Sometimes we will have some information about a problem & will need to create the equation to solve the problem. In the below, we have worked through three examples, one for finding the equation for costing the manufacture of a box; one for the equation for total mileage used across two vehicles with different miles per gallon figures and one for calculating the time spent falling, given other variable values.

Let’s say we have a problem – we need to figure out how much it would cost for us to make a box out of wood.

We know that the base costs $10/m squared and the sides cost $6/m squared. We also know that the length is double the length of the width.

So then, we know that the base is going to be the length * the width * the cost per metre squared. The formula therefore is: $10 * 2w * w or $10*2w squared.

We have two ends, so we have 2 * width * height or 2(wh).

Finally, we know that the two sides are double the length of the ends, so we do 2 * 2 * width * height or 2(2wh).

Overall, we can put the formulas together to create the equations written in red at the bottom.

Now, the issue is, we want to be able to simply put the width into the equation & be able to tell the overall cost of the box. So we need to get rid of h.

To do that, we take the formula of 2w * w * h = 10 (10 metres cubed) and simplify to calculate h. which equals 5/w squared.

We can then plug this new calculation into the formula, removing h.

In this scenario, we have a car and a truck. Each have their own MPG figures. We know that the car has travelled 400 miles more than the truck & we want to know the total gallons used across both vehicles. We end up with the formula: *(c/27)+(c-400/18); *which is: *(total_car_miles / car MPG) + (total_car_miles – 400 / truck MPG)*.

We can simplify this formula as below. First, we need to standardise the denominator, which we can do by multiplying the left fraction by 2 and the right by 3; we can then add the two standardised fractions together.

Let’s say we need to figure out t (the time spent falling) for a ball based on the speed it travels & the height from which it is dropped.

We have the below information – the speed & height of the ball. We also know, that distance is equal to* (height*time)squared + (speed * time). *We also know that *average speed = distance / time*.

To find t, we need to first plug in the values we have. So, we know that v(average speed) is equal to distance / time & we know the formula for that. Let’s replace the letters for the numbers we know. We get *((16*t) squared + (12 * t))/t. *

We must now remove t from the equation, because it also appears above as multiplication, we can effectively cancel that out with the divisions, so we get *((16 * t) + 12).* If we now divide by 16 to get t on its own, we get* (v/16) = 5 + 12*. Now, we just minus 12 to get *(v -12 / 16) = t.*

So now, we know the formula for calculating the time that the ball spent falling.

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