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Coi PT thứ nhất là PT(1) và PT thứ 2 là PT(2)
a)
Từ PT$(2)\Rightarrow y=18-5x$
Thế vào PT$(1)$: $3x-2(18-5x)=5$
$\Leftrightarrow 13x=41\Leftrightarrow x=\frac{41}{13}$
\(y=18-5x=18-5.\frac{41}{13}=\frac{29}{13}\)
Vậy.......
b)
PT\((1)\Rightarrow y=2x-8\)
Thế vào $PT(2)\Rightarrow$ \(x+3(2x-8)=10\)
$\Leftrightarrow 7x=34\Rightarrow x=\frac{34}{7}$
$y=2x-8=2.\frac{34}{7}-8=\frac{12}{7}$
Vậy........
c)
HPT \(\Leftrightarrow \left\{\begin{matrix} 12x-9y=6\\ 12x-16y=-8\end{matrix}\right.\)
Từ PT$(1)\Rightarrow 12x=9y+6$
Thế vào PT$(2)\Rightarrow 9y+6-16y=-8$
$\Leftrightarrow y=2$
$x=\frac{9y+6}{12}=\frac{9.2+6}{12}=2$
Vậy.........
d)
HPT \(\Leftrightarrow \left\{\begin{matrix} 10x+25y=65\\ 10x-6y=-28\end{matrix}\right.\)
Từ PT$(1)\Rightarrow 10x=65-25y$
Thế vào PT$(2)\Rightarrow 65-25y-6y=-28$
$\Leftrightarrow y=3$
$x=\frac{65-25y}{10}=\frac{65-25.3}{10}=-1$
Vậy........
a)\(\left\{{}\begin{matrix}8x+2y=4\\8x+3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\4x+1=2\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}y=1\\x=\frac{1}{4}\end{matrix}\right.\)b)
\(\left\{{}\begin{matrix}12x-8y=44\\12x-15y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7y=35\\4x-5y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\4x-5.5=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\x=7\end{matrix}\right.\)c)\(\left\{{}\begin{matrix}9x=-18\\4x+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\4.\left(-2\right)+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=7\end{matrix}\right.\)
1) \(\left\{{}\begin{matrix}3x-2y=4\\4x+2y=10\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}3x-2y=4\\7x=14\end{matrix}\right.< =>\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
2)\(\left\{{}\begin{matrix}2x+3y=5\\4x+6y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+6y=10\\4x=6y=10\end{matrix}\right.\)
=> Hệ có vô số nghiệm.
3)\(\left\{{}\begin{matrix}3x-4y=-2\\10x+4y=28\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}3x-4y=-2\\13x=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
4)\(\left\{{}\begin{matrix}6x+15y=9\\6x-4y=28\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}6x+15y=9\\19y=19\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=-1\end{matrix}\right.\)
1) hpt \(\Leftrightarrow\left\{{}\begin{matrix}x+4y=2\\6x+4y=8\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{2-x}{4}\\5x=6\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{5}\\x=\dfrac{6}{5}\end{matrix}\right.\)
Kl: x=6/5 và y=1/5
2) hpt \(\Leftrightarrow\left\{{}\begin{matrix}-2x-2y=4\\-2x-4y=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-2-y\\2y=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=2\end{matrix}\right.\)
Kl...
