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Simulataneous Equation A System of Equations

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Fc Alan Walker
19/02/2018 at 21:15
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solution of the equation :

\(\dfrac{x-35}{21}+\dfrac{x-36}{20}>\dfrac{x-37}{19}+\dfrac{x-38}{18}\)

Simulataneous Equation A System of Equations

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    ¤« 19/02/2018 at 21:28

    We have :

    \(\dfrac{x-35}{21}+\dfrac{x-36}{20}>\dfrac{x-37}{19}+\dfrac{x-38}{18}\)

    \(\left(\dfrac{x-35}{21}-1\right)+\left(\dfrac{x-36}{20}-1\right)>\left(\dfrac{x-37}{19}-1\right)+\left(\dfrac{x-38}{18}-1\right)\)

    \(\dfrac{x-56}{21}+\dfrac{x-56}{20}>\dfrac{x-56}{19}+\dfrac{x-56}{18}\)

    \(\dfrac{x-56}{21}+\dfrac{x-56}{20}-\dfrac{x-50}{19}-\dfrac{x-50}{18}>0\)

    \(\left(x-56\right)\left(\dfrac{1}{21}+\dfrac{1}{20}-\dfrac{1}{19}-\dfrac{1}{18}\right)>0\)

    To \(\dfrac{1}{21}+\dfrac{1}{20}-\dfrac{1}{19}-\dfrac{1}{18}\) different 0

    \(x-56< 0\)

    \(x< 56\)
     


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justin bieber
19/03/2017 at 12:56
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Given a, b, c and d are positive integers

If (a - b - d)(a - c - d)(a + c - d)(a + b + c - d) = 210, find a, b, c and d

Simulataneous Equation A System of Equations


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justin bieber
19/03/2017 at 12:55
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If a, b, c and d statisfy a system of equations 

\(\left\{{}\begin{matrix}3a+2b+c+d=-7\\a+3b+2c+d=12\\a+b+3c+2d=62\\2a+b+c+3d=45\end{matrix}\right.\)

find the value of 10c + 5d

Simulataneous Equation A System of Equations


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justin bieber
19/03/2017 at 12:53
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It is given \(\left\{{}\begin{matrix}2r-4s+2t=0\\r-3r+4t=0\end{matrix}\right.\) Find r:s:t

Simulataneous Equation A System of Equations


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justin bieber
19/03/2017 at 12:52
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If \(\left\{{}\begin{matrix}a+2b+3c+4d=8\\a-2b+4c+3d=5\end{matrix}\right.\), find the value of a + 10b + c + 6d

Simulataneous Equation A System of Equations


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justin bieber
19/03/2017 at 12:49
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If \(\left\{{}\begin{matrix}3a+7b+c=103\\4a+10b+c=143\end{matrix}\right.\), what is a + b + c ?

Simulataneous Equation A System of Equations


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justin bieber
19/03/2017 at 12:48
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If \(\left\{{}\begin{matrix}3x+2y=12\\3x-y=3\end{matrix}\right.\), find the value of a in 4x - 8y + 2a = 0

Simulataneous Equation A System of Equations

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    FA KAKALOTS 06/02/2018 at 12:32

    {3x+2y=123x−y=3⇒{(3x+2y)−(3x−y)=12−33x=y+3

    ⇒⎧⎨⎩3y=9x=y3+1⇒{y=3x=2

    => 4x - 8y = -16 => 2a = 16 => a = 8

  • ...
    Phan Thanh Tinh Coordinator 22/03/2017 at 21:40

    \(\left\{{}\begin{matrix}3x+2y=12\\3x-y=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left(3x+2y\right)-\left(3x-y\right)=12-3\\3x=y+3\end{matrix}\right.\)

    \(\Rightarrow\left\{{}\begin{matrix}3y=9\\x=\dfrac{y}{3}+1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=3\\x=2\end{matrix}\right.\)

    => 4x - 8y = -16 => 2a = 16 => a = 8


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justin bieber
19/03/2017 at 12:47
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Find the value of 5 . (2x + y) +2 . (3y - 3x) + x + y, if x and y satisfy the equations:

