\(a,\left\{{}\begin{matrix}3x-y=5\\4x+2y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-y=5\\2x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\\ b,\left\{{}\begin{matrix}5x+2y=9\\x+5y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x+2y=9\\5x+25y=55\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x+2y=9\\23y=46\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

\(c,\left\{{}\begin{matrix}3x+y=10\\4x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x+3y=30\\4x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}13x=39\\4x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\\ d,\left\{{}\begin{matrix}4x+3y=22\\5x+3y=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\5x+3y=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=2\end{matrix}\right.\)

\(e,\left\{{}\begin{matrix}4x-3y=5\\5x+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x=18\\5x+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

a. \(\left\{{}\begin{matrix}3x-y=5\\4x+2y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x-2y=10\\4x+2y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}10x=20\\6x-2y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

b. \(\left\{{}\begin{matrix}5x+2y=9\\x+5y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x+2y=9\\5x+25y=55\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}23y=46\\5x+2y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=1\end{matrix}\right.\)

c. \(\left\{{}\begin{matrix}3x+y=10\\4x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x+3y=30\\4x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}13x=39\\4x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)

d. \(\left\{{}\begin{matrix}4x+3y=22\\5x+3y=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\4x+3y=22\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=2\end{matrix}\right.\)

e. \(\left\{{}\begin{matrix}4x-3y=5\\5x+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x=18\\4x-3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

a) \(\begin{cases} 3x -y=5\\ 4x +2y=10 \end{cases} \)

    \(\begin{cases} 12x - 4y= 20\\ 12x +6y= 30 \end{cases} \)

    \(\begin{cases} -10y=-10\\ 3x-y=5 \end{cases} \)

    \(\begin{cases} y=1\\ 3x-1=5 \end{cases} \)

    \(\begin{cases} y=1\\ 3x=6 \end{cases} \)

    \(\begin{cases} y=1\\ x=2 \end{cases} \)

Hpt có nghiệm duy nhất: {1;2}

b)\(\begin{cases} 5x +2y=9\\ x+5y=11 \end{cases} \)

  \(\begin{cases} 5x+2y=9\\ 5x+25y=55 \end{cases} \)

  \(\begin{cases} -23y=-46\\ x+5y=11 \end{cases} \)

  \(\begin{cases} y=2\\ x+ 5*2=11 \end{cases} \)

  \(\begin{cases} y=2\\ x+10=11 \end{cases} \)

Hpt có nghiệm duy nhất:{1;2}

c)\(\begin{cases} 3x+y=10\\ 4x-3y=9 \end{cases} \)

  \(\begin{cases} 12x+4y=40\\ 12x-9y=27 \end{cases} \)

 \(\begin{cases} 13y=13\\ 3x+y=10 \end{cases} \)

 \(\begin{cases} y=1\\ 3x+1=10 \end{cases} \)

 \(\begin{cases} y=1\\ 3x=9 \end{cases} \)

hpt có nghiệm duy nhất:{1;3}

d)\(\begin{cases} 4x+3y=22\\ 5x+3y=26 \end{cases} \)

  \(\begin{cases} 20x+15y=110\\ 20x+12y=104 \end{cases} \)

 \(\begin{cases} 3y=6\\ 4x+3y=22 \end{cases} \)

 \(\begin{cases} y=2\\ 4x+3*2=22 \end{cases} \)

 \(\begin{cases} y=2\\ 4x+6=22 \end{cases} \)

hệ phương trình có nghiệm duy nhất:{2;4}

e)\(\begin{cases} 4x-3y=5\\ 5x+3y=13 \end{cases} \)

  \(\begin{cases} 20x-15y=25\\ 20x+12y=52 \end{cases} \)

 \(\begin{cases} -27y=-27\\ 4x-3y=5 \end{cases} \)

  \(\begin{cases} y=1\\ 4x-3*1=5 \end{cases} \)

   \(\begin{cases} y=1\\ 4x-3=5 \end{cases} \)

Hệ phương trình có nghiệm duy nhất là:{1;2}

a: Ta có: \(\left\{{}\begin{matrix}3x+2y=14\\5x+3y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}15x+10y=70\\15x+9y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=67\\3x=14-2y=14-2\cdot67=-120\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-40\\y=67\end{matrix}\right.\)

b: Ta có: \(\left\{{}\begin{matrix}-x+2y-6=0\\5x-3y-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-x+2y=6\\5x-3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-5x+10y=30\\5x-3y=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}7y=35\\2y-x=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=4\end{matrix}\right.\)

⇒4y+6=6+5

⇒4y=5x

⇒y=\(\dfrac{5x}{4}\)=1,25x

⇒x=\(\dfrac{4y}{5}\)=0.8y

⇔ 2 x - 5 y + z = 10 x + 2 y - 3 z = 10 5 y - z = - 6 → ( 1 ) + ( 3 ) 2 x - 5 y + z = 10 x + 2 y - 3 z = 10 2 x = 4 ⇔ 4 - 5 y + z = 10 2 + 2 y - 3 z = 10 x = 2 ⇔ - 5 y + z = 6 2 y - 3 z = 8 x = 2 ⇔ - 15 y + 3 z = 18 2 y - 3 z = 8 x = 2 ⇔ - 13 y = 26 2 y - 3 z = 8 x = 2 ⇔ y = - 2 z = - 4 x = 2

Vậy nghiệm của hệ phương trình đã cho là ( 2; -2; -4).

Chọn C,

Lời giải:
a. Thay $y=x+1$ vào điều kiện ban đầu có:

$3x+5(x+1)=13$
$8x+5=13$

$8x=8$

$x=1$

$y=x+1=2$
b. Thay $x=y+5$ vô điều kiện đầu thì:

$2(y+5)-3y=4$

$-y+10=4$

$-y=-6$

$y=6$

$x=6+5=11$

c. Thay $y=x-2$ vô điều kiện đầu thì:

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

$4x-10=-6$

$4x=10+(-6)=4$

$x=1$

$y=x-2=1-2=-1$

a) Ta có: \(\left\{{}\begin{matrix}3x+5y=13\\x+1=y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x+5y=13\\x-y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+5y=13\\3x-3y=-3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}8y=16\\x+1=y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=y-1=2-1=1\end{matrix}\right.\)

b) Ta có: \(\left\{{}\begin{matrix}2x-3y=4\\x=y+5\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}-y=-6\\x=y+5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=6\\x=11\end{matrix}\right.\)

c) Ta có: \(\left\{{}\begin{matrix}-x+5y=-6\\y=x-2\end{matrix}\right.\)

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

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