Practice Problems

Answers

  1. Understanding the type of statement

    We are given an \text{AND} statement. A certain x‍-value is a solution of the statement if it satisfies both of the inequalities.

    Therefore, the solution of this statement is the intersection of the solutions of both inequalities. In other words, the overlap of the solutions of both inequalities is the solution of the statement.

    Finding the solutions to the two inequalities

    The solution of the first inequality, 12x-39\leq 9, is ‍x\leq 4.

    The solution of the second inequality, -4x+3 < -6, is x>\dfrac{9}4.

    The inequalities {x\leq4} and  {x>\dfrac{9}4} are graphed below.

    A number line from negative four to nine by ones. There is an open circle at nine-fourths with an arrow to the right. There i

    The intersection of the solutions of both inequalities is \dfrac{9}4 < x\leq4.

    The answer

    The solution is \dfrac{9}4 < x\leq4.


  2. Understanding the type of statement

    We are given an \text{OR} statement. A certain x‍-value is a solution of the statement if it satisfies either of the inequalities.

    Therefore, the solution of this statement is the union of the solutions of both inequalities. In other words,the entire collection of the solutions of both inequalities forms the solution of this statement.

    Finding the solutions to the two inequalities

    The solution of the first inequality, 12x+7 < -11, is ‍x < -\dfrac{3}2.

    The solution of the second inequality, 5x-8>40, is x>\dfrac{48}{5}.

    The inequalities {x < -\dfrac{3}2} and  {x>\dfrac{48}{5}} are graphed below.


    A number line from negative four to fourteen by ones. There is an open circle at negative three-halves with an arrow to the l
     

    The union of the solutions of both inequalities is all x-values that are less than -\dfrac{3}2 along with all ‍x-values that are greater than \dfrac{48}5.

    This can be represented mathematically as x < -\dfrac{3}2 OR x>\dfrac{48}{5}.

    The answer

    The solution is x < -\dfrac{3}2 OR x>\dfrac{48}{5}.


  3. Understanding the type of statement

    We are given an \text{AND} statement. A certain x‍-value is a solution of the statement if it satisfies both of the inequalities.

    Therefore, the solution of this statement is the intersection of the solutions of both inequalities. In other words, the overlap of the solutions of both inequalities is the solution of the statement.

    Finding the solutions to the two inequalities

    The solution of the first inequality, 4x-39 > -43, is ‍x\leq 4.

    The solution of the second inequality, 8x+31 < 23, is x < -1.

    The inequalities {x > -1} and  {x < -1} are graphed below.

    A number line from negative four to four by ones. There is an open circle at negative one with an arrow to the left. There is

    The intersection of the solutions of both inequalities is empty.

    The answer

    There are no solutions.


  4. Understanding the type of statement

    We are given an \text{OR} statement. A certain x‍-value is a solution of the statement if it satisfies either of the inequalities.

    Therefore, the solution of this statement is the union of the solutions of both inequalities. In other words,the entire collection of the solutions of both inequalities forms the solution of this statement.

    Finding the solutions to the two inequalities

    The solution of the first inequality, 5x-4\geq 12, is ‍x\geq \dfrac{16}{5}.

    The solution of the second inequality, 12x+5\leq-4, is x\leq-\dfrac{3}{4}.

    The inequalities {x \geq \dfrac{16}{5}} and  {x\leq-\dfrac{3}{4}} are graphed below.

    A number line from negative four to four by ones. There is a closed circle at negative three-fourths with an arrow to the lef

    The union of the solutions of both inequalities is all x-values that are less than \dfrac{16}5 along with all ‍x-values that are greater than -\dfrac{3}4.

    This can be represented mathematically as x\geq\dfrac{16}{5} OR x\leq-\dfrac{3}{4}.

    The answer

    The solution is x\geq\dfrac{16}{5} OR x\leq-\dfrac{3}{4}.