Ten Points for whoever works this out for me?

I need to check my workings (and also if the answer is correct, that would help!).....





Mr Jones is ordering tea and coffee for his cafe.


He spends exactly 拢108 on these each month.





One month he orders 4kg of tea and 6kg of coffee.


The next month he changes his order to 8kg of tea and 3kg of coffee.





By forming a pair of equations in two unknowns, find how much he pays per kg for tea AND for coffee.








Thanks guys :) First answer that tallies with mine gets the 10 points :D


Ten Points for whoever works this out for me?
4x + 6y = 108





8x + 3y = 108


8x + 12y = 216


------------------------


-9y = -108


y = 12





4x + 6y = 108


4x = 108 - 72


x = 36/4


x = 9





he paid tea for 9$/kg and coffee for 12$/kgTen Points for whoever works this out for me?
4t+6c=108


8t+3c=108


these are the equations,use the substitution:


4t=108-6c t=1/2(54-3c)


I change the tea with his correspondent with Coffee in the second equation:


8*1/2(54-3c)+3c=108


4*54-12c+3c=108


-9c=-108


c=108/9=12


now i find the tea:


t=1/2(54-3*12)=1/2(18)=9



4T + 6C = 108


8T + 3C = 108


8T + 12C = 216 (multiply eqn 1 by 2)


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9C = 108 (subtract eqn 2 from the new eqn)


C = 12, so coffee costs 拢12 per kg.





4T + 6(12) = 108


4T = 108-72 = 36


T = 9, so tea costs 拢9 per kg.
Make Coffee X and Tea Y





Then:





4y+6x=108


and the following month


8y+3x=108





By multiplying the second equation by 2





16y+6x=216


then subtracting the first from the second


4y+6x=108





12y=108


y=9





substituting:





4y+6x=108


4*9+6x=108


6x=72


x=12
4T + 6C = 108 ............A


8T + 3C = 108.............B





1........ T = 108/4 - (3/2)C............F


Replace into B:


8*{(108/4 - (3/20)C} + 3C = 108


C = 90 Replace into A or B or F then


T = 108/4 -(3/2)*90


T = -108
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