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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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
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\(\left\{{}\begin{matrix}3x+2y=12\\3x-y=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left(3x+2y\right)-\left(3x-y\right)=12-3\\3x=y+3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3y=9\\x=\dfrac{y}{3}+1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=3\\x=2\end{matrix}\right.\)
=> 4x - 8y = -16 => 2a = 16 => a = 8
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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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\(\left\{{}\begin{matrix}2x+3y=5\\y-3x=9\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x+3\left(3x+9\right)=5\\y=3x+9\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}11x+27=5\\y=3x+9\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-2\\y=3\end{matrix}\right.\)
=> 5(2x + y) + 2(3y - 3x) + x + y = 5.(-1) + 2.15 - 2 + 3 = 26
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\(\left\{{}\begin{matrix}3x+4y=5\\4x-2y=14\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3x+4y=5\\4y-8x=-28\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}11x=33\\y=\dfrac{5-3x}{4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\)
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\(\left\{{}\begin{matrix}3x+y=9\\5x-4y=32\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=9-3x\\5x-4\left(9-3x\right)=32\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y=9-3x\\17x-36=32\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=-3\\x=4\end{matrix}\right.\)
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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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Nguyệt Nguyệt 19/03/2017 at 12:07
3x - 2y = 5
justin bieber selected this answer.
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 -
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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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
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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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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.
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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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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.
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→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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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.
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→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.