Simultaneous Algreba problem dealing with two simulaneous equations being p = unknown cost of plastic mug and d=unknown cost of dairy
3p=6+2d
d=p+1
substitute d into first equation
3p=6+2(p+1)
3p=6+2p+2
3p-2p=6+2p-2p+2
p=6+2
p=8
from the first part
d=p+1
p=d-1
take 3p=6+2d
3(d-1)=6+2d
3d-3=6+2d
1d=9
putting p and d back into 3p=6+2d to prove answer correct
3*8=6+2*9
24=6+18
24=24
putting into second equation to prove answer correct
p=d-1
8=9-1
8=8
so p=8 and d=9
so plastic mugs cost 8
and diaries cost 9
R42
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