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Math Help - 2 more world problems (please help!!)

  1. #1
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    2 more world problems (please help!!)

    a) Two cement blocks and three bricks weigh 102lb, as do one cement block and ten bricks. What does one brick weigh?


    b) A refrigerator repairman charges a fixed amount for a service call in addition to his hourly rate. If a two-hour repair costs $50 and a four-hour repair costs $74, what is the repairman's hourly rate?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by pinkmath135 View Post
    a) Two cement blocks and three bricks weigh 102lb, as do one cement block and ten bricks. What does one brick weigh?
    Let c be the weight of a cement block
    Let b be the weight of a brick

    then we have the system:

    2c + 3b = 102 ....................(1)
    c + 10b = 102 ....................(2)

    solve this system for b and you will have your answer. can you continue?

    b) A refrigerator repairman charges a fixed amount for a service call in addition to his hourly rate. If a two-hour repair costs $50 and a four-hour repair costs $74, what is the repairman's hourly rate?
    Let m be the hourly rate
    Let x be the number of hours the repair takes
    Let b be the fixed charge.

    then the cost of the repair is given by:

    c = mx + b

    since a two hour repair costs $50, we have:

    2m + b = 50 ..................(1)

    since a four-hour repair costs $74, we have:

    4m + b = 74 ...................(2)


    thus we need to solve the following system

    2m + b = 50 ..................(1)
    4m + b = 74 ..................(2)

    for m

    can you continue?
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    Let c be the weight of a cement block
    Let b be the weight of a brick

    then we have the system:

    2c + 3b = 102 ....................(1)
    c + 10b = 102 ....................(2)

    solve this system for b and you will have your answer. can you continue?

    Let m be the hourly rate
    Let x be the number of hours the repair takes
    Let b be the fixed charge.

    then the cost of the repair is given by:

    c = mx + b

    since a two hour repair costs $50, we have:

    2m + b = 50 ..................(1)

    since a four-hour repair costs $74, we have:

    4m + b = 74 ...................(2)


    thus we need to solve the following system

    2m + b = 50 ..................(1)
    4m + b = 74 ..................(2)

    for m

    can you continue?


    need more explaining please.
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