My name is Dario, I am a postgraduate student in Computer Science at The University of Manchester. Previously, I have attained a First Class BSc in Computer Science at Sheffield University.Â I offer tutoring inÂ Computing/Computer Science (and any related programming discipline) andÂ MathsÂ at GCSE, A-Levels and IB.
I am also bilingual inÂ Italian andÂ EnglishÂ and can therefore offer language tutoring and help develop linguistic skills.
I have previous experience teaching in Sixth-Form colleges and primary schools and have previously tutored students at different levels.
Please get in touch and we can discuss how best we can structure your program. I am very flexible with times and sure we can find slots that integrate well into your schedule.
|Computing||A Level||£36 /hr|
|(IB) Higher Level Mathematics||Baccalaureate||5|
|(IB) Higher Level Physics||Baccalaureate||5|
|(IB) Higher Level Biology||Baccalaureate||5|
|(IB) Higher Level English||Baccalaureate||7|
|(IB) Standard Level Italian||Baccalaureate||6|
|(IB) Standard Level History||Baccalaureate||5|
|BSc Computer Science||Bachelors Degree||FIRST CLASS|
|Before 12pm||12pm - 5pm||After 5pm|
Please get in touch for more detailed availability
Joseph (Student) June 20 2016
Jilesh (Student) May 21 2016
Jilesh (Student) May 20 2016
Joseph (Parent) April 24 2016
For which values of x does the following inequality hold true:
(|x|+2)/(|x|-3) < 4
The first thing we want to do is eliminate the fraction on the left-hand side (LHS) of the equation. We do so by multiplying both side by |x|-3, leading to:
|x|+2 < 4(|x|-3)
We can simplify the right-hand side (RHS):
|x| + 2 < 4|x| - 12
We now want to group like terms together. As such, we will move all |x| terms to the RHS, and all integer terms to the LHS.
2 + 12 < 4|x| - |x|
14 < 3|x|
We now divide both sides by 3 to eliminate the coefficient of |x|
14/3 < |x|
|x| > 14/3
Now, imagine a number line...
<---- -x ------- 0 ----- x --------->
I would like you to remember the definition of |x| (absolute value) which indicates the distance of a value x from 0. (In other words, it makes any negative number positive and leaves any positive number positive)
This suggests that any x > 14/3 will satisfy |x| > 14/3.
However, we also have to remember that any value inferior to -14/3 will also satisfy the inequality.
As such, we solve the solution as
x > 14/3 OR x < -14/3