SOLUTION: Using the system you selected above, solve the system of equations. Show/ upload all work. Jesse and Adam are both selling items to raise money for Baseball uniforms. They are e

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Question 1171212: Using the system you selected above, solve the system of equations.
Show/ upload all work.
Jesse and Adam are both selling items to raise money for Baseball uniforms. They are each selling
two items: chocolate bars and candles. Jesse sells 12 chocolate bars and 7 candles and makes a
total of $94. Adam sells 24 chocolate bars and 5 candles and makes a total of $98. How much does
one chocolate bar and one candle cost?

Answer by ikleyn(52783) About Me  (Show Source):
You can put this solution on YOUR website!
.
Jesse and Adam are both selling items to raise money for Baseball uniforms. They are each selling
two items: chocolate bars and candles. Jesse sells 12 chocolate bars and 7 candles and makes a
total of $94. Adam sells 24 chocolate bars and 5 candles and makes a total of $98. How much does
one chocolate bar and one candle cost?
~~~~~~~~~~~~~~


    12H + 7C = 94       (1)   (H stands for chocolate bars;  C stands for candles)

    24H + 5C = 98       (2)


Multiply equation (1) by 2 (both sides).  Keep equation (2) as is.  You will get


    24H + 14C = 188     (1') 

    24H +  5C =  98     (2')


Now subtract equation (2') from equation (1').  

The terms "24H" will cancel each other, and you will get a single equation in one unknown C

          14C - 5C = 188 - 98

             9C    =  90

              C    =  90/9 = 10.


Thus the cost of each candle is $10.


To find another unknown, H, substitute the found valu C into equation (1).  You will get

    12*H + 7*10 = 94

    12H         = 94 - 70 = 24

      H                   = 24/12 = 2.


ANSWER.  $2 per chocolate bar and $10 for each candle.

Solved.

From my solution, learn on how the ELIMINATION method works.

............

I remember that in my previous post I showed you, how to solve this problem MENTALLY.

So, now you know TWO methods, which is good (!)

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On solving systems of linear equations in two unknowns see the lessons
    - Solution of a linear system of two equations in two unknowns by the Substitution method
    - Solution of a linear system of two equations in two unknowns by the Elimination method
    - Solution of a linear system of two equations in two unknowns using determinant
    - Geometric interpretation of a linear system of two equations in two unknowns
    - Useful tricks when solving systems of 2 equations in 2 unknowns by the Substitution method
    - Solving word problems using linear systems of two equations in two unknowns

    - Word problems that lead to a simple system of two equations in two unknowns
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    - Using systems of equations to solve problems on tickets
    - Three methods for solving standard (typical) problems on tickets
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    - HOW TO algebreze and solve this problem on 2 equations in 2 unknowns
    - One unusual problem to solve using system of two equations
    - Solving mentally word problems on two equations in two unknowns
in this site.

Also,  you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic "Systems of two linear equations in two unknowns".


Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.