Giải các hệ bất phương trình :
a. \(\left\{{}\begin{matrix}6x+\dfrac{5}{7}< 4x+7\\\dfrac{8x+3}{2}< 2x+5\end{matrix}\right.\)
b. \(\left\{{}\begin{matrix}15x-2>2x+\dfrac{1}{3}\\2\left(x-4\right)< \dfrac{3x-14}{2}\end{matrix}\right.\)
a) 6x + < 4x + 7 <=> 6x - 4x < 7 -
<=> x <
< 2x +5 <=> 4x - 2x < 5 -
<=> x <
Tập nghiệm của hệ bất phương trình:
Y = ∩
=
.
b) 15x - 2 > 2x + <=> x >
2(x - 4) < <=> x < 2
Tập nghiệm S = ∩ (-∞; 2) =
giải hệ phương trình
1)\(\left\{{}\begin{matrix}3x+4y=11\\2x-y=-11\end{matrix}\right.\) 2)\(\left\{{}\begin{matrix}3x+2y=0\\2x+y=-1\end{matrix}\right.\) 3)\(\left\{{}\begin{matrix}3x+\dfrac{5}{2}y=9\\2x+\dfrac{1}{3}y=2\end{matrix}\right.\)
4)\(\left\{{}\begin{matrix}-x+3y=16\\2x+y=3\end{matrix}\right.\) 5)\(\left\{{}\begin{matrix}\dfrac{-3}{x-y}+\dfrac{5}{2x+y}=-2\\\dfrac{4}{x-y}-\dfrac{10}{2x+y}=2\end{matrix}\right.\) 6)\(\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{1}{y}=1\\\dfrac{3}{x}+\dfrac{4}{y}=5\end{matrix}\right.\)
1. \(\left\{{}\begin{matrix}3x+4y=11\\2x-y=-11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+4y=11\\8x-4y=-44\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+4y=11\\11x=-33\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\x=-3\end{matrix}\right.\)
2. \(\left\{{}\begin{matrix}3x+2y=0\\2x+y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+2y=0\\4x+2y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=3\\x=-2\end{matrix}\right.\)
3.\(\left\{{}\begin{matrix}3x+\dfrac{5}{2}y=9\\2x+\dfrac{1}{3}y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+5y=18\\6x+y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4y=12\\6x+y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=3\\x=\dfrac{1}{2}\end{matrix}\right.\)
Giải các hệ phương trình sau bằng phương pháp thế:
a)\(\left\{{}\begin{matrix}3x-2y=11\\4x-5y=3\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\dfrac{x}{2}-\dfrac{y}{3}=1\\5x-8y=3\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}3x+5y=1\\2x-y=-8\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}\dfrac{x}{y}=\dfrac{2}{3}\\x+y-10=0\end{matrix}\right.\)
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: \(\left\{{}\begin{matrix}\dfrac{x}{y}=\dfrac{2}{3}\\x+y-10=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y\\x+y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{3}y+y=10\\x=\dfrac{2}{3}y\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{5}{3}y=10\\x=\dfrac{2}{3}y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=6\\x=\dfrac{2}{3}\cdot6=4\end{matrix}\right.\)
Giải các hệ phương trình :
a) \(\left\{{}\begin{matrix}5\left(x+2y\right)=3x-1\\2x+4=3\left(x-5y\right)-12\end{matrix}\right.\);
b) \(\left\{{}\begin{matrix}4x^2-5\left(y+1\right)=\left(2x-3\right)^2\\3\left(7x+2\right)=5\left(2y-1\right)-3x\end{matrix}\right.\);
c) \(\left\{{}\begin{matrix}\dfrac{2x+1}{4}-\dfrac{y-2}{3}=\dfrac{1}{12}\\\dfrac{x+5}{2}=\dfrac{y+7}{3}-4\end{matrix}\right.\);
d) \(\left\{{}\begin{matrix}\dfrac{3s-2t}{5}+\dfrac{5s-3t}{3}=s+1\\\dfrac{2s-3t}{3}+\dfrac{4s-3t}{2}=t+1\end{matrix}\right.\).
