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Math Help - More problems

  1. #1
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    More problems

    plz help solve these, thank you



    1.Find two numbers who Difference is 10 and 1 over 6 whoses sum is 11.



    2. The sum of two numbers is 33. Twice the first number minus the second number Equals 18. Find both numbers.



    3. The units digit of a two digit numeral is 5 more than the tens' digit. The number is 3 times the sum of its digits. Find the numeral



    4. The sum of the digits of a two digit numeral is 8. If the digits are reversed, the new number is 18 greater then the original number. Find the original numeral
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  2. #2
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    Quote Originally Posted by Aarxn
    2. The sum of two numbers is 33. Twice the first number minus the second number Equals 18. Find both numbers.
    I do one for you but you have to promise me to try the other three.

    These are your equations,
    \left\{ \begin{array}{c}x+y=33 \\ 2x-y=18

    There five ways to solve these problem.
    I will only mention two which I believe you should know: addition method, substitution method.

    Addition method
    Add the two equations, note that x+2x=0, y+(-y)=0, 33+18=51
    Thus,
    3x=51 from here divide both sides by three,
    x=17, thus, y=16
    The entire purpose of the addition method is when you add your equation you eliminate one of the variables and are left with one.

    Substitution method
    Solve and equation for any variable, i.e. first equation for 'x' thus,
    x=33-y
    Now substitute that into the second equation,
    2(33-y)-y=18, you substituted 'x' for 'y' and hence eliminated a variable.
    Open parantheses,
    66-2y-y=18
    Subtract 66 and combine 'y' to get,
    -3y=-48 divide by (-3) to get,
    y=16 thus, x=33-16=17
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  3. #3
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    Quote Originally Posted by ThePerfectHacker
    I do one for you but you have to promise me to try the other three.

    These are your equations,
    \left\{ \begin{array}{c}x+y=33 \\ 2x-y=18

    There five ways to solve these problem.
    I will only mention two which I believe you should know: addition method, substitution method.

    Addition method
    Add the two equations, note that x+2x=0, y+(-y)=0, 33+18=51
    Thus,
    3x=51 from here divide both sides by three,
    x=17, thus, y=16
    The entire purpose of the addition method is when you add your equation you eliminate one of the variables and are left with one.

    Substitution method
    Solve and equation for any variable, i.e. first equation for 'x' thus,
    x=33-y
    Now substitute that into the second equation,
    2(33-y)-y=18, you substituted 'x' for 'y' and hence eliminated a variable.
    Open parantheses,
    66-2y-y=18
    Subtract 66 and combine 'y' to get,
    -3y=-48 divide by (-3) to get,
    y=16 thus, x=33-16=17
    Alright, I'll try to do the other 3. thx
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  4. #4
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    I got 14 and 9 for the second one, I think thats right?
    I can't figure out the other two
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  5. #5
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    Quote Originally Posted by Aarxn
    4. The sum of the digits of a two digit numeral is 8. If the digits are reversed, the new number is 18 greater then the original number. Find the original numeral
    Okay, let me do this one.
    This question tests your knowledge of what numbers mean. For example the number 69 means 6 tens and 9 ones. That means, 6\cdot 10+9\cdot 1.

    Thus, given a two digit number,
    XY it is equivalent to 10X+Y
    When reverse gives YX=10Y+X

    Sum of digits is X+Y=8

    When reversed is greater than is 18 greater then original meaning,
    YX=XY+18
    Thus,
    10Y+X=10X+Y+18
    Rewrite this as,
    9Y-9X=18 and divide by 9,
    Y-X=2 thus, Y=X+2
    substitute that into equation above to get,
    X+(X+2)=8 combine,
    2X+2=8 subtract by two,
    2X=6 and divide by 2,
    X=3 then, Y=3+2=5,
    thus the required number is 35
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  6. #6
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    [HTML]3. The units digit of a two digit numeral is 5 more than the tens' digit. The number is 3 times the sum of its digits. Find the numeral[/HTML]

    This is a problem which requires you to convert the problem statement into an algebra equation. You must first find 2 simultaneous equations:

    y= x+5
    10x+y= 3(x+y)
    10x+x+5= 3(x+x+5) <--------- since y= x+5, substitute x+5 and get rid of y to make the equation solvable.
    11x+5= 3x+3x+15 <--------- expand the brackets then add like terms
    5x= 10 <--------- put one set of like terms on one side and other set of like terms on other.. be consistent with this (i tend to put the pronumeral on left hand side).
    .: x=2
    y=7
    Now 10x+y= 3(x+y) referring back to the problem, therefore x must be the tens digit and y be the units.

    .: The number is 27

    These problems are not as hard as they first sound... try converting these algebraicly, and it will help tremendously. If you get used to it, it's like solving simple algebra questions, just take your time.
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