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Original post by g_a1
Sorry why is there 4 days left

8 machines = 5 days to do all of the work
so the 4 machines will take 10 days to complete all of the work.

They were turned on for 2 days = 2/10 of the work is done (2 being the time they were on and 10 being the total time it takes 4 machines to do all of the work) 2/10 = 1/5

1/5 of the work is done. So 4/5 of the work still needs to be done.

4/5 of 5 (time it takes 8 machines to do the work) = 4 days (4/5 * 5/1 = 20/5 = 4/1 = 4)

total time = 4 (time spent by the 8 machines) + 2 (time spent by the 4 machines) = 6 days

Hope this helped.
Original post by _gcx
Divide both sides by 3. You will then be able to plot it like any other line (using a table etc.)


Ok so 3/3 is y and 6/3 is 2 so y=2x+2. Is that correct?
Original post by _gcx
Divide both sides by 3. You will then be able to plot it like any other line (using a table etc.)

Also if I come across another equation like this, is the rule the same? Do you divided both sides by the first part ?
Original post by Zonyali123
Also if I come across another equation like this, is the rule the same? Do you divided both sides by the first part ?


You do whatever you need to do to get y= on the left side. Have you done changing the subject of a formula? You may need to collect like terms, expand brackets / factorise or add and minus stuff from both sides as well as dividing and multiplying.

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Reply 64
Original post by g_a1
Sorry why is there 4 days left

You have two stages to make the bottles:

Stage 1 is done by 4 machines in 2 days
Stage 2 is done by 8 machines in n days

And we know that 8 machines take 5 days in total to make the bottles.

4 machines in 2 days do the same amount of work as 8 machines in 1 day. So you can imagine that stage 1 is actually done by 8 machines so you have:

Stage 1 is done by 8 machines in 1 day
Stage 2 is done by 8 machines in n days

Since it takes 5 days in total for 8 machines to make the bottles, stage 2 must take 5 - 1 = 4 days.

Then since stage 1 actually took 2 days, the total process takes 2 + 4 = 6 days.
Original post by Zonyali123
Ok so 3/3 is y and 6/3 is 2 so y=2x+2. Is that correct?


You have forgot to divide the 2x by 3, unless that was a typo in your previous post. (3y = 2x + 6)
Original post by _gcx
You have forgot to divide the 2x by 3, unless that was a typo in your previous post. (3y = 2x + 6)

Sorry it wasn't a typo, but if I divide 2x by 3 I get 0.6.
Reply 67
Original post by Frazzle27
8 machines = 5 days to do all of the work
so the 4 machines will take 10 days to complete all of the work.

They were turned on for 2 days = 2/10 of the work is done (2 being the time they were on and 10 being the total time it takes 4 machines to do all of the work) 2/10 = 1/5

1/5 of the work is done. So 4/5 of the work still needs to be done.

4/5 of 5 (time it takes 8 machines to do the work) = 4 days (4/5 * 5/1 = 20/5 = 4/1 = 4)

total time = 4 (time spent by the 8 machines) + 2 (time spent by the 4 machines) = 6 days

Hope this helped.


Yes!!! I get it thanks :smile:)

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Original post by Zonyali123
Sorry it wasn't a typo, but if I divide 2x by 3 I get 0.6.

I have studied the slope intercept form, and you leave 2/3 as it is, so it's y=2/3+2. However, I want to know whether there is another way of doing this, as this way is confusing me. I know I have to start on 0 on the y axis then I go 2 up (vertically)as I should add 2. Then 2 up again (vertically)and 3 horizontally right, but after that I don't know what to do.
Original post by Zonyali123
Sorry it wasn't a typo, but if I divide 2x by 3 I get 0.6.


You should leave it as a fraction, don't try to round it.
Original post by Zonyali123
I have studied the slope intercept form, and you leave 2/3 as it is, so it's y=2/3+2. However, I want to know whether there is another way of doing this, as this way is confusing me. I know I have to start on 0 on the y axis then I go 2 up (vertically)as I should add 2. Then 2 up again (vertically)and 3 horizontally right, but after that I don't know what to do.


If that method is confusing you, use a table with xx in one column and yy in the other, and plot the points like that.
Original post by _gcx
If that method is confusing you, use a table with xx in one column and yy in the other, and plot the points like that.

Can you please tell me how I should make a table? I have made tables for other equations but I was given the numbers, but in this question, I've not been given any numbers.
Original post by Zonyali123
Can you please tell me how I should make a table? I have made tables for other equations but I was given the numbers, but in this question, I've not been given any numbers.


In the xx column, put the values that you're asked for. If you're given the range "between 0 and 3", for example, write the natural numbers between 0 and 3.

To get the yy value, substitute each corresponding xx value into the original equation. For example, given x=3x=3, the equation would be 23(3)+2\frac{2}{3}(3)+2, which would evaluate to y=63+2y=\frac{6}{3}+2 (which is 2+22+2), which further evaluates to y=4y=4. Therefore, when x=3x=3, y=4y=4, and you would add it to the table as such.

Original post by _gcx

In the xx column, put the values that you're asked for. If you're given the range "between 0 and 3", for example, write the natural numbers between 0 and 3.

To get the yy value, substitute each corresponding xx value into the original equation. For example, given x=3x=3, the equation would be 23(3)+2\frac{2}{3}(3)+2, which would evaluate to y=63+2y=\frac{6}{3}+2 (which is 2+22+2), which further evaluates to y=4y=4. Therefore, when x=3x=3, y=4y=4, and you would add it to the table as such.



So I multiply the numbers in the x column by 2/3 then I add 2 which gives me the answer to y. Is this correct?
Following thread 😊
Original post by Zonyali123
So I multiply the numbers in the x column by 2/3 then I add 2 which gives me the answer to y. Is this correct?


Yes :smile:
Original post by _gcx
Yes :smile:


Thanks for your help. You explained everything very clearly 😁
Original post by _gcx
Yes :smile:


Just to make sure that I am on the right lines, in the equation: 2y=5x-8 do I divide all the numbers by 2? For example 2/2 is y, 5/2 stays as it is and 8/2 is 4 so I'm left with y=5/2-4.
Original post by Zonyali123
Just to make sure that I am on the right lines, in the equation: 2y=5x-8 do I divide all the numbers by 2? For example 2/2 is y, 5/2 stays as it is and 8/2 is 4 so I'm left with y=5/2-4.


yes
do past papers forget about everything else

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