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Questions on understanding inequality notation and solving inequalities

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The answer is either 'True' or 'False'

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Identify the truth value of an inequality (T/F) between two numbers

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Inequality Notation

", "advice": "

The symbol is a comparitor that $<$ says that the object to its left is less than the object to its right. Similarly, $>$ says the object to its left is greater than the object to its right.

In each case the sentence can be read in two ways e.g. $a<b$ can be read as \"$a$ is less than $b$\" OR \"$b$ is greater than $a$\".

The symbol $\\leq$ just changes the sentence to include \"...or equal to...\". This can be of particular relevance when dealing with integers (whole numbers) e.g. $x \\geq 4$ and $x$ is a whole number means that $x$ could be $4, 5, 6,$ or $7$ and so on. Whereas, $x>4$ and $x$ is a whole number means that $x$ could be $5,6,$... etc. Notably in the second case $x$ cannot be $4$.

Use this link to find some resources which will help you revise this topic.

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Is the following statement true or false?

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Is the following statement true or false?

\$\\var{disp_values2[0]}\\var{disp_op2}\\var{disp_values2[1]}\$

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Solving an inequality of the form x+a < b

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Solve in the inequality

\\[\\simplify{x+{a1}}\\var{Symb[0]}\\var{a2}.\\]

", "advice": "

We treat inequalities similarly to solving to an equation. We want to get $x$ on its own on one side, in this case we can achieve this by subtracting $\\var{a1}$ from both sides which gives us the following.

\\[\\simplify{x+{a1}}\\var{Symb[0]}\\var{a2}\\]

\\[\\simplify[basic]{x+{a1}-{a1}}\\var{Symb[0]}\\simplify[basic]{{a2}-{a1}}\\]

\\[x\\var{Symb[0]}\\var{a2-a1}\\]

Use this link to find some resources which will help you revise this topic.

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x[[1]][[0]]

Solving an inequality of the form ax < b

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Solve in the inequality

\\[\\simplify{{a1}x}\\var{Symb[0]}\\var{a1*a2}.\\]

", "advice": "

We treat inequalities similarly to solving to an equation. We want to get $x$ on its own on one side, in this case we can achieve this by dividing $\\var{a1}$ from both sides which gives us the following.

\\[\\simplify{{a1}x}\\var{Symb[0]}\\var{a1*a2}\\]

\\[\\simplify[!all]{{a1}x/{a1}}\\var{Symb[0]}\\simplify[!all]{{a1*a2}/{a1}}\\]

\\[x\\var{Symb[0]}\\var{a2}\\]

Use this link to find some resources which will help you revise this topic.

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x[[1]][[0]]

Solving an inequality of the form x/a < b

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Solve in the inequality

\\[\\simplify{x/{a1}}\\var{Symb[0]}\\var{a2}.\\]

", "advice": "

We treat inequalities similarly to solving to an equation. We want to get $x$ on its own on one side, in this case we can achieve this by multiplying $\\var{a1}$ on both sides which gives us the following.

\\[\\simplify{x/{a1}}\\var{Symb[0]}\\var{a2}\\]

\\[\\simplify[!all,alwaysTimes]{x/{a1}*{a1}}\\var{Symb[0]}\\simplify[!all,alwaystimes]{{a2}*{a1}}\\]

\\[x\\var{Symb[0]}\\var{a2*a1}\\]

Use this link to find some resources which will help you revise this topic.

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x[[1]][[0]]

Solving an inequality of the form ax+b < c

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Solve in the inequality

\\[\\simplify{{a1}x+{a3}}\\var{Symb[0]}\\var{a1*a2+a3}.\\]

", "advice": "

We treat inequalities similarly to solving to an equation. We want to get $x$ on its own on one side, in this case we can achieve this by first subtracting $\\var{a3}$ and then dividing by $\\var{a1}$ which gives us the following.

\\[\\simplify{{a1}x+{a3}}\\var{Symb[0]}\\var{a1*a2+a3}\\]

\\[\\simplify[basic]{{a1}x+{a3}-{a3}}\\var{Symb[0]}\\simplify[basic]{{a1*a2+a3}-{a3}}\\]

\\[\\simplify{{a1}x}\\var{Symb[0]}\\simplify{{a1*a2}}\\]

\\[\\simplify[!all]{{a1}x/{a1}}\\var{Symb[0]}\\simplify[!all]{{a1*a2}/{a1}}\\]

\\[x\\var{Symb[0]}\\var{a2}\\]

Use this link to find some resources which will help you revise this topic.

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x[[1]][[0]]

Solving an inequality of the form ax+b < c

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Solve in the inequality

\\[\\simplify{(x+{a3})/{a1}}\\var{Symb[0]}\\var{a2}.\\]

", "advice": "

We treat inequalities similarly to solving to an equation. We want to get $x$ on its own on one side, in this case we can achieve this by first multiplying by $\\var{a1}$ and then subtracting by $\\var{a3}$ which gives us the following.