\(\left\{{}\begin{matrix}2x+3y=5\\y-3x=9\end{matrix}\right.\)

Simulataneous Equation A System of Equations

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    FA KAKALOTS 06/02/2018 at 12:32

    {2x+3y=5y−3x=9⇒{2x+3(3x+9)=5y=3x+9

    ⇒{11x+27=5y=3x+9⇒{x=−2y=3

    => 5(2x + y) + 2(3y - 3x) + x + y = 5.(-1) + 2.15 - 2 + 3 = 26

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    Phan Thanh Tinh Coordinator 22/03/2017 at 21:50

    \(\left\{{}\begin{matrix}2x+3y=5\\y-3x=9\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x+3\left(3x+9\right)=5\\y=3x+9\end{matrix}\right.\)

    \(\Rightarrow\left\{{}\begin{matrix}11x+27=5\\y=3x+9\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-2\\y=3\end{matrix}\right.\)

    => 5(2x + y) + 2(3y - 3x) + x + y = 5.(-1) + 2.15 - 2 + 3 = 26


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justin bieber
19/03/2017 at 12:43
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The solution for \(\left\{{}\begin{matrix}x+ay=11\\x-y=2\end{matrix}\right.\)are positive integers . Find the value of a.

Simulataneous Equation A System of Equations


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justin bieber
19/03/2017 at 12:42
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Solve for the values of x and y in the equations:

\(\left\{{}\begin{matrix}3x+4y=5\\4x-2y=14\end{matrix}\right.\)

 

Simulataneous Equation A System of Equations

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    FA KAKALOTS 06/02/2018 at 12:33

    {3x+4y=54x−2y=14⇒{3x+4y=54y−8x=−28

    ⇒⎧⎨⎩11x=33y=5−3x4⇒{x=3y=−1

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    Phan Thanh Tinh Coordinator 22/03/2017 at 21:56

    \(\left\{{}\begin{matrix}3x+4y=5\\4x-2y=14\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3x+4y=5\\4y-8x=-28\end{matrix}\right.\)

    \(\Rightarrow\left\{{}\begin{matrix}11x=33\\y=\dfrac{5-3x}{4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\)


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justin bieber
19/03/2017 at 12:41
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If \(\left\{{}\begin{matrix}2x+3y=11\\ax-by=11\end{matrix}\right.\)and \(\left\{{}\begin{matrix}ax+by=-7\\3x-5y=-12\end{matrix}\right.\)have the same solution, then the values of a and b are

Simulataneous Equation A System of Equations


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justin bieber
19/03/2017 at 12:08
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The solution (x,y) to \(\left\{{}\begin{matrix}3x+y=9\\5x-4y=32\end{matrix}\right.\)is

Simulataneous Equation A System of Equations

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    FA KAKALOTS 06/02/2018 at 12:33

    {3x+y=95x−4y=32⇒{y=9−3x5x−4(9−3x)=32

    ⇒{y=9−3x17x−36=32⇒{y=−3x=4

  • ...
    Phan Thanh Tinh Coordinator 22/03/2017 at 22:17

    \(\left\{{}\begin{matrix}3x+y=9\\5x-4y=32\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=9-3x\\5x-4\left(9-3x\right)=32\end{matrix}\right.\)

    \(\Rightarrow\left\{{}\begin{matrix}y=9-3x\\17x-36=32\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=-3\\x=4\end{matrix}\right.\)


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justin bieber
19/03/2017 at 12:08
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x and y are integers that satisfy the equations:

\(\left\{{}\begin{matrix}2x-ay=12\\4x+y=2\end{matrix}\right.\)

If a is a possible integer, its value is

Simulataneous Equation A System of Equations


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justin bieber
19/03/2017 at 12:06
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How many sets of solution are there in 7x + 4y = 100?