giải hệ sau bằng phương pháp thế
a)\(\left\{{}\begin{matrix}2x-y=4\\x+5y=3\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}-2x+3y=-1\\x+2y=3\end{matrix}\right.\)
giải hệ sau:
a)\(\left\{{}\begin{matrix}x+y=-1\\2x+y=1\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{5}\\\dfrac{3}{x}+\dfrac{4}{y}=2\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}2\dfrac{5}{x-1}+\dfrac{3}{3y-2}=1\\\dfrac{2}{2x-1}+\dfrac{1}{3y-2}=1\end{matrix}\right.\)
Giải hệ sau :
Câu a :
\(\left\{{}\begin{matrix}x+y=-1\\2x+y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=-1\\-x=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=-1\\x=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y=-3\\x=2\end{matrix}\right.\)
Vậy ...........................
Câu b :
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}=a\\\dfrac{1}{y}=b\end{matrix}\right.\) . Ta có :
\(\left\{{}\begin{matrix}a+b=\dfrac{1}{5}\\3a+4b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a+3b=\dfrac{3}{5}\\3a+4b=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-b=-\dfrac{7}{5}\\3a+4b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{7}{5}\\a=-\dfrac{6}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{7}{5}\\\dfrac{1}{y}=-\dfrac{6}{5}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{7}\\y=-\dfrac{5}{6}\end{matrix}\right.\)
Vậy..................
\(a,\left\{{}\begin{matrix}2x-y=4\\x+5y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-y=4\\2x+10y=6\end{matrix}\right.\left\{{}\begin{matrix}11y=2\\2x+10y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{2}{11}\\2x+10.\dfrac{2}{11}=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{2}{11}\\2x=\dfrac{46}{11}\end{matrix}\right.\left\{{}\begin{matrix}y=\dfrac{2}{11}\\x=\dfrac{23}{11}\end{matrix}\right.\)
giải hệ pt:
9) \(\left\{{}\begin{matrix}\dfrac{7}{2x+y}+\dfrac{4}{2x-y}=74\\\dfrac{3}{2x+y}+\dfrac{2}{2x-y}=32\end{matrix}\right.\)
10) \(\left\{{}\begin{matrix}x=2y-1\\2x-y=5\end{matrix}\right.\)
11) \(\left\{{}\begin{matrix}3x-6=0\\2y-x=4\end{matrix}\right.\)
12) \(\left\{{}\begin{matrix}2x+y=5\\x+7y=9\end{matrix}\right.\)
13) \(\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{4}{y}=2\\\dfrac{4}{x}-\dfrac{5}{y}=3\end{matrix}\right.\)
14) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)
15) \(\left\{{}\begin{matrix}2\sqrt{x-1}-\sqrt{y-1}=1\\\sqrt{x-1}+\sqrt{y-1}=2\end{matrix}\right.\)
giúp mk vs ạ mai mk học rồi
9) \(\left\{{}\begin{matrix}\dfrac{7}{2x+y}+\dfrac{4}{2x-y}=74\\\dfrac{3}{2x+y}+\dfrac{2}{2x-y}=32\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{21}{2x+y}+\dfrac{12}{2x-y}=222\\\dfrac{21}{2x+y}+\dfrac{14}{2x-y}=224\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{2x-y}=2\\\dfrac{7}{2x+y}+\dfrac{4}{2x-y}=74\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=\dfrac{1}{10}\\2x-y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-2y=\dfrac{9}{10}\\2x+y=\dfrac{1}{10}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{9}{20}\\x=\dfrac{11}{40}\end{matrix}\right.\)
10) \(\left\{{}\begin{matrix}x=2y-1\\2x-y=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x-4y=-2\\2x-y=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2y-1\\3y=7\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{3}\\y=\dfrac{7}{3}\end{matrix}\right.\)
11) \(\left\{{}\begin{matrix}3x-6=0\\2y-x=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3x=6\\y=\dfrac{x+4}{2}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