\\[\\simplify{(x+{a3})/{a1}}\\var{Symb[0]}\\var{a2}\\]

\\[\\simplify[basic,alwaysTimes]{(x+{a3})/{a1}*{a1}}\\var{Symb[0]}\\simplify[alwaysTimes]{{a2}*{a1}}\\]

\\[\\simplify[basic,alwaysTimes]{x+{a3}}\\var{Symb[0]}\\simplify[alwaysTimes]{{a2*a1}}\\]

\\[\\simplify[basic]{x+{a3}-{a3}}\\var{Symb[0]}\\simplify[basic]{{a1*a2}-{a3}}\\]

\\[x\\var{Symb[0]}\\var{a2*a1-a3}\\]

Use this link to find some resources which will help you revise this topic.

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x[[1]][[0]]

Solving an inequality of the form ax+b < c where a is negative.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Solve in the inequality

\\[\\simplify{{a1}x+{a3}}\\var{Symb[0]}\\var{a1*a2+a3}.\\]

", "advice": "

\\[\\simplify{{a1}x+{a3}}\\var{Symb[0]}\\var{a1*a2+a3}\\]

\\[\\simplify[basic]{{a1}x+{a3}-{a3}}\\var{Symb[0]}\\simplify[basic]{{a1*a2+a3}-{a3}}\\]

\\[\\simplify{{a1}x}\\var{Symb[0]}\\simplify{{a1*a2}}\\]

\\[\\simplify[!all]{{a1}x/{a1}}\\var{SymbFlipped[0]}\\simplify[!all]{{a1*a2}/{a1}}\\]

\\[x\\var{SymbFlipped[0]}\\var{a2}\\]

Use this link to find some resources which will help you revise this topic.

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x[[1]][[0]]

Solving an inequality of the form ax+b < c where a is negative.

", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Solve in the inequality

\\[\\simplify{{a1}x+{a3}}\\var{Symb[0]}\\simplify{{a2}x+{(a1-a2)*a4+a3}}.\\]

", "advice": "

{Advice}

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We treat inequalities similarly to solving to an equation. We want to get $x$ on its own on one side, in this case we can achieve this by first subtracting $\\\\var{a3}$ and $\\\\var{a2}x$ from both sides and and then dividing by $\\\\var{a1}-\\\\var{a2}$. Remember when multiplying or dividing by a negative number we need to flip the inequality symbol. This gives us the following.

\\n

\\\\[\\\\simplify{{a1}x+{a3}}\\\\var{Symb[0]}\\\\simplify{{a2}x+{(a1-a2)*a4+a3}}\\\\]

\\n

\\\\[\\\\simplify[basic]{{a1}x+{a3}-{a3}}\\\\var{Symb[0]}\\\\simplify[basic]{{a2}x+{(a1-a2)*a4+a3}-{a3}}\\\\]

\\n

\\\\[\\\\simplify{{a1}x}\\\\var{Symb[0]}\\\\simplify{{a2}x+{(a1-a2)*a4}}\\\\]

\\n

\\\\[\\\\simplify[basic]{{a1}x-{a2}x}\\\\var{Symb[0]}\\\\simplify{{(a1-a2)*a4}}\\\\]

\\n

\\\\[\\\\simplify{{a1-a2}x}\\\\var{Symb[0]}\\\\simplify{{(a1-a2)*a4}}\\\\]

\\n

\\\\[\\\\simplify[!all]{{a1-a2}x/{a1-a2}}\\\\var{SymbFlipped[0]}\\\\simplify[!all]{{(a1-a2)*a4}/{a1-a2}}\\\\]

\\n

\\\\[x\\\\var{SymbFlipped[0]}\\\\var{a4}.\\\\]

\\n

Use this link to find some resources which will help you revise this topic.

\"", "description": "", "templateType": "long string", "can_override": false}, "AdviceNotFlipped": {"name": "AdviceNotFlipped", "group": "Ungrouped variables", "definition": "\"

\\n

\\\\[\\\\simplify{{a1}x+{a3}}\\\\var{Symb[0]}\\\\simplify{{a2}x+{(a1-a2)*a4+a3}}\\\\]

\\n

\\\\[\\\\simplify[basic]{{a1}x+{a3}-{a3}}\\\\var{Symb[0]}\\\\simplify[basic]{{a2}x+{(a1-a2)*a4+a3}-{a3}}\\\\]

\\n

\\\\[\\\\simplify{{a1}x}\\\\var{Symb[0]}\\\\simplify{{a2}x+{(a1-a2)*a4}}\\\\]

\\n

\\\\[\\\\simplify[basic]{{a1}x-{a2}x}\\\\var{Symb[0]}\\\\simplify{{(a1-a2)*a4}}\\\\]

\\n

\\\\[\\\\simplify{{a1-a2}x}\\\\var{Symb[0]}\\\\simplify{{(a1-a2)*a4}}\\\\]

\\n

\\\\[\\\\simplify[!all]{{a1-a2}x/{a1-a2}}\\\\var{Symb[0]}\\\\simplify[!all]{{(a1-a2)*a4}/{a1-a2}}\\\\]

\\n

\\\\[x\\\\var{Symb[0]}\\\\var{a4}.\\\\]

\\n

Use this link to find some resources which will help you revise this topic.

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x[[1]][[0]]

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The answer is either 'True' or 'False'