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    bigdee1313's Avatar
    bigdee1313 Posts: 3, Reputation: 1
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    #1

    Mar 29, 2017, 03:01 PM
    Basic algerba
    the sum of two numbers is 14. The total of 3 times the smaller and twice the larger is 33. Assume x is the smaller number. I need to form the equation and find the smaller and larger numbers.
    smoothy's Avatar
    smoothy Posts: 25,492, Reputation: 2853
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    #2

    Mar 29, 2017, 03:21 PM
    What do you have so far?
    bigdee1313's Avatar
    bigdee1313 Posts: 3, Reputation: 1
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    #3

    Mar 29, 2017, 03:29 PM
    3x+2=33 I don't think that's even right I am drawing a blank
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    bigdee1313 Posts: 3, Reputation: 1
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    #4

    Mar 29, 2017, 03:43 PM
    I figured it out using 3x+28-2x=33. 5 is the smaller and 9 is the larger. The numbers are correct but its marking the equation wrong

    I figured it out lol. 3x+2(14-x)=33 x=5 the larger number is 9
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #5

    Mar 30, 2017, 05:57 AM
    Quote Originally Posted by bigdee1313
    I figured it out lol. 3x+2(14-x)=33 x=5 the larger number is 9
    Very good! However, some advice for future problem like this: when given information about two unknown quantities it is recommended that you take the time to set up the two equations that describe what's going on, then simplify. In this case we can let x = one unknown number and y = the other. From "the sum of two numbers is 14" we then get:

    x+y=14.

    Then from "The total of 3 times the smaller and twice the larger is 33" we get:

    3x + 2y = 33

    The next step is to figure a way to combine these two equations and eliminate one of the variables. Since yo are looking for the smaller of the two numbers, and we are told that is the 'x' term, we want to figure a way to susbtitue for y in one of the equations with an expression that involves x. From the first equation you can rearrange to get:

    y = 15-x

    Now substitute this value for 'y' into the second equation:

    3x+2(14-x)=33

    So now we are where you got to, but by going through these steps in a consistent way we can apply this same technique to much more complicated problems that may not be quite so easy. If you simply try to jump to the final equation without rigorously going through these steps you may introduce errors.

    I would also point out that you could have started with rearranging the second equation to get it in terms of y, then substitute back into the first:

    3x+2y = 33, therefore y = (33-3x)/2. Then from x+y=14 we get x +(33-3x)/2 = 14. This can be rearranged to x/2 = 5/2, and hence x=5. I wonder if your homework answer tool would accept x + (33-2x)/2=14 as a correct answer?

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