Solving linear inequalities

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Christian Lawson-Perfect	said	Ready to use	7 years, 8 months ago
Bradley Bush	said	Needs to be tested	7 years, 8 months ago
Lauren Richards	said	Has some problems	7 years, 8 months ago
Vicky Hall	said	Has some problems	7 years, 8 months ago
Chris Graham	said	Has some problems	7 years, 9 months ago

History

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Elliott Fletcher 7 years, 8 months ago

Published this.

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Christian Lawson-Perfect 7 years, 8 months ago

Gave some feedback: Ready to use

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Christian Lawson-Perfect 7 years, 8 months ago

Saved a checkpoint:

I reworded the advice a little bit.

This is a good question - there's a nice progression from simple inequalities to more complicated.

Restore this checkpoint

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Bradley Bush 7 years, 8 months ago

Gave some feedback: Needs to be tested

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Bradley Bush commented 7 years, 8 months ago

Thank you Lauren, some good points here, I've made all of these changes.

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Lauren Richards commented 7 years, 8 months ago

I don't think you should ask everyone to put their answers as fractions when they're not needed to be in that form. Maybe add a bit saying "where applicable" at the end.

Remember to have full stops at the end of each section of your advice.

I don't think you need the section describing that c) is a mix of a) and b). Just state what you have to do. Also, as you have it right now, there is a typo - in instead of an.

Good to highlight dividing by a negative number in part d).

Typo in part g) of the advice - "rearraniging" instead of rearranging.

The advice for part g) is not quite right. the integer does not have the correct sign for the middle two sections of your workings then suddenly has corrected itself by the last line.

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Lauren Richards 7 years, 8 months ago

Gave some feedback: Has some problems

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Bradley Bush 7 years, 8 months ago

Gave some feedback: Needs to be tested

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Bradley Bush commented 7 years, 8 months ago

Thank you for the feedback. I have simplified the answers and added a drop down for the inequality signs (great idea by the way). I've also fixed the advice error.

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Vicky Hall 7 years, 8 months ago

Gave some feedback: Has some problems

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Vicky Hall commented 7 years, 8 months ago

In parts b), c) and d) the expected answer isn't simplified. I got $1$ for part b) but the answer displays as $\displaystyle\frac{3}{3}$ . I agree that it's important to include questions that involve flipping the inequality sign as you have in parts d) and g), but my concern is that you have entered the correct inequality for the student and they may not notice that it has been switched around as their answer would be the same either way. I know you've highlighted this in the advice but if the student gets their answer marked right, they'll not read the advice. It could be good to change a couple of the questions (one where the inequality remains the same, the other where it has changed) so they have the choose the correct inequality from a drop down as well as enter a numerical value.

In part g) of the advice, the wrong variable has been entered in the first line of the solution.

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Bradley Bush 7 years, 9 months ago

Gave some feedback: Needs to be tested

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Bradley Bush commented 7 years, 9 months ago

Thank you for the feedback, I think all the errors which you've highllighted should have been resolved by now.

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Christian Lawson-Perfect 7 years, 9 months ago

Gave some feedback: Has some problems

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Christian Lawson-Perfect commented 7 years, 9 months ago

Parts
Part e has "x<..." in the expected answer. It should just be a number.

In part f, I think you've forgotten the coefficient of $a$ .

Advice
I've corrected "on it's own" to "on its own".

Why were there sometimes brackets around the numbers? I've removed them.

Part d: minus signs have gone missing in the second line. You say you've highlighted something in red, but I can't see it because I'm colourblind. Find a way to refer to things in text.

Something's gone wrong in part e: the right-hand $x$ term goes missing in the second line. And I wouldn't think of this as pulling out a factor of $x$ anyway: you're collecting like terms. The thrid line seems to be nonsense too.

Part f: I think the second line is supposed to have collected all the $x$ terms on the left, but the coefficient is wrong. And then I got a third line showing division by $15-5$ . Why wouldn't you collect together first and just divide by $10$ ?

