// Numbas version: exam_results_page_options {"name": "Inequalities", "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "duration": 0, "percentPass": 0, "showQuestionGroupNames": false, "shuffleQuestionGroups": false, "showstudentname": true, "question_groups": [{"name": "Group", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": ["", "", "", "", "", ""], "variable_overrides": [[], [], [], [], [], []], "questions": [{"name": "Inequalities: symbol names", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ben Brawn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/605/"}], "functions": {}, "ungrouped_variables": [], "tags": ["inequalities", "inequality", "inequations"], "advice": "", "rulesets": {}, "parts": [{"stepsPenalty": "1", "maxAnswers": 0, "prompt": "

Match the following inequality symbols with their meanings.

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$<$

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$\\leq$

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$>$

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$\\ge$

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less than

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less than or equal to

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greater than

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greater than or equal to

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The statement $x<3$ is read as '$x$ is less than 3'. Notice there is a small side and a big side to the symbol $<$. The small part of the symbol goes on the side with the small number and the large side goes on the side with the large number.

\n

For example, given $y>2$, we see the large side of the symbol is on the left and the small side is on the right. Therefore, the statement is saying $y$ is greater than 2. 

\n

When there is an extra line on the bottom, e.g. $\\ge$, this gives the look of an 'equals sign' and so $w\\ge 5$ means '$x$ is greater than or equal to 5'. Likewise, $z\\le 0$, means '$z$ is less than or equal to 0'.

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Which of the following values for {variable} satisfy the inequality {question}?

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The statement {question} means {variable} {words} {numtex}.

\n

\n

When dealing with negative numbers, it is important to remember that 'less' means further to the left on the number line, and 'greater' means further to the right on the number line. So $-10$ is less than $-5$, since $-10$ is five more places to the left than $-5$ on the number line.

\n

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[less, num1, middle, num2, more]

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Which of the following values for {variable} satisfy the inequality {question}?

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The statement {question} means '{num1tex} {words1[0]} {variable} and {variable} {words2[0]} {num2tex}'. Another way to say this is '{variable} is between {num1tex} and {num2tex}, {words1[1]} {num1tex} and {words2[1]} {num2tex}'.

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The condition that {question2}, is equivalent to which of the following?  

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'{question2}' requires that both inequalities are satisfied. Both will be satisfied when {solution}.

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{variable2+ineqs[0]+num3}

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{variable2+ineqs[1]+num4}

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[less, num1, middle, num2, more]

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Which of the following values for {variable} satisfy the inequalities {question}?

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The statement {question} means '{variable} {words1[0]} {num1tex} or {variable} {words2[0]} {num2tex}'.

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{num1}

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{middle}

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{num2}

", "

{more}

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The condition that {question2}, is equivalent to which of the following?  

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'{question2}' only requires that one of the inequalities are satisfied (but both can be satisified as well). This condition will be satisfied when {solution}.

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{variable2+ineqs[0]+num3}

", "

{variable2+ineqs[1]+num4}

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Given $\\var{a}x+\\var{b}<\\var{c}$, solving for $x$ gives $x<$ [[0]].

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Solve like an equation except when dividing or multiplying by a negative we must swap the direction of the inequality.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$\\var{a}x+\\var{b}$$<$$\\var{c}$ 
 
$\\var{a}x+\\var{b}-\\var{b}$$<$$\\var{c}-\\var{b}$
 
$\\var{a}x$$<$$\\var{c-b}$
 
$\\displaystyle{\\frac{\\var{a}x}{\\var{a}}}$ $<$$\\displaystyle{\\frac{\\var{c-b}}{\\var{a}}}$
 
$x$$<$$\\displaystyle{\\simplify{{c-b}/{a}}}$
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To satisfy $\\var{d}-\\var{f}y\\le\\var{g}$, we require $y\\ge$ [[0]].

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Solve like an equation except when dividing or multiplying by a negative we must swap the direction of the inequality.

\n

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$\\var{d}-\\var{f}y$$\\le$$\\var{g}$ 
 
$\\var{d}-\\var{f}y-\\var{d}$$\\le$$\\var{g}-\\var{d}$
 
$-\\var{f}y$$\\le$$\\var{g-d}$
 
$\\displaystyle{\\frac{\\var{-f}y}{\\var{-f}}}$ $\\ge$$\\displaystyle{\\frac{\\var{g-d}}{\\var{-f}}}$
 
$y$$\\ge$$\\displaystyle{\\simplify{{g-d}/{-f}}}$
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Rearrange $\\displaystyle{-\\frac{z}{\\var{h}}}-\\var{j}<\\var{k}$ to determine the value of $z$.

\n

$z>$ [[0]]

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Solve like an equation except when dividing or multiplying by a negative we must swap the direction of the inequality.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$\\displaystyle{-\\frac{z}{\\var{h}}}-\\var{j}$$<$$\\var{k}$ 
 
$\\displaystyle{-\\frac{z}{\\var{h}}}-\\var{j}+\\var{j}$$<$$\\var{k}+\\var{j}$
 
$\\displaystyle{-\\frac{z}{\\var{h}}}$$<$$\\var{k+j}$
 
$\\displaystyle{-\\frac{z}{\\var{h}}\\times\\var{h}}$ $<$$\\var{k+j}\\times \\var{h}$
 
$-z$$<$$\\var{-ans3}$
 
$z$$>$$\\var{ans3}$
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Solve $\\displaystyle{\\frac{a-\\var{l}}{\\var{m}}}\\le\\var{n}$ for $a$.

