the numbers that brand either inequality truthful.
    3. Write the solution in interval notation.
    Instance \(\PageIndex{13}\)

    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.
    .
    \((−\infty,−32]\cup[two,\infty)\)
    Example \(\PageIndex{14}\)

    Solve the compound inequality. Graph the solution and write the solution in interval notation: \(1−2x\leq −3\) or \(seven+3x\leq 4\).

    Answer

    The solution is x is greater than or equal to 2 or x is less than or equal to 1. The graph of the solutions on a number line has a closed circle at negative 1 and shading to the left and a closed circle at 2 with shading to the right. The interval notation is the union of negative infinity to negative 1 within a parenthesis and a bracket and 2 and infinity within a bracket and a parenthesis.

    Example \(\PageIndex{xv}\)

    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

    The solution is x is greater than or equal to 1 or x is less than or equal to negative 1. The graph of the solutions on a number line has a closed circle at negative 1 and shading to the left and a closed circle at 1 with shading to the right. The interval notation is the union of negative infinity to negative 1 within a parenthesis and a bracket and 1 and infinity within a bracket and a parenthesis.

    Example \(\PageIndex{16}\)

    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
    \(\frac{2}{three}x−4\leq 3\) 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.
    \((−\infty ,\infty )\)
    Instance \(\PageIndex{17}\)

    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

    The solution 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.

    Example \(\PageIndex{eighteen}\)

    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

    The solution 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.

    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.

    Example \(\PageIndex{19}\)

    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
    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.
    Popular:   Suppose That You Have a Mass of 457 Kg
    Case \(\PageIndex{20}\)

    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.

    Instance \(\PageIndex{21}\)

    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”

      1. Solve each inequality.
      2. Graph each solution. Then graph the numbers that make
        both
        inequalities truthful. This graph shows the solution to the compound inequality.
      3. 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: \[\begin{align*} a<x<b & & \text{is equivalent to} & & a<x\;\text{and}\;x<b \\
        a≤10≤b & & \text{is equivalent to} & & a≤x\;\text{and}\;x≤b \\
        a>x>b & & \text{is equivalent to} & & a>x\;\text{and}\;x>b \\
        a≥ten≥b & & \text{is equivalent to} & & a≥x\;\text{and}\;x≥b \end{marshal*}\]

    • 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.

    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