3) hpt \(\Leftrightarrow\left\{{}\begin{matrix}2x-3y=2\\2x-3y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2+3y}{2}\\0=3\left(vô-lý\right)\end{matrix}\right.\)
kl: hpt vn
1/
\(\left\{{}\begin{matrix}3x+2y=6\\x-y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+2y=6\\3x-3y=6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}5y=0\\x-y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=2\end{matrix}\right.\)
Vậy hệ phương trình đã cho có nghiệm duy nhất \(\left(x;y\right)=\left(2;0\right)\)
2/
\(\left\{{}\begin{matrix}2x-3y=1\\-4x+6y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x-6y=2\\-4x+6y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}0x=4\\-4x+6y=2\end{matrix}\right.\)
Vì 0x=4 vô nghiệm \(\Rightarrow-4x+6y=2\) vô nghiệm
Vậy hệ phương trình đã cho vô nghiệm
3/ \(\left\{{}\begin{matrix}2x+3y=5\\5x-4y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}10x+15y=25\\10x-8y=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}23y=23\\5x-4y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\5x-4=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=1\end{matrix}\right.\)
Vậy hệ phương trình đã cho có nghiệm duy nhất (x;y) = (1;1)
a, Ta có : \(\left\{{}\begin{matrix}3x-y=5\\2x+3y=18\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=3x-5\\2x+3\left(3x-5\right)=18\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=3x-5\\2x+9x-15=18\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=3x-5\\11x=33\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=3.3-5=4\\x=\frac{33}{11}=3\end{matrix}\right.\)
Vậy phương trình có nghiệm duy nhất là ( x;y ) = ( 3;4 )
b, Làm tương tự a
c, Ta có : \(\left\{{}\begin{matrix}\frac{14}{x-y+2}-\frac{10}{x+y-1}=9\\\frac{3}{x-y+2}+\frac{2}{x+y-1}=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\frac{14}{x-y+2}-\frac{10}{x+y-1}=9\\\frac{15}{x-y+2}+\frac{10}{x+y-1}=20\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\frac{29}{x-y+2}=29\\\frac{3}{x-y+2}+\frac{2}{x+y-1}=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x-y+2=1\\\frac{3}{x-y+2}+\frac{2}{x+y-1}=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=y-1\\\frac{3}{y-1-y+2}+\frac{2}{y-1+y-1}=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=y-1\\3+\frac{2}{2y-2}=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=y-1\\\frac{2}{2y-2}=1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=y-1\\2y-2=2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=2-1=1\\y=2\end{matrix}\right.\)
Vậy phương trình có nghiệm duy nhất là ( x;y ) = ( 1;2 )
1) \(\left\{{}\begin{matrix}4x+y=2\\8x+3y=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=2-4x\\8x+3\left(2-4x\right)=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{4}\\y=1\end{matrix}\right.\)
2) 2 pt 3 ẩn không giải được.
3) \(\left\{{}\begin{matrix}3x+2y=6\\x-y=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=x-2\\3x+2\left(x-2\right)=6\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)
4) \(\left\{{}\begin{matrix}2x-3y=1\\-4x+6y=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{3y+1}{2}\\-4\cdot\frac{3y+1}{2}+6y=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\varnothing\\x=\varnothing\end{matrix}\right.\)
5) \(\left\{{}\begin{matrix}2x+3y=5\\5x-4y=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{-3y+5}{2}\\5\cdot\frac{-3y+5}{2}-4y=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=1\end{matrix}\right.\)
6) \(\left\{{}\begin{matrix}3x-y=7\\x+2y=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=3x-7\\x+2\left(3x-7\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)
7) \(\left\{{}\begin{matrix}x+4y=2\\3x+2y=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2-4y\\3\left(2-4y\right)+2y=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\frac{1}{5}\\x=\frac{6}{5}\end{matrix}\right.\)
8) \(\left\{{}\begin{matrix}-x-y=2\\-2x-3y=9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-x-2\\-2x-3\left(-x-2\right)=9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-5\end{matrix}\right.\)
9) \(\left\{{}\begin{matrix}2x-3y=2\\-4x+6y=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{3y+2}{2}\\-4\cdot\frac{3y+2}{2}+6y=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\varnothing\\x=\varnothing\end{matrix}\right.\)
Câu a: Thế y=5-2x rồi giải pt bậc2
Câu b : từ pt thứ 2, tương đương (x-3)(y-3)=0, xét 2 TH rồi thế vào pt thứ 1
Câu c: từ pt 1 suy ra 2x = 2-3y
Nhân 2 vào pt 2 rồi thế vào
\(\left\{{}\begin{matrix}2x-3y=19\\3x+4y=-14\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}8x-12y=76\\9x+12y=-42\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}17x=34\\2x-3y=19\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\2.2-3y=19\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-5\end{matrix}\right.\)
Vậy nghiệm của hệ phương trình là: \(\left\{{}\begin{matrix}x=2\\y=-5\end{matrix}\right.\)