\n

$a\\le$ [[0]]

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Solve like an equation except when dividing or multiplying by a negative we must swap the direction of the inequality.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$\\displaystyle{\\frac{a-\\var{l}}{\\var{m}}}$$\\le$$\\var{n}$ 
 
$\\displaystyle{\\frac{a-\\var{l}}{\\var{m}}}\\times \\var{m}$$\\le$$\\var{n}\\times\\var{m}$
 
$a-\\var{l}$$\\le$$\\var{n*m}$
 
$a-\\var{l}+\\var{l}$ $\\le$$\\var{n*m}+\\var{l}$
 
$a$$\\le$$\\var{ans4}$
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Solve $\\var{p}>\\var{q}(\\var{r}+b)$.

\n

$b>$ [[0]]

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Solve like an equation except when dividing or multiplying by a negative we must swap the direction of the inequality.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$\\var{p}$$>$$\\var{q}(\\var{r}+b)$ 
 
$\\displaystyle{\\frac{\\var{p}}{\\var{q}}}$$<$$\\displaystyle{\\frac{\\var{q}(\\var{r}+b)}{\\var{q}}}$
 
$\\displaystyle{\\simplify{{p}/{q}}}$$<$$\\var{r}+b$
 
$\\displaystyle{\\simplify{{p}/{q}}}-\\var{r}$ $<$$\\var{r}+b-\\var{r}$
 
$\\displaystyle{\\simplify{{p-r*q}/{q}}}$$<$$b$
 
$b$$>$$\\displaystyle{\\simplify{{p-r*q}/{q}}}$
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Solve $\\displaystyle{\\frac{\\var{s}w}{\\var{t}}}\\ge\\var{u}$.

\n

$w\\le$ [[0]]

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Solve like an equation except when dividing or multiplying by a negative we must swap the direction of the inequality.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$\\displaystyle{\\frac{\\var{s}w}{\\var{t}}}$$\\ge$$\\var{u}$ 
 
$\\displaystyle{\\frac{\\var{s}w}{\\var{t}}}\\times\\var{t}$$\\ge$$\\var{u}\\times\\var{t}$
 
$\\var{s}w$$\\ge$$\\var{u*t}$
 
$\\displaystyle{\\frac{\\var{s}w}{\\var{s}}}$ $\\le$$\\displaystyle{\\frac{\\var{u*t}}{\\var{s}}}$
 
$w$$\\le$$\\displaystyle{\\simplify{{u*t}/{s}}}$
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Given that $x>\\var{left_ba}$ and $x<\\var{right_ba}$, we can write [[0]] $<x<$ [[1]].

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$\\var{left_ba} <x< \\var{right_ba}$ means $\\var{left_ba} <x$ and $x< \\var{right_ba}$. You could read this as \"$x$ is between $\\var{left_ba}$ and $\\var{right_ba}$\".

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Given $\\var{left_bb}<\\frac{x}{\\var{c}}<\\var{right_bb}$, solving for $x$ gives [[0]]$<x<$ [[1]].

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Solving inequalities is similar to solving equations, ensure you do the same thing to all sides. Note that the operations we do are to get $x$ by itself.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$\\var{left_bb}$$<$$\\displaystyle \\frac{x}{\\var{c}}$$<$$\\var{right_bb}$
 
$\\var{left_bb}\\times \\var{c}$$<$$\\displaystyle \\frac{x}{\\var{c}}\\times \\var{c}$$<$$\\var{right_bb}\\times \\var{c}$
 
$\\var{bleft}$$<$$x$$<$$\\var{bright}$
\n

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Given $\\var{left_bc}<\\frac{\\simplify{{a}x+{b}}}{\\var{c}}<\\var{right_bc}$, solving for $x$ gives [[0]]$<x<$ [[1]].

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Solving inequalities is similar to solving equations, ensure you do the same thing to all sides. Note that the operations we do are to get $x$ by itself.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$\\var{left_bc}$$<$$\\displaystyle \\frac{\\simplify{{a}x+{b}}}{\\var{c}}$$<$$\\var{right_bc}$
  
$\\var{left_bc}\\times \\var{c}$$<$$\\displaystyle \\frac{\\simplify{{a}x+{b}}}{\\var{c}}\\times \\var{c}$$<$$\\var{right_bc}\\times \\var{c}$
  
$\\var{left_bc*c}$$<$$\\simplify{{a}x+{b}}$$<$$\\var{right_bc*c}$
 
$\\simplify[basic]{{left_bc*c}-{b}}$$<$$\\simplify[basic]{{a}x+{b}-{b}}$ $<$$\\simplify[basic]{{right_bc*c}-{b}}$
 
$\\var{left_bc*c-b}$$<$$\\var{a}x$$<$$\\var{right_bc*c-b}$
 
$\\displaystyle\\frac{\\var{left_bc*c-b}}{\\var{a}}$$<$$\\displaystyle\\frac{\\var{a}x}{\\var{a}}$$<$$\\displaystyle\\frac{\\var{right_bc*c-b}}{\\var{a}}$
 
$\\displaystyle\\simplify{{left_bc*c-b}/{a}}$$<$$x$$<$$\\displaystyle\\simplify{{right_bc*c-b}/{a}}$
\n

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