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 \(\Leftrightarrow\left\{{}\begin{matrix}15x-6y=-27\\8x+6y=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2y=5x+9\\23x=-23\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(-1;2\right)\)

2)

HPT \(\Leftrightarrow\left\{{}\begin{matrix}2x+y=4\\2x+4y=10\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-3y=-6\\x=5-2y\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=1\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(1;2\right)\)

3)

HPT \(\Leftrightarrow\left\{{}\begin{matrix}4x+6y=14\\3x+6y=12\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\2y=4-x\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(2;1\right)\)

4) 

HPT \(\Leftrightarrow\left\{{}\begin{matrix}5x+6y=17\\54x-6y=42\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}59x=59\\y=9x-7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(1;2\right)\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}8y=0\\2x+3y=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\y=0\end{matrix}\right.\)

Lời giải:

Phương hướng giải là bạn sử dụng phương pháp thế, biểu diễn $x$ theo $y$ qua 1 trong 2 PT, sau đó thế vô PT còn lại giải PT 1 ẩn $y$
a) \(\left\{\begin{matrix} x-6y=17\\ 5x+y=23\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=17+6y\\ 5x+y=23\end{matrix}\right.\)

\(\Rightarrow 5(17+6y)+y=23\)

\(\Leftrightarrow 31y=-62\Leftrightarrow y=-2\)

$x=17+6y=17+6(-2)=5$

Vậy $(x,y)=(5,-2)$

Các phần còn lại bạn giải tương tự

b) $(x,y)=(\frac{1}{4}, 0)$

c) $(x,y)=(3, 4)$

d) $(x,y)=(\frac{79}{21}, \frac{44}{21})$

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

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

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

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

(Các câu khác tương tự nhé.)

a, Ta có : \(\left\{{}\begin{matrix}3x+2y=-2\\-x+4y=3\end{matrix}\right.\)

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

=> \(\left\{{}\begin{matrix}12y-9+2y=-2\\x=4y-3\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}14y=7\\x=4y-3\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=\frac{1}{2}\\x=\frac{4.1}{2}-3=-1\end{matrix}\right.\)

Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(-1;\frac{1}{2}\right)\)

b, Ta có : \(\left\{{}\begin{matrix}x+2y=11\\5x-3y=3\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=11-2y\\5\left(11-2y\right)-3y=3\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=11-2y\\55-10y-3y=3\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=11-2y\\-13y=-52\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=11-2.4=3\\y=4\end{matrix}\right.\)

Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(3;4\right)\)

c, Ta có : \(\left\{{}\begin{matrix}10x-9y=1\\15x+21y=36\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}30x-27y=3\\30x+42y=72\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}10x-9y=1\\-69y=-69\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}10x-9=1\\y=1\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(1;1\right)\)

d, Ta có : \(\left\{{}\begin{matrix}2x+y=3\\x+y=2\end{matrix}\right.\)

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

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

=> \(\left\{{}\begin{matrix}y=3-2.0=3\\x=0\end{matrix}\right.\)

Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(0;3\right)\)

e, Ta có : \(\left\{{}\begin{matrix}x+y=2\\2x-3y=9\end{matrix}\right.\)

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

=> \(\left\{{}\begin{matrix}x=2-y\\4-2y-3y=9\end{matrix}\right.\)

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

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

Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(3;-1\right)\)

f, Ta có : \(\left\{{}\begin{matrix}x-2y=11\\5x+3y=3\end{matrix}\right.\)

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

=> \(\left\{{}\begin{matrix}x=11+2y\\55+10y+3y=3\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=11+2y\\13y=-52\end{matrix}\right.\)​

Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(3;-4\right)\)

g, Ta có : \(\left\{{}\begin{matrix}3x-y=5\\2x+3y=18\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=3x-5\\2x+3\left(3x-5\right)=18\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=3x-5\\2x+9x-15=18\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=3x-5\\11x=33\end{matrix}\right.\)

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

Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(3;4\right)\)

h, Ta có : \(\left\{{}\begin{matrix}5x+3y=-7\\3x-y=-8\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}5x+3\left(3x+8\right)=-7\\y=3x+8\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}5x+9x+24=-7\\y=3x+8\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}14x=-31\\y=3x+8\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=-\frac{31}{14}\\y=3.\left(-\frac{31}{14}\right)+8=\frac{19}{14}\end{matrix}\right.\)

Vậy hệ phương trình có duy nhất 1 nghiệm là \(\left(x;y\right)=\left(-\frac{31}{14};\frac{19}{14}\right)\)

...????

a, Ta có : \(\dfrac{4}{6}=-\dfrac{2}{-3}\ne\dfrac{5}{5}=1\)

vậy hpt vô nghiệm 

b, Ta có \(\dfrac{2}{4}=\dfrac{3}{6}=\dfrac{5}{10}\)-> hệ pt có vô số nghiệm 

a: \(\left\{{}\begin{matrix}3x-2y=11\\4x-5y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x=11+2y\\4x-5y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y+\dfrac{11}{3}\\4\left(\dfrac{2}{3}y+\dfrac{11}{3}\right)-5y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y+\dfrac{11}{3}\\\dfrac{8}{3}y+\dfrac{44}{3}-5y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}y+\dfrac{11}{3}\\-\dfrac{7}{3}y=3-\dfrac{44}{3}=-\dfrac{35}{3}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=5\\x=\dfrac{2}{3}\cdot5+\dfrac{11}{3}=\dfrac{10}{3}+\dfrac{11}{3}=\dfrac{21}{3}=7\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}\dfrac{x}{2}-\dfrac{y}{3}=1\\5x-8y=3\end{matrix}\right.\)

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

=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y+2\\\dfrac{10}{3}y+10-8y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{14}{3}y=3-10=-7\\x=\dfrac{2}{3}y+2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=7:\dfrac{14}{3}=7\cdot\dfrac{3}{14}=\dfrac{3}{2}\\x=\dfrac{2}{3}\cdot\dfrac{3}{2}+2=3\end{matrix}\right.\)

c: \(\left\{{}\begin{matrix}3x+5y=1\\2x-y=-8\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2x+8\\3x+5\left(2x+8\right)=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2x+8\\3x+10x+40=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2x+8\\13x=-39\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-3\\y=2\cdot\left(-3\right)+8=8-6=2\end{matrix}\right.\)

d: \(\left\{{}\begin{matrix}\dfrac{x}{y}=\dfrac{2}{3}\\x+y-10=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y\\x+y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{3}y+y=10\\x=\dfrac{2}{3}y\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{5}{3}y=10\\x=\dfrac{2}{3}y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=6\\x=\dfrac{2}{3}\cdot6=4\end{matrix}\right.\)