Simulataneous Equation A System of Equations

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    Vũ Mạnh Hùng 06/12/2017 at 20:04

    đề sai hay sao ý bạn còn nếu mà hỏi có bao nhiêu cách thì trả lời của mình là 0 cách


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justin bieber
19/03/2017 at 12:05
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If integers x and y satisfy 2008x = 16y, the smallest possible value of x + y is

Simulataneous Equation A System of Equations


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justin bieber
19/03/2017 at 12:03
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Find the value of xyz if x,y and z satisfy the equations:

\(\left\{{}\begin{matrix}3x-2y=5\\2y+3x=1\\x-z=4\end{matrix}\right.\)

Simulataneous Equation A System of Equations

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    Nguyệt Nguyệt 19/03/2017 at 12:07

    3x - 2y = 5
    2y + 3z = 1
    x - z = 4
    We have :
    ( 3x - 2y ) - ( 2y + 3z ) = 5 - 1
    3x - 2y - 2y - 3z = 4
    3x - 4y  - 3z = 4
    3x - 3z - 4y = 4
    3. ( x - z ) - 4y = 4
    3 . 4 - 4y  = 4   ( replace x - z = 4 )
    12 - 4y  = 4
           4y   = 12 - 4
          4y     = 8
            y     = 8 : 4
           y      =  2
    => 3x - 2.2 = 5    ( replace y = 2 )
          3x - 4 = 5
          3x       = 5 + 4
          3x       = 9
            x       = 9 : 3
            x       = 3
    => 3 - z = 4  ( replace x = 3 )
               z  = 3 - 4
                z = -1
        So y = 2, x = 3 and z = -1

    justin bieber selected this answer.
  • ...
    FA KAKALOTS 06/02/2018 at 12:33

    3x - 2y = 5
    2y + 3z = 1
    x - z = 4
    We have :
    ( 3x - 2y ) - ( 2y + 3z ) = 5 - 1
    3x - 2y - 2y - 3z = 4
    3x - 4y  - 3z = 4
    3x - 3z - 4y = 4
    3. ( x - z ) - 4y = 4
    3 . 4 - 4y  = 4   ( replace x - z = 4 )
    12 - 4y  = 4
           4y   = 12 - 4
          4y     = 8
            y     = 8 : 4
           y      =  2
    => 3x - 2.2 = 5    ( replace y = 2 )
          3x - 4 = 5
          3x       = 5 + 4
          3x       = 9
            x       = 9 : 3
            x       = 3
    => 3 - z = 4  ( replace x = 3 )
               z  = 3 - 4
                z = -1
        So y = 2, x = 3 and z = -1


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justin bieber
19/03/2017 at 12:01
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Given x = 5 and y = 4 is the solution to the euqations:

\(\left\{{}\begin{matrix}kx-my=7\\mx+ky=22\end{matrix}\right.\)

what is the value of k? What id the value of m?

Simulataneous Equation A System of Equations

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    FA KAKALOTS 06/02/2018 at 12:33

    Replace x = 5 and y = 4 into the given equations,we have :

    {5k−4m=75m+4k=22⇒{20k−16m=2820k+25m=110

    ⇒⎧⎨⎩41m=82k=4m+75⇒{m=2k=3

  • ...
    Phan Thanh Tinh Coordinator 22/03/2017 at 22:08

    Replace x = 5 and y = 4 into the given equations,we have :

    \(\left\{{}\begin{matrix}5k-4m=7\\5m+4k=22\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}20k-16m=28\\20k+25m=110\end{matrix}\right.\)

    \(\Rightarrow\left\{{}\begin{matrix}41m=82\\k=\dfrac{4m+7}{5}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m=2\\k=3\end{matrix}\right.\)


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Bill gates
18/03/2017 at 10:54
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Compute:

\(\left(123456\right)^2-123457\times123455\)

Simulataneous Equation A System of Equations

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    mathlove 18/03/2017 at 11:19

    Put    \(x=123456\),so \(123456^2-123457.123455=x^2-\left(x+1\right)\left(x-1\right)=x^2-\left(x^2-1\right)=1\).

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  • ...
    FA KAKALOTS 06/02/2018 at 12:34

    Put    x=123456,so 1234562−123457.123455=x2−(x+1)(x−1)=x2−(x2−1)=1.

