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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.
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    (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?
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    (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 ?
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    (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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    (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.
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    (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)
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    (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.
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    (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 )

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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.
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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.
    You should leave it as a fraction, don't try to round it.
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    (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 x in one column and y in the other, and plot the points like that.
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    (Original post by _gcx)
    If that method is confusing you, use a table with x in one column and y 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.
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    (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 x 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 y value, substitute each corresponding x value into the original equation. For example, given x=3, the equation would be \frac{2}{3}(3)+2, which would evaluate to y=\frac{6}{3}+2 (which is 2+2), which further evaluates to y=4. Therefore, when x=3, y=4, and you would add it to the table as such.
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    (Original post by _gcx)
    • In the x 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 y value, substitute each corresponding x value into the original equation. For example, given x=3, the equation would be \frac{2}{3}(3)+2, which would evaluate to y=\frac{6}{3}+2 (which is 2+2), which further evaluates to y=4. Therefore, when x=3, y=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?
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    Following thread 😊
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    (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
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    (Original post by _gcx)
    Yes
    Thanks for your help. You explained everything very clearly 😁
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    (Original post by _gcx)
    Yes
    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.
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    (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
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    do past papers forget about everything else
 
 
 
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