Simulataneous Equation A System of Equations

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

\(x-56< 0\)

\(x< 56\)
 

{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

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

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

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

{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

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

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

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

{3x+4y=54x−2y=14⇒{3x+4y=54y−8x=−28

⇒⎧⎨⎩11x=33y=5−3x4⇒{x=3y=−1

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

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

{3x+y=95x−4y=32⇒{y=9−3x5x−4(9−3x)=32

⇒{y=9−3x17x−36=32⇒{y=−3x=4

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

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

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

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

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

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

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

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

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

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

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

(123456)² -123457 x 123455

= (123456)² - (123456-1) x (123456+1)

=(123456)² - (123456²-1)

=1

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    \(6a+4b=110\).

Multiply the first equation with 2, the second with 1, the thirst with -1, the fourth with -1, then add all, we have   . Therefore    .

Multiply the first equation with 2, the second with 1, the thirst with -1, the fourth with -1, then add all, we have   . Therefore    .

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  \(r=1,s=2,t=3,u=4\).

Add all equations we have  .  Compare this equation with the second equation we deduce that  . Substitute  for the thirst equation we get   .

By the first equation we have  . Substitute  for  the fourth equation we deduce that , and the second equation give us   . So  .

Add all equations we have  .  Compare this equation with the second equation we deduce that  . Substitute  for the thirst equation we get   .

By the first equation we have  . Substitute  for  the fourth equation we deduce that , and the second equation give us   . So  .