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Math Help - [SOLVED] solving linear equations in one variable????

  1. #1
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    [SOLVED] solving linear equations in one variable????

    here are the problems i need help with. can u tell me the steps to solve these problems?? thank you!

    first problem: a(x+b)=c

    second problem: 9x+2a=-3a+4x
    Last edited by student897; August 31st 2009 at 05:40 AM.
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  2. #2
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    Hi student897!

    Quote Originally Posted by student897 View Post
    i got a summer packet and i need some help with these math problems. I'm not very good at math

    ok so here are the problems i need help with can u tell me the steps to solve these problems?? thank you!

    first problem: a(x+b)=c

    second problem: 9x+2a=-3a+4x
    What do you mean? Solving for x?

    In this case

    1) a(x+b) = c

    a*x+a*b = c

    minus a*b

    a*x = c-a*b

    divide by a (a is not equal to 0 )

    x = \frac{c-a*b}{a}

    x = \frac{c}{a}-\frac{a*b}{a}

    x = \frac{c}{a}-b

    2) 9x+2a=-3a+4x

    minus 2a

    9x = -3a+4x - 2a

    9x = -3a-2a + 4x

    9x = -5a + 4x

    minus 4x

    9x-4x = -5a

    5x = -5a

    divide by 5

    x = -a

    Do you understand?

    Yours
    Rapha
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  3. #3
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    i dont like math its always been my worst subjectt! but thnx for the help i sort of get it noww
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  4. #4
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    thnx i understand but i sort of did it in a different way ... the way they taught me in school much less complicated i think.. but thnxxx i understand now
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  5. #5
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    " divide by a (a is not equal to 0 ) "

    This is extremely important and a cause for many errors when solving equations.

    That being said, the trick is to try to get X alone on one side. Divide, add, remove as much as you have to until it's just X = .....

    Might be helpful to repeat some of the basic rules;
    <br />
a(b + c) = a \cdot b + a \cdot c<br />
    <br />
d(e - f) = d \cdot e - d \cdot f<br />
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  6. #6
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    Nice work

    Good........
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  7. #7
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    Quote Originally Posted by student897 View Post
    i got a summer packet and i need some help with these math problems. I'm not very good at math

    ok so here are the problems i need help with can u tell me the steps to solve these problems?? thank you!

    first problem: a(x+b)=c

    second problem: 9x+2a=-3a+4x
    The basic idea is to "undo" whatever has been done to x. In the first problem, I see that two things have been done to x: first b is added to it, then that sum is multiplied by a. We can "undo" that by doing the opposite, in the opposite order: The opposite of "add b" is to subtract b and the opposite of "multiply by a" is to divide by a. And, since I do this in the opposite order, I first divide by a then subtract b. And, of course, whatever I do on one side of the equation, I do on the other to keep them "balanced".
    Starting from a(x+ b)= c and dividing both sides by a, a(x+b)/a= c/a or x+ b= c/a since "a/a"= 1. Subtracting b from both sides, x+ b- b= c/a- b or x= c/a- b since b- b= 0.

    You may notice that this is not exactly what Rapha did. He chose to "multiply" out a(x+b)= ax+ ab first. But the answer is the same, of course.

    The second, 9x+2a=-3a+4x, is slightly more complicated because it has "x" on both sides. We can fix that by getting rid of the "x" on the right- subtract 4x from both sides: 9x+ 2a- 4x= -3a+ 4x- 4x is the same as 5x+ 2a= -3a. Now that has x multiplied by 5 and then 2a added. The opposite of that is "subtract 2a" and then "divide by 5": 5x+ 2a- 2a= -3a- 2a is the same as 5x= -5a. Now divide both sides by 5: 5x/5= -5a/5 is the same as x= -a.
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  8. #8
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    thnx alot for the help now i think i understand it much better !!!i did it the way u said but at the end i ended up just putting 5x=-5a and then i didnt know what to pput!
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