SOLUTION: A customer buys five cans of baked beans, four packets of brown sugar and two packets of biscuits from a supermarket. He gives the cashier R150.00 and gets R26.00 in change. A can

Algebra ->  Human-and-algebraic-language -> SOLUTION: A customer buys five cans of baked beans, four packets of brown sugar and two packets of biscuits from a supermarket. He gives the cashier R150.00 and gets R26.00 in change. A can       Log On


   



Question 1030122: A customer buys five cans of baked beans, four packets of brown sugar and two packets of biscuits from a supermarket. He gives the cashier R150.00 and gets R26.00 in change. A can of baked beans costs half as much as a packet of biscuits whilst a packet of sugar is three rand cheaper than a packet of biscuits.
Calculate the cost of a can of baked beans, a packet of sugar and a packet of biscuits

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +a+ = the cost of a can of baked beans
Let +b+ = the cost of a packet of sugar
Let +c+ = the cost of a packet of biscuits
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(1) +5a+%2B+4b+%2B+2c+=+150+-+26+
(2) +a+=+%281%2F2%29%2Ac+
(3) +b+=+c+-+3+
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There are 3 equations and 3 unknowns, so it's solvable
Substitute (2) and (3) into (1)
(1) +5%2A%28+%281%2F2%29%2Ac+%29+%2B+4%2A%28+c-3+%29+%2B+2c+=+150+-+26+
(1) +%285%2F2%29%2Ac+%2B+4c+-+12+%2B+2c+=+124+
(1) +%28+5%2F2+%29%2Ac+%2B+6c+=+136+
(1) +%28+5%2F2+%29%2Ac+%2B+%28+12%2F2+%29%2Ac+=+136+
(1) +%2817%2F2%29%2Ac+=+136+
(1) +17c+=+272+
(1) +c+=+16+
and
(2) +a+=+%281%2F2%29%2A16+
(2) +a+=+8+
and
(3) +b+=+16+-+3+
(3) +b+=+13+
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R8.00 = the cost of a can of baked beans
R13.00 = the cost of a packet of sugar
R16.00 = the cost of a packet of biscuits
------------------
check:
(1) +5a+%2B+4b+%2B+2c+=+150+-+26+
(1) +5%2A8+%2B+4%2A13+%2B+2%2A16+=+150+-+26+
(1) +40+%2B+52+%2B+32+=+124+
(1) +124+=+124+
OK