Math Is Fun Forum

Yes.  Just make y the subject of the equation.  If you're not sure how to do that here's a step by step.

1)  Multiply out all brackets and multiply by the lowest common denominator to eliminate any fractions.

2) Move terms about so that all the y containing terms are on the left and  everything else is on the right.

3) If there's more than one y term, factorise it out so you have y(......) on the left.

4) Divide by the bracket from the line above so that y is on its own.

eg.

2x/3 + 4(x+y) = x - y + 5

1) clear that bracket and times all by 3

2x + 12x + 12y = 3x - 3y + 15

2) move terms

12y + 3y = 3x - 12x - 2x + 15

3) factorise the y. In this case we can do a lot of simplifying as well.

y(12+ 3) = 15y = -11x + 15

4) y = -11x/15 + 1

Bob

check put x = 15 in the final line so y = -11 + 1 = -10

Substitute these values into what we had at the beginning. If it works then I've probably not made an error.

LHS = 10 + 4 times (15-10) = 10 + 4 times 5 = 30

RHS = 15 --10 +5 = 15 + 10 + 5 = 30

This method of checking can be very useful. The rules of algebra are the same as the rules for numbers so if you choose some numbers to fit an equation at one stage in the working, those same numbers should 'fit' at every line in the working.  If they don't you've found an error with your working.

The forumula for the area of a circle, radius r is

We are told diameters so we need to divide by 2 to get the right radius.  ie 4 and 3

If a ring is a circle with a hole in it you can get the area of the ring by calculating the area of the big circle and subtracting the area of the inside hole.

You're told to take pi as 22/7 which suggests to me there's a simplification in the working involving cancelling that 7.  Let's see:

pi is a common factor so you can do the subtraction and times the answer by pi

Bob

ps.  This example uses substitution into a formula, factorising and cancelling fractions.  Would you like some easier examples of any of these?

https://www.mathsisfun.com/data/function-grapher.php

B

It happens to us all. I wasted a large chunk of an exam trying to eliminate a variable and failing.  Afterwards I spoke to a friend who had done it with one simple thing that I'd missed. Ggggrrr!

Bob

I've found time to look at the video.  The measurements are different from the problem I met years ago and the working comes out more easily, so I thought it would be simplest to make a new picture.

If the barn wasn't there then the goat could reach all the grass inside a circle radius 10, so (pi times 10 times 10) in area.

But the barn gets in the way of the rope and there's no grass to eat inside the barn anyway. The orange area is the bit that the goat can reach without any rope snagging problems.  Hence 3/4 x pi x 10 squared.

When the goat is on the line AD produced, the point D acts as a new tether point with new radius 10 - 7.  Once again the goat cannot reach the whole of the new circle; only 1/4 of a circle radius 3.

Similarly when the goat is on the line AB produced the point B cats as a new tether point with new radius 10 - 9.  So the extra area here is 1/4 of a circle radius 1.

I agree with the video answer.

Bob

Ok. So try one here.

Bob

It sounds like you have to find out how many seconds in that year (not a leap year) and divide by 43.

But you also have to consider what variance there might be in making that assessment.  I'd certainly want to round off my answer ... decimal place figures after the point would have no meaning ... but even the whole number answer probably needs rounding further.  So two people could get different answers as a result of different amounts of rounding I suppose.  But all answers should be in the same ball park.  Someone must have known the total in order to get the 'every 43 seconds' claim.

Bob

Yes! I'm assuming it means he ran  a total of T km in 5 days and his mean daily distance was 13.2 km. Work out T.

Without the word 'daily' it could mean all sorts of things.  Maybe that's his hourly rate!  (OK that's unlikely for a runner but math questions need to be worded clearly.)

Maybe the terrain was very rough and the temperature very hot and so the questioner thought it was mean for the organisers to set such a challenge.

Bob

hi locallycompact

Welcome to the forum.

Whoops, it looks like you are right.  I tried this on a dummy account and got no email.  I can do some things as an administrator but not fix that. Sorry.

Do you have another username that you want to resurrect?  I can probably fix that for you. Tell me the old membership name and I'll see what I can do. I will have to be convinced that you're not trying to steal someone else's account of course.

Bob

hi Chenfeng Liu

Welcome to the forum.

Also you may post questions here for help.

Bob

I tried googling that quote again and most hits attribute it to Galileo but a similar one  "I cannot teach anybody anything, I can only make them think" is attributed to Socrates.  I think you're right.  Galileo would have read Socrates and the two quotes are very similar especially when you take account of translation variations.

A modern version is "You can lead a horse to water, but you can't make him drink."

You can also use the method to find the highest common factor.

eg. Find the HCF of 45 and 72

45 =                 3 x 3 x 5
72 = 2 x 2 x 2 x 3 x 3

This time extract all the primes that occur in both.  So 3 x 3

If you create a Venn diagram with all the primes and each circle enclosing those for one of the  numbers then the lowest common multiple (LCM) is the union of the sets and the HCF is the intersection.

Bob

Reduce each demoninator to its prime factors

3 = 3
9 = 3x3
4 = 2x2
16 = 2x2x2x2
10 = 2x5

Write the largest power of each prime and multiply them

(2x2x2x2) x (3x3) x 5

That's the only way to make sure that 16 and 9 go into it.

Bob