  • ...
    Nguyễn Trần Thành Đạt 18/03/2017 at 16:34

    (123456)2 -123457 x 123455

    = (123456)2 - (123456-1) x (123456+1)

    =(123456)2 - (1234562-1)

    =1


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Bill gates
18/03/2017 at 10:46
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a,b,c and d satisfy a system of equations:

\(\left\{{}\begin{matrix}2a+b+c+d=15\\a+2b+c+d=30\\a+b+2c+d=5\\a+b+c+2d=0\end{matrix}\right.\)

Find value 6a+4b

Simulataneous Equation A System of Equations

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    mathlove 18/03/2017 at 11:28


    Multiply the first equation with 2, the second with 1, the thirst with -1, the fourth with -1, then add all, we have   \(3a+2b=30+30-5-0=55\). Therefore    \(6a+4b=110\).


     

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  • ...
    →இے๖ۣۜQuỳnh 23/03/2017 at 06:10

    Multiply the first equation with 2, the second with 1, the thirst with -1, the fourth with -1, then add all, we have   3a+2b=30+30−5−0=553a+2b=30+30−5−0=55. Therefore    6a+4b=1106a+4b=110.

  • ...
    →இے๖ۣۜQuỳnh 23/03/2017 at 06:04

    Multiply the first equation with 2, the second with 1, the thirst with -1, the fourth with -1, then add all, we have   3a+2b=30+30−5−0=553a+2b=30+30−5−0=55. Therefore    6a+4b=1106a+4b=110.


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Bill gates
18/03/2017 at 10:43
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Solve for the values of r,s,t and u in the follwing equations:

\(\left\{{}\begin{matrix}2r+s=4\\2t+u=10\\2s+t=7\\2u+r=9\end{matrix}\right.\)

Simulataneous Equation A System of Equations

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    mathlove 18/03/2017 at 12:04

    Add all equations we have  \(3\left(r+s+t+u\right)=30\Rightarrow r+s+t+u=10\).  Compare this equation with the second equation we deduce that  \(t=r+s\). Substitute \(t=r+s\) for the thirst equation we get   \(r+3s=7\).

    By the first equation we have  \(\left\{{}\begin{matrix}2r+s=4\\r+3s=7\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}r=1\\s=2\end{matrix}\right.\). Substitute \(r=1\) for  the fourth equation we deduce that \(u=4\), and the second equation give us   \(t=3\). So  \(r=1,s=2,t=3,u=4\).

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  • ...
    →இے๖ۣۜQuỳnh 23/03/2017 at 06:09

    Add all equations we have  3(r+s+t+u)=30⇒r+s+t+u=103(r+s+t+u)=30⇒r+s+t+u=10.  Compare this equation with the second equation we deduce that  t=r+st=r+s. Substitute t=r+st=r+s for the thirst equation we get   r+3s=7r+3s=7.

    By the first equation we have  {2r+s=4r+3s=7{2r+s=4r+3s=7⇔{r=1s=2⇔{r=1s=2. Substitute r=1r=1 for  the fourth equation we deduce that u=4u=4, and the second equation give us   t=3t=3. So  r=1,s=2,t=3,u=4r=1,s=2,t=3,u=4.

  • ...
    →இے๖ۣۜQuỳnh 23/03/2017 at 06:04

    Add all equations we have  3(r+s+t+u)=30⇒r+s+t+u=103(r+s+t+u)=30⇒r+s+t+u=10.  Compare this equation with the second equation we deduce that  t=r+st=r+s. Substitute t=r+st=r+s for the thirst equation we get   r+3s=7r+3s=7.

    By the first equation we have  {2r+s=4r+3s=7{2r+s=4r+3s=7⇔{r=1s=2⇔{r=1s=2. Substitute r=1r=1 for  the fourth equation we deduce that u=4u=4, and the second equation give us   t=3t=3. So  r=1,s=2,t=3,u=4r=1,s=2,t=3,u=4.


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