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Math Help - solve by substitution

  1. #1
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    solve by substitution

    I tried many times to solve the following 2 problems but cannot get the answer. It looks simple enough yet so tough.

    Solve each system of equation by substitution.

    Problem 1.

    2c - d + 2 = 0
    3c + 2d + 10 = 0

    Problem 2.

    a + 4b = 3
    5b = -2a + 3

    No matter how I substitute, my answers could not match the answer key. The answer key for problem 1 is (-2,-2) and problem 2 is (-1,1). Please help. Thanks.
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  2. #2
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by shenton View Post
    I tried many times to solve the following 2 problems but cannot get the answer. It looks simple enough yet so tough.

    Solve each system of equation by substitution.

    Problem 1.

    2c - d + 2 = 0
    3c + 2d + 10 = 0
    You have equation: 2c - d + 2 = 0

    add d to both sides: 2c + 2 = d

    You have equation: 3c + 2d + 10 = 0

    substitute: 3c + 2(2c + 2) + 10 = 0

    multiply: 3c + 4c + 4 + 10 = 0

    add: 7c + 14 = 0

    subtract 14 from both sides: 7c = -14

    divide both sides by 7: c = -2

    go back to equation: 2c - d + 2 = 0

    substitute: 2(-2) - d + 2 = 0

    multiply: -4 - d + 2 = 0

    add: -2 - d = 0

    add 2 to both sides: -d = 2

    multiply by -1: d = -2

    can you do the other one now?
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  3. #3
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    Thanks Quick for coming to the rescue. This one is not as hard as it should be.

    I was able to solve the second problem with the encouragement from your workings.

    Thanks.
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  4. #4
    Senior Member OReilly's Avatar
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    Quote Originally Posted by shenton View Post
    I tried many times to solve the following 2 problems but cannot get the answer. It looks simple enough yet so tough.

    Solve each system of equation by substitution.

    Problem 1.

    2c - d + 2 = 0
    3c + 2d + 10 = 0

    Problem 2.

    a + 4b = 3
    5b = -2a + 3

    No matter how I substitute, my answers could not match the answer key. The answer key for problem 1 is (-2,-2) and problem 2 is (-1,1). Please help. Thanks.
    Solution of problem 2:

    From first equation we get a = 3-4b

    We substitute a in second equation:
    5b=-2(3-4b)+3
    5b=-6+8b+3
    5b=-3+8b
    3=3b
    b=1

    We can now substitute b in first equation:
    a=3-4(1)
    a=-1

    So, solution is a=-1 and b=1
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  5. #5
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by shenton View Post
    Thanks Quick for coming to the rescue. This one is not as hard as it should be.

    I was able to solve the second problem with the encouragement from your workings.

    Thanks.
    It's always nice to see someone learn from this site rather than just get all the answers without being able to do them themselves.
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  6. #6
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    Thanks OReilly for your help and providing the workings. This is very useful in understanding the how the final answer is derived.

    I think I will try some equation solving by "elimination". Hopefully, that won't cause me any headache.

    Thanks.
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