Posted by Austin on Thursday, November 13, 2008 at 9:19pm.
Luis bought eight bags of candy at the store for a total price of $21.54. He bought some bags of candy that cost $2.22 each and the rest cost $3.48 each. How many bags of the $3.48 candy did Luis buy?
Responses
Math - qwerty, Thursday, November 13, 2008 at 9:23pm
A=2.22
B=3.48
A + B = 8
2.22A x 3.48B = 21.54
solve for B
Math - Austin, Thursday, November 13, 2008 at 10:08pm
I got this far, but how do I solve for B?
Math - bobpursley, Thursday, November 13, 2008 at 9:24pm
C means cheap candy bag, E means expensive candy bag.
C*2.22+E*3.48=21.54
C+E=8
A=8-B
Put that into the other equation for A, and then solve for B.
To solve for the number of bags of the $3.48 candy that Luis bought, you need to use the given equations:
C*2.22 + E*3.48 = 21.54
C + E = 8
In these equations, C represents the number of bags of $2.22 candy that Luis bought, and E represents the number of bags of $3.48 candy that Luis bought.
To solve for E, we can use the second equation to solve for C in terms of E:
C = 8 - E
Now substitute this value of C into the first equation:
(8 - E)*2.22 + E*3.48 = 21.54
Simplify the equation:
17.76 - 2.22E + 3.48E = 21.54
1.26E = 3.78
Divide both sides by 1.26:
E = 3
Therefore, Luis bought 3 bags of the $3.48 candy.