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30 tháng 3 2022

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

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

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}5x=10\\x-y=3\end{matrix}\right.\)

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

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)

 

3 tháng 7 2023

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

\(\Leftrightarrow\left\{{}\begin{matrix}4x+6y=10\\4x-5y=1\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}2x+3\cdot\dfrac{9}{11}=5\\y=\dfrac{9}{11}\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}2x=\dfrac{28}{11}\\y=\dfrac{9}{11}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{14}{11}\\y=\dfrac{9}{11}\end{matrix}\right.\)

Vậy: \(x=\dfrac{14}{11};y=\dfrac{9}{11}\)

18 tháng 2 2022

\(\left\{{}\begin{matrix}2x+3y=4\\2x+2y=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=10\\x=-3-10=-13\end{matrix}\right.\)

18 tháng 2 2022

\(\left\{{}\begin{matrix}4-2x=3y\\x+y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+3y=4\\x+y=-3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x+3y=4\\2x+2y=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=10\\x+y=-3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=10\\x=-7\end{matrix}\right.\)

12 tháng 11 2015

PT (1) <=> x = 3y + 3. Thay  x = 3y + 3 vào PT (2) ta có: \(\left(3y+3\right)^2+y^2-2\left(3y+3\right)-2y-9=0\Leftrightarrow10y^2+10y-6=0\Leftrightarrow y=\frac{-5+\sqrt{85}}{10}\)hoặc \(y=\frac{-5-\sqrt{85}}{10}\)

- Nếu \(y=\frac{-5+\sqrt{85}}{10}\) \(\Rightarrow x=3y+3=\frac{15+3\sqrt{85}}{10}\)

- Nếu \(y=\frac{-5-\sqrt{85}}{10}\Rightarrow x=3y+3=\frac{15-3\sqrt{85}}{10}\) 

18 tháng 6 2020

ĐK: \(x\ge\frac{1}{2}\)

\(\hept{\begin{cases}x\left(2x-2y-1\right)=3\left(y+2\right)\left(1\right)\\3y+6\sqrt{2x-1}=y^2-x+23\left(2\right)\end{cases}}\)

pt (1) <=> \(2x^2-2xy-x-3y-6=0\)

<=> \(2x^2-x\left(2y+1\right)-\left(3y+6\right)=0\)

có \(\Delta=\left(2y+1\right)^2+4\left(3y+6\right)=4y^2+28y+49=\left(2y+7\right)^2\)

=> (1) có hai nghiệm: \(\orbr{\begin{cases}x_1=\frac{\left(2y+1\right)-\left(2y+7\right)}{4}=-\frac{3}{2}\left(loai\right)\\x_2=\frac{\left(2y+1\right)+\left(2y+7\right)}{4}=y+2\end{cases}}\)

+) Với \(x=y+2\) thế vào (2) ta có: 

\(3y+6\sqrt{2\left(y+2\right)-1}=y^2-\left(y+2\right)+23\)

<=> \(6\sqrt{2y+3}=y^2-4y+21\)

ĐK: \(y\ge-\frac{3}{2}\)

\(6\sqrt{2y+3}=y^2-4y+21\)

<=> \(6\sqrt{2y+3}-2y-12=y^2-6y+9\)

<=> \(\frac{2\left(9\left(2y+3\right)-\left(y+6\right)^2\right)}{3\sqrt{2y+3}+y+6}-\left(y-3\right)^2=0\)

<=> \(\frac{-2\left(y-3\right)^2}{3\sqrt{2y+3}+y+6}-\left(y-3\right)^2=0\)

<=> \(\left(y-3\right)^2\left(\frac{-2}{3\sqrt{2y+3}+y+6}-1\right)=0\)

<=> y - 3 = 0 

<=> y = 3 thỏa mãn 

khi đó x = y + 2 = 3 + 2 = 5 thỏa mãn

Kết luận:...

=>10x+15y=5m và -10x+2y=-2

=>17y=5m-2 và -5x+y=-1

=>y=5/17m-2/17 và 5x-y=1

=>y=5/17m-2/17 và 5x=1+y=5/17m+15/17

=>y=5/17m-2/17 và x=1/17m+5/17

x>0; y>0

=>5m-2>0 và m+5>0

=>m>2/5

7 tháng 10 2021

\(1,\Leftrightarrow\left\{{}\begin{matrix}x=3-y\\3-y+2y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3-y\\y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}x-2x-1=3\\y=2x+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=2\left(-2\right)+1=-3\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}2x+3x-6=4\\y=x-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\\ 4,\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\y+2=3y+8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-3\end{matrix}\right.\\ 5,\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1+y}{2}\\\dfrac{3+3y}{2}-4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1+y}{2}\\3+3y-8y=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{y+1}{2}\\y=-\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\y=-\dfrac{1}{5}\end{matrix}\right.\)

27 tháng 2 2020

a, Thay m=3 vào hpt ta có :

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

\(\Leftrightarrow\left\{{}\begin{matrix}2x+3y=3\\-15x+3y=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+3y=3\\17x=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{4}{17}\\y=\frac{43}{51}\end{matrix}\right.\)