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Math Help - sim equations

  1. #1
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    sim equations

    Solve the following equations algebraically
    2x + y = 0
    −2x −3y = 8 im stuck i can do the easy ones when in the format y=mx+c
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by realistic View Post
    Solve the following equations algebraically
    2x + y = 0
    −2x −3y = 8 im stuck i can do the easy ones when in the format y=mx+c
    Thee's generally two ways to do these, but I'll hint towards the addition method because it is alreaddy set up

    Add the two equations like term by like term

    2x+(-2x)=0

    y+(-3y)=-2y

    0+8=8

    Now set up what's left

    -2y=8

    Can you finish?
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  3. #3
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    Quote Originally Posted by VonNemo19 View Post
    Thee's generally two ways to do these, but I'll hint towards the addition method because it is alreaddy set up

    Add the two equations like term by like term

    2x+(-2x)=0

    y+(-3y)=-2y

    0+8=8

    Now set up what's left

    -2y=8

    Can you finish?
    yeah that would be -4
    what is the other way cause i think thats what we are doing at school

    3x-5y=1
    x +5y=7

    =
    3x+x =4x
    -5y+5y=0
    1+7=8

    0=4x+8?
    x=-2?

    however the answer is 2 what did i do wrong?
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by realistic View Post
    yeah that would be -4
    what is the other way cause i think thats what we are doing at school
    It's called the substitution method.

    EG

    We have the sim equations

    x+y=1 and 3x-2y=7

    Then we solve for a variable in one of the equations and substitute for the variable in the other

    x+y=1
    y=1-x

    Now we sub for y in the other

    3x-2\overbrace{(1-x)}^y=7

    solve for x and we can determine the solution
    3x-2(1-x)=7
    3x-2+2x=7
    5x=9
    x=\frac{9}{5}

    Now we put \frac{9}{5} back in the original for x.

    Do you see?
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  5. #5
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    Quote Originally Posted by VonNemo19 View Post
    It's called the substitution method.

    EG

    We have the sim equations

    x+y=1 and 3x-2y=7

    Then we solve for a variable in one of the equations and substitute for the variable in the other

    x+y=1
    y=1-x

    Now we sub for y in the other

    3x-2\overbrace{(1-x)}^y=7

    solve for x and we can determine the solution
    3x-2(1-x)=7
    3x-2+2x=7
    5x=9
    x=\frac{9}{5}

    Now we put \frac{9}{5} back in the original for y.

    Do you see?
    yeah thats what we are doing but your first way is alot easier

    can you view the one i just tried? plz

    3x-5y=1
    x +5y=7

    =
    3x+x =4x
    -5y+5y=0
    1+7=8

    0=4x+8?
    x=-2?[/B]

    also how would you tackle this one

    2x=11 -3y
    0=6x-2y +11
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  6. #6
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by realistic View Post
    yeah that would be -4
    what is the other way cause i think thats what we are doing at school

    3x-5y=1
    x +5y=7
    Adding these two I'm left with

    4x+0=8

    Then x=2

    If we substitute back into the original we get

    3(2)-5y=1
    6-5y=1
    -5y=-5
    y=1

    So the equations have a common solution of

    (2,1)

    ...
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  7. #7
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    Quote Originally Posted by VonNemo19 View Post
    ...
    ah i c my mistake with that one

    2x = 11-3y
    0= 6x-2y+11

    =

    2x=-3xy+11
    -6x= -2y+11

    would you times the top by 3 then just add them ?
    6x= -9y+33

    +
    -6x=2y +11

    0= -11y +44
    11y= 44
    y= 4 ?

    then just substitute y =4 into one of above equations to get x?
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  8. #8
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by realistic View Post
    ah i c my mistake with that one

    2x = 11-3y
    0= 6x-2y+11

    =

    2x=-3xy+11
    -6x= -2y+11

    im stuck
    Think of it like a game, with the object being to line thx's and y's up underneath of each other and then add or subtract them to get rid of one of them.
     2x = 11-3y
    0= 6x-2y+11

    You can see here that the x's and y's are not right underneath of each other, but wait...

     2x =-3y+11
    -6x=-2y+11

    Now the x's and y's are lined up. But remember that the object is to add (or subtract) so that I eliminate one of the variables. To do this I have to do a little trick with multiplication...

     (-3)(2x) =(-3y+11)(-3)
    -6x=-2y+11

     -6x =9y-33
    -6x=-2y+11

    Now look! The x's and y's are still lined up, but now if I subtract the two equations, the x's will dissapear and I can solve for y!

    Do you see?
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  9. #9
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    Quote Originally Posted by VonNemo19 View Post
    Think of it like a game, with the object being to line thx's and y's up underneath of each other and then add or subtract them to get rid of one of them.
     2x = 11-3y
    0= 6x-2y+11

    You can see here that the x's and y's are not right underneath of each other, but wait...

     2x =-3y+11
    -6x=-2y+11

    Now the x's and y's are lined up. But remember that the object is to add (or subtract) so that I eliminate one of the variables. To do this I have to do a little trick with multiplication...

     (-3)(2x) =(-3y+11)(-3)
    -6x=-2y+11

     -6x =9y-33
    -6x=-2y+11

    Now look! The x's and y's are still lined up, but now if I subtract the two equations, the x's will dissapear and I can solve for y!

    Do you see?
    thanks a bunch i now understand that i can either subtract or plus i always just thought i could just minus.

    this has helped me alot and i thank you for you time
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  10. #10
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by realistic View Post
    thanks a bunch i now understand that i can either subtract or plus i always just thought i could just minus.

    this has helped me alot and i thank you for you time
    No problem!
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  11. #11
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    been doing alot of them and just got stuck at this one

    3y +2x +10 = 0
    2y = x −2

    =
    3y +2x +10 = 0
    2y - x +2 = 0

    x by 2? then add both? i tried but got wrong answer
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  12. #12
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by realistic View Post
    been doing alot of them and just got stuck at this one

    3y +2x +10 = 0
    2y = x −2

    =
    3y +2x +10 = 0
    2(2y - x +2) = (0)2

    3y +2x +10 = 0
    4y -2x +4 = 0

    Now add the two equations to eliminate the x-terms
    .
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  13. #13
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    Quote Originally Posted by VonNemo19 View Post
    .
    actually i did that i got 6 but it was wrong?

    what did you get?
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  14. #14
    Super Member Matt Westwood's Avatar
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    Add them, don't subtract them.
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    Quote Originally Posted by Matt Westwood View Post
    Add them, don't subtract them.
    ah yah silly me

    -2
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