I buy five t-shirts and eight pairs of shoes for $450. I really like shoes. In the same shop, my friend buys two t-shirts and two pairs of shoes for $120. How much do the t-shirts cost?
This is a 'simultaneous equation' problem. Simultaneous equations are just two or more equations that are created using the same variables. In this case, we will use 't' for t-shirts and 's' for shoes. First we will have a brief look at how to turn this sentence into the equations. Then we will look at how to solve a simpler version, to help us understand how the method we will use works. Lastly, we will solve this question. Scroll down to below the *** to go straight to solving this question. The first sentence could be expressed as: 5t + 8s = 450 The second sentence could be expressed as: 2t + 2s = 120 The values of 't' and 's' are consistent across both equations, just as the cost of a t-shirt or a pair of shoes is the same both times. To understand how to work these out, let's consider a simpler example. Imagine we buy one cake and one sandwich for $3 c + s = 3 and a friend buys one cake and two sandwiches for $4 c + 2s = 4 Can you work out the price of the sandwich now? There is an extra $1 difference, and an extra sandwich, so a sandwich is $1. If a sandwich is $1, then the cake must be $2. Now we work out the 'difference' between these two equations by subtracting one from the other. The first equation has a smaller value, so lets subtract that from the second. To do this, we subtract each term by its equivalent term, one by one. First we write the equations to have the larger value at the top. We don't have to do this, but it makes it easier, and so makes a mistake less likely. c + 2s = 4 c + s = 3 c - c = 0 2s - s = s 3 - 2 = 1 We are left with: 0 + s = 1 Or: s = 1 We can then 'substitute' this value for 's' into one of our equations: c + s = 3 becomes c + 1 = 3 take away 1 from both sides c + 1 - 1 = 3 - 1 becomes c + 0 = 2 or c = 2 Now we understand how these work, I will take you through the more complicated question from the top. In maths, it always helps to simplify a problem in order to understand the underlying concept and the method needed to solve this type of problem. It's easy when you know how! *** From the initial question, we are able to deduce the following simultaneous equations: 5t + 8s = 450 2t + 2s = 120 If we simply try to find the difference between the two equations, we run into a problem. Subtracting all the terms from the equivalents leaves us with this: 3t + 6s = 300 Another equation with two variables! We won't be able to solve this on it's own either. (Sometimes creating these new equations make solving the problem easier, so don't be afraid to try it. However, there is a quicker way.) The problem was that there was not an equal number of 't' or 's' in the two equations. If there were, then one of the variables is reduced to 0, or cancelled out. Happily, though, we can scale the equations up or down. This isn't cheating - just think about what would happen in the shop. If you scaled up your order by 2x, buying twice as many t-shirts and pairs of shoes, then the bill would be twice as big too. The same if you scaled it up by 3x or 100x. Looking at our equations: 5t + 8s = 450 2t + 2s = 120 I notice that if we scale the second equation up 4x, then we will have 8s in both equations. Let's do this. 4 x (2t + 2s = 120) = 8t + 8s = 480 Now we can take our new equation, and the original equation that also had 8s, and find the difference. 8t + 8s = 480 5t + 8s = 450 = 3t + 0 = 30 or 3t = 30 divide both sides by 3 t = 10 So t-shirts are $10 each. *** To find the price of the shoes, we would just need to substitute '10' for 't' into one of our equations, then solve for 's'. 2t + 2s = 120 becomes 2 x 10 + 2s = 120 20 + 2s = 120 2s = 120 - 20 2s = 100 s = 50 *** If we are asked something like, how much would 2 pairs of shoes and 10 t shirts cost, then we simply take our costs and multiply up. Algebraically, it could be expressed as follows: 2s + 10t = c (where c is the cost of the items) s = 50 t = 10 2 x 50 + 10 x 10 = c 100 + 100 = c 200 = c
How can I improve my creative writing?
There are many different techniques for improving creative writing, usually related with working out how to make your creation more real, or vivid, for the reader. Consider what makes an experience real. Thoughts, feelings and sensations are all important. When writing, you must convey these to the reader. If you find yourself writing action packed narratives, you might want to consider slowing the pace down and allowing a focus on the world the action is taking place in. A simple, effective device is to try to 'feed the senses'. Think about what your reader would be seeing, most importantly, but also hearing, smelling, tasting or touching. Is it blisteringly hot, or is there a bone chilling wind whistling through the trees? Additionally, what are the characters thinking, and what are they feeling? Remember that it is often better to show this through the character's behaviour, rather than just telling the reader outright. "Joe's head dropped, and he seemed to shrink an inch" is often better than, "Joe was disappointed". It is a combination of lots of different ways of communicating to your reader that works best. Learning the many different ways you can communicate to your reader is important to hep make your writing naturally more varied and interesting.
Explain the descent of a sky diver, in terms of the forces acting on them.
1. When the sky diver jumps from the plane, gravity pulls them towards the ground with a force relatively large compared to any opposing force, so they accelerate rapidly. 2. The rate of acceleration decreases in proportion to their velocity, until they reach terminal velocity. 3. This is because air resistance, caused by collision with air particles, increases with velocity. 4. The forces of gravity and air resistance eventually balance, when the sky diver is at their 'terminal velocity'. 5. Whilst travelling at terminal velocity, the forces on the sky diver are balanced. 5. When the sky diver releases their parachute, they massively increase their surface area, unbalancing the forces as the air resistance acting on them increases. 6. This decelerates them to a safe velocity, until they hit the ground. 7. Once stationary on the ground, the forces on them are once again balanced.