Homework 1

Released: 4pm Monday Week 3. Due: 4pm Monday Week 4.

General feedback

Thank you to everyone who completed this work!

Your work is now marked with personalised feedback. Please take some time to view it on Crowdmark and compare with the more general comments and model solution below. If you are unsure about why you have received the comments or marks that you have, please get in touch.

I’m pleased to see nearly all of the works I marked had the correct numerical solution. The mark scheme myself and Simon used to mark is roughly the following:

• Up to 6 marks for the elimination/finding the first variable step

• Up to 2 marks for back-substitution and finding the second variable

• 1 mark for a conclusion statement

• 1 mark for the check

High marks were awarded for using a correct method, correct mathematical notation, with appropriate explanation, written in a clear, easy to follow fashion. Method marks were lost if we couldn’t follow your argument at any point, and/or for notational issues. A full model solution with mark scheme can be viewed below.

Being the very first homework of the year, there were quite a few presentational issues, leading to lost marks. This happens every year, and typically people improve substantially each time they submit work.

The key thing to do now is to review your work on Crowdmark, think about the comments, get in touch with Rosie if you would like to talk about anything, and work on applying any feedback to your submission for Homework 3 (due Monday 10th November).

Quick links:

• Principles of good presentation (lecture notes)

• What a valid check looks like

• Where to get help

Question

Solve the following simultaneous equation:

\[\begin{split} \begin{aligned} \left.\begin{array}{rcl} \displaystyle 2x &=& \displaystyle 4-3y \\ \displaystyle 7x-8 &=& \displaystyle 5y \end{array}\right\}\end{aligned} \end{split}\]

Your answer should include a check. (10 marks)

Model solution with mark scheme

Hidden in case you want to re-attempt the question as revision.

Click to view solution

We label the equations as

\[\begin{split} \begin{aligned} \left.\begin{array}{rclr} 2x &=& 4 - 3y &\text{(1)} \\ 7x-8 &=& 5y &\text{(2)} \end{array}\right\}. \end{aligned} \end{split}\]

(1 mark)

Adding \(3y\) to both sides of (1) and \(8-5y\) to both sides of (2), these equations can be expressed in the more familiar form

\[\begin{split} \begin{aligned} \left.\begin{array}{rclr} 2x+3y &=& 4 &\text{(3)} \\ 7x-5y &=& 8 &\text{(4)} \end{array}\right\}. \end{aligned} \end{split}\]

(2 marks)

We aim to eliminate \(x\). Multiplying equation (3) by \(7\) and equation (4) by \(2\) gives

\[\begin{split} \begin{aligned} \left.\begin{array}{rclr} 14x +21y &=& 28 &\text{(5)} \\ 14x-10y &=& 16 &\text{(6)} \end{array}\right\}. \end{aligned} \end{split}\]

(1 mark)

Now subtracting equation (6) from equation (5) leads to \(31y = 12\). (1 mark)

So, \(y = \frac{12}{31}\). (1 mark)

We can substitute this into equation (1) to find that \(2x= 4 - 3\left(\frac{12}{31}\right)\). (1 mark)

Dividing by \(2\), we get

\[ x = 2 - \frac{18}{31} = \frac{62 - 18}{31} = \frac{44}{31}. \]

(1 mark)

Thus the solution to the above simultaneous equations is \(x = \frac{44}{31}\) and \(y = \frac{12}{31}\). (1 mark).

Check: We check our solution using equation (2) (note we only just used (1) to find \(x\)).

When \(\displaystyle x = \frac{44}{31}\) and \(y = \frac{12}{31}\), then

\[\begin{split} \begin{aligned} \text{LHS(2)} = 7x-8 &= 7\left(\frac{44}{31}\right)-8 \\ &= \frac{7\times 44}{31} - \frac{8 \times 31}{31} \\ &= \frac{308-248}{31} \\ &= \frac{60}{31} = 5\left(\frac{12}{31}\right) = 5y = \text{RHS(2)}. \end{aligned} \end{split}\]

(1 mark)