I dispute that it's easier to divide first in part $g$ ! I got an $h$ term and the advice says to divide by $2$ first. Maybe make sure that you're dividing by the gcd of all the coefficients, and the gcd is greater than 1.

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Bradley Bush 7 years, 9 months ago

Gave some feedback: Needs to be tested

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Bradley Bush commented 7 years, 9 months ago

Thank you for the feedback. I have now changed the questions into gaps to make them less hackable and reworded some of the feedback to be, hopefully, more eloquent and contain less typos.

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Chris Graham 7 years, 9 months ago

Gave some feedback: Has some problems

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Chris Graham commented 7 years, 9 months ago

You should pull $x<$ outside of the gap (or $x>$ , as appropriate) and adjust the correct answers accordingly. The way the marking works is that the part's correct answer is evaluated at a number of test points, and in this case will return logical values: let's say it evaluates to true at all of the test points. If that's the case then any student answer that also evaluates to true at the test points will be marked as correct. Have a go yourself in part (a) with, say, x<100 or x=x.

Some of the wording in the advice needs to be carefully considered. For example, in part (a), are you really "subtracting the constant"? And there is a "constant" on both sides (thoug I appreciate that you give the value). I think I would be more direct and offer motivation for the step, with something like "to obtain x on its own on the left hand side, first add 8 to both sides"

There are a few typos in the advice, which is otherwise laid out nicely.

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Bradley Bush 7 years, 9 months ago

Gave some feedback: Needs to be tested

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Bradley Bush 7 years, 9 months ago

Created this.

Name	Status	Author	Last Modified
Solving linear inequalities	Ready to use	Bradley Bush	20/11/2019 14:42
Solve quadratic inequalities	Ready to use	Bradley Bush	20/11/2019 14:42
Solving linear inequalities	Should not be used	Adrian Jannetta	31/08/2023 17:00
Ida's copy of Solve quadratic inequalities	draft	Ida Landgärds	13/04/2018 09:16
Ida's copy of Solving linear inequalities	draft	Ida Landgärds	13/04/2018 09:16
Solving linear inequalities [L5 Randomised]	Needs to be tested	Matthew James Sykes	25/07/2018 10:51
cormac's copy of Solving linear inequalities	draft	cormac breen	24/10/2018 13:44
cormac's copy of Solve quadratic inequalities	draft	cormac breen	24/10/2018 13:46
Solving linear inequalities	draft	Xiaodan Leng	11/07/2019 01:43

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Question statement

Solve the following linear inequalities by finding the set of possible values for $x$ . State your answers as fractions where applicable.

Give any introductory information the student needs.

			Name	Type	Generated Value

a	list	List of 10 items
b	list	List of 10 items
c	integer	2
x

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xxxxxxxxxx
 
repeat(random(2..10),10)

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Generated value: `list`

[ 3, 4, 7, 10, 2, 9, 6, 2, 5, 3 ]

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Parts

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Ask the student a question, and give any hints about how they should answer this part.

$\simplify{x-{a[0]}<{a[1]}}$

$x<$ Unnamed gap

Advice

As with regular linear equations, we aim to isolate the variable by subtracting any constants when dividing by the $x$ coefficient. The only major difference is that when we divide or multiply by a negative number, the inequality sign is reversed.

For example, the following inequality is true:

$-3 \lt -2$

When we multiply both sides by $-2$ , the inequality sign must reverse:

$6 \gt 4$

a)

To put $x$ on its own, we need to add $\var{a[0]}$ to both sides of the inequality.

$\begin{align} \simplify{x-{a[0]}}&<\var{a[1]}\\[1em] \var{x}&<\simplify[]{{a[1]}+{a[0]}}\\[1em] x&<\simplify{({a[1]}+{a[0]})}\text{.} \end{align}$

b)

In this example we find $x$ by dividing both sides by the coefficient of $x$ , $\var{a[2]}$ .

