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Why do we equate the two values of x?

Actually I put the two ys equal and solved for x. But I could have eliminated the xs and solved for y. That would have worked as well.

If the lines cross then, at the intersection point they both have the same x and y coordinates. You have two equations so it becomes a simultaneous equations problem. What are the values of x and y that fit both equations simultaneously. So eliminate one unknown and solve for the other. As both equations are in the form y = function of x, the quickest way to an answer is to make the two functions of x equal and find the one x that works for both. Once you have that you can substitite that x value into either equation to get y. It works whichever equation you choose because that y is the one that fits in both equations.

Many routes to the same answer:

y = 2x -2

y = x/2 -1/2

Subtract the left hand sides and the right hand sides:

0 = 3x/2 - 3/2 so 3x/2 = 3/2 so x=1

substitute in y = 2x -2 ..... y = 2times 1 -2 = 0

substitute in y = x/2 -1/2 ...... y = 1/2 - 1/2 = 0

Make x the subject of each:

x = (y+2)/2

x = 2(y+1/2)

Set these equal

Substitiute in y = 2x - 2 ....... 0 = 2x - 2 ....... x = 1

Substitite in y = x/2 - 1/2 ....... 0 = x/2 - 1/2 ........ x/2 = 1/2 ....... x = 1

Bob

Equating those values of y gives

In general

Bob

It's all in the name. HCF is highest common factor. The common factors must be in the intersection because that's what the intersection means. We want the highest so multiply them together.

Splitting the factors into their primes components helps us to identify the common ones. You could put {1,2,3,6,7,14,21,42} all in one circle and {1,2,4,7,8,14,28,56} in the other, and then you could spot that 14 is the highest in both sets but it's a lot more work.

Bob

To make a Venn diagram, draw one circle for each number. Put the prime factors inside the circles so that common primes are in the intersecting part of the diagram.

The union gives the LCM.

Bob

The coordinates of two points before and after the enlargement is enough. Let's say they are (x1,y1) and (x2,y2) in the first shape and (X1,Y1) and (X2,Y2) in the transformed shape.

Find the equation of the line joining (x1,y1) to (X1,Y1) and also the other similar line. These are the rays but expressed algebraically.

Find where they cross. that's the centre.

You could probably construct a formula for this and then you're independent of a graph entirely.

Bob

If the centre is C and A = (x1,y1) and B = (X1,Y1) then CB/CA gives the scale factor.

I think it's odd that the question refers to small bars and large bars. Does that mean there are two sizes of bar? I think the question just means bars that are put in small packs and same size bars that are piut in large packs. Altogether a very badly worded and thought out question. It would be easy to change the numbers a bit so that only whole packs are sold and still test the same skills.

Bob

Starting with the hexagon divided into 6 equilateral triangles.

Let side = a and apothem = h. Note this is the height of a triangle.

Area of a triangle = half base times height = 0.5 ah.

So area of hexagon = 6 times 0.5 ah = 3ah. But perimenter = P = 6a so area = 0.5 times 6ah = 0.5 hP.

Using Pythag on a half triangle h = √ [a^2 - (0.5a)^2] = √3 a/2

so area = 3ah = 3a .√3.a/2 = 3√3 a^2 /2.

Bob

My working is a little different but comes to the same answer, 50 small bars leading to 12.5 packs. you can get the answer even though the answer makes no sense. I've re-checked this several ways and still get to the same conclusion.

Odd

Bob

The grower bags up the pots and weighs them. He puts the rounded down amount on the bag so no customer can complain they are being undersold.

Bob

You're talking about a regular hexagon I think; so the answer is yes. You can divide the hexagon into 6 equilateral triangles each with side = the same as for the hexagon itself. Let's say side = a. Then perimeter = 6a and diagonal = 2a.

Bob

Correct!

Bob

5x= 6 means x = 1.2

An equation that gives rise to a line is why the word linear is used.

It's the absence of any powers eg x^2 that allows this to happen.

This usage has spilt over into equations generally so in your examples linear is used because there's no powers .

Bob

hi paulb203

It looked to me like this should be 'provable' using algebra but I've come unstuck with it. I don't think I'm properly following what you're suggesting.

If a square has side 'a' then the 'diameter' = a and so the 'radius' = a/2 and the perimeter = 4a

sqi = the ratio of the ‘diameter’ of the square to the square’s ‘circumference’

So the ratio of diameter/circumference = a/4a = 1/4. Ah! Think I've just spotted what to do.

Ratio of circumference/diameter = 4

Then 'area' = 4 x (a/2)^2 = 4 (a^2)/4 = a^2

Bob

Can all linear equations be rearranged to the gradient-intercept form?

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.

I got that too, at first. Then I re-read the question!

outside diameter=8m and inside diameter=6m.

The area formula needs the radius not the diameter. So you have to half the numbers given. Here's another way to think about it.

Take that 8m circle. You could box it in with a 8m by 8m square and that would have an area of 8x8 = 64. So the answer must be a lot less than that.

Bob

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?

hi Dombo,

Welcome to the forum.

If you need help then you're at the right place.

What's the problem?

Bob

When you're learning about y = mx + c it's a big help to have equal scales but there going to come a time when you have to deviate from this.

eg y = 1000x + 100.

The MIF function grapher has equal scales.

My ideal would default to equal scales but give the user the option so rescale each axis.

Bob

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

hi Irene

Algebra is a vast topic, so it's hard for me to know where to begin.

Please give me an example of something you're stuck on?

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

If I've got to do this without a calculator then I'd do some carefully chosen rounding to make the sums easy. But if I'm allowed then I'd always prefer to keep to the exact values until the end and only then round off.

365 x 24 x 60 x 60 / 43 is 733395.349 so the 'book' answer of 720000 is pretty good. I'd say 600000 to 800000 is reasonable.

Bob

Ok. So try one here.

Bob

I've got a busy day so I won't be able to respond properly for a while. A version of the tethered goat problem came up on the forum in 2015 and I made a series of pictures to show what's going on. You won't find it in a search because this feature doesn't go back before the forum upgrade.

But you can find it by searching my posts. Click on my name and then on show all posts. There's a lot but you can home in on the right date. I think my pictures are clearer than the ones on the vid.

The basic idea is that the goat can reach grass using the full extent of the rope until the rope catches a corner of the barn. Then the available radius is reduced by the length of the side of the barn.

The measurements may be new. I'll try to look again tomorrow.

Bob