SOLUTION: Hey, my name is Charity. Can you please help me on this problem>>
Two hamburgers and a coke cost $5.10. Three hamburgers and two cokes cost $8.40. Let h= cost of hamburgers and
Algebra ->
Systems-of-equations
-> SOLUTION: Hey, my name is Charity. Can you please help me on this problem>>
Two hamburgers and a coke cost $5.10. Three hamburgers and two cokes cost $8.40. Let h= cost of hamburgers and
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Question 1021776: Hey, my name is Charity. Can you please help me on this problem>>
Two hamburgers and a coke cost $5.10. Three hamburgers and two cokes cost $8.40. Let h= cost of hamburgers and c= cost of a coke
Thanks in advance :) Answer by macston(5194) (Show Source):
You can put this solution on YOUR website! .
2H+C=$5.10
C=$5.10-2H . Use this to substitute for C below.
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3H+2C=$8.40 . Substitute for C from above.
3H+2($5.10-2H)=$8.40
3H+$10.20-4H=$8.40
-H=-$1.80
H=$1.80
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C=$5.10-2H=$5.10-2($1.80)=$5.10-$3.60=$1.50
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ANSWER: Hamburgers cost $1.80 and cokes cost $1.50.
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CHECK:
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2H+C=$5.10
2($1.80)+$1.50=$5.10
$3.60+$1.50=$5.10
$5.10=$5.10
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3H+2C=$8.40
3($1.80)+2($1.50)=$8.40
$5.40+$3.00=$8.40
$8.40=$8.40
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