$\begin{align} \simplify{{a[2]}}x&<\var{a[3]}\\[1em] x&<\simplify{{a[3]}/{a[2]}}\text{.} \end{align}$

c)

$\begin{align} \simplify{{a[6]}x-{a[4]}}&<\var{a[5]}\\[1em] \var{a[6]}x&<\var{a[5]}+\var{a[4]} & \text{Add } 8 \text{ to get } x \text{ on its own.}\\[1em] x&<\simplify[]{({a[5]}+{a[4]})/{a[6]}} & \text{ Divide by } \var{a[6]} \text{.} \\[1em] x&<\simplify{({a[5]}+{a[4]})/{a[6]}}\text{.} \end{align}$

d)

In this example, take the constants to one side, and keep the $x$ term on the other. Divide through by the negative $x$ -coefficient to find an inequality for $x$ . Notice that where you divide (or multiply) an equality by a negative value, the inequality sign is reversed.

$\begin{align} \simplify{{-a[6]}x - {a[4]}} &< \var{a[5]} \\[1em] \var{-a[6]}x &< \var{a[5]} + \var{a[4]} & \text{Add } \var{a[4]} \text{ to both sides.} \\[1em] x &> \simplify[]{({a[5]}+{a[4]})/-{a[6]}} \text{ Divide by } \var{-a[6]} \text{. The inequality is reversed.} \\[1em] x &> \simplify{({a[5]}+{a[4]})/-{a[6]}}\text{.}\\ \end{align}$

e)

In this example, separate the constants and the $x$ -term, then divide by the $x$ -coefficient to find an inequality for $x$ .

$\begin{align} \simplify{{b[0]}x-{b[1]}}&<\simplify{{b[3]}-{b[2]}x}\\[1em] \simplify{({b[0]}+{b[2]})x}&<\simplify{{b[3]}+{b[1]}}\\[1em] x&<\simplify{({b[3]}+{b[1]})/({b[0]}+{b[2]})}\text{.}\\[1em] \end{align}$

f)

In this example, separate the $x$ -term from all other terms and remember to reverse the inequality when dividing by $\simplify{{a[7]}-{b[4]}}$ .

$\begin{align} \simplify{-{b[4]}x+{a[8]}a}&>\simplify{{b[5]}+b-{a[7]}x}\\[1em] \simplify{{a[7]}-{b[4]}}x&>\simplify{{b[5]}+b-{a[8]}a}\\[1em] x&<\simplify{(-{b[5]}-b+{a[8]}a)/({b[4]}-{a[7]})}\text{.}\\[1em] \end{align}$

In this example, a simple way to solve for $x$ is to divide by $-\var{c}$ before rearranging the rest of the equation by subtracting $g$ from both sides.

$\begin{align} \simplify{-{c}(x+g)}&>\simplify{6h-{c}{a[0]}}\\[1em] \simplify{(x+g)}&<\simplify[]{6h/-{c}+{a[0]}}\\[1em] x&<\simplify[]{6h/-{c}+{a[0]}-g}\\[1em] x&<\simplify{{a[0]}-6h/{c}-g}\text{.} \end{align}$

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Elliott Fletcher 7 years, 8 months ago

Christian Lawson-Perfect 7 years, 8 months ago

Christian Lawson-Perfect 7 years, 8 months ago

Bradley Bush 7 years, 8 months ago

Bradley Bush commented 7 years, 8 months ago

Lauren Richards commented 7 years, 8 months ago

Lauren Richards 7 years, 8 months ago

Bradley Bush 7 years, 8 months ago

Bradley Bush commented 7 years, 8 months ago

Vicky Hall 7 years, 8 months ago

Vicky Hall commented 7 years, 8 months ago

Bradley Bush 7 years, 9 months ago

Bradley Bush commented 7 years, 9 months ago

Christian Lawson-Perfect 7 years, 9 months ago

Christian Lawson-Perfect commented 7 years, 9 months ago

Bradley Bush 7 years, 9 months ago

Bradley Bush commented 7 years, 9 months ago

Chris Graham 7 years, 9 months ago

Chris Graham commented 7 years, 9 months ago

Bradley Bush 7 years, 9 months ago

Bradley Bush 7 years, 9 months ago

Error in variable testing condition

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Generated value: list

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Generated value: `list`