12) \(\left\{{}\begin{matrix}2x+y=5\\x+7y=9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=5\\2x+14y=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=5\\13y=13\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
13) \(\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{4}{y}=2\\\dfrac{4}{x}-\dfrac{5}{y}=3\end{matrix}\right.\)(ĐKXĐ: \(x,y\ne0\))
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{12}{x}-\dfrac{16}{y}=8\\\dfrac{12}{x}-\dfrac{15}{y}=9\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{4}{y}=2\\\dfrac{1}{y}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\y=1\left(tm\right)\end{matrix}\right.\)
14) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)(ĐKXĐ: \(x,y\ne0\))
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{8}{x}+\dfrac{8}{y}=\dfrac{2}{3}\\\dfrac{8}{x}+\dfrac{15}{y}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{12}\\\dfrac{7}{y}=\dfrac{1}{3}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=28\left(tm\right)\\y=21\left(tm\right)\end{matrix}\right.\)
15) \(\left\{{}\begin{matrix}2\sqrt{x-1}-\sqrt{y-1}=1\\\sqrt{x-1}+\sqrt{y-1}=2\end{matrix}\right.\)(ĐKXĐ: \(x\ge1,y\ge1\))
\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-1}=3\\\sqrt{x-1}+\sqrt{y-1}=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{y-1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y-1=1\end{matrix}\right.\)\(\Leftrightarrow x=y=2\left(tm\right)\)
Giải các hệ phương trình :
a) \(\left\{{}\begin{matrix}5x+3y=-7\\2x-4y=6\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}7x+14y=17\\2x+4y=5\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}\dfrac{2}{5}x+\dfrac{3}{7}y=\dfrac{1}{3}\\\dfrac{5}{3}x-\dfrac{5}{7}y=\dfrac{2}{3}\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}-0,2x+0,5y=1,7\\0,3x+0,4y=0,9\end{matrix}\right.\)
a) \(\left\{{}\begin{matrix}5x+3y=-7\\2x-4y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x+3y=-7\\x-2y=3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}5x+3y=-7\\x=3+2y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5.\left(3+2y\right)+3y=-7\\x=3+2y\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}13y=-22\\x=3+2y\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-22}{13}\\x=3+2.\dfrac{-22}{13}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-22}{13}\\x=\dfrac{-5}{13}\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm là: \(\left\{{}\begin{matrix}y=\dfrac{-22}{13}\\x=\dfrac{-5}{13}\end{matrix}\right.\).
b)\(\left\{{}\begin{matrix}7x+14y=17\\2x+4y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}14x+28y=34\\14x+28y=35\end{matrix}\right.\) (vô nghiệm)
Vậy hệ phương trình vô nghiệm.
c) \(\left\{{}\begin{matrix}\dfrac{2}{5}x+\dfrac{3}{7}y=\dfrac{1}{3}\\\dfrac{5}{3}x-\dfrac{5}{7}=\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{5}x+\dfrac{3}{7}y=\dfrac{1}{3}\\x-\dfrac{3}{7}y=\dfrac{2}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{5}x+\dfrac{3}{7}y=\dfrac{1}{3}\\x=\dfrac{3}{7}y+\dfrac{2}{5}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{5}.\left(\dfrac{3}{7}y+\dfrac{2}{5}\right)+\dfrac{3}{7}y=\dfrac{1}{3}\\x=\dfrac{3}{7}y+\dfrac{2}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{5}y=\dfrac{13}{75}\\x=\dfrac{3}{7}y+\dfrac{2}{5}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{13}{45}\\x=\dfrac{11}{21}\end{matrix}\right.\).
Vậy hệ phương trình có nghiệm là: \(\left\{{}\begin{matrix}y=\dfrac{13}{45}\\x=\dfrac{11}{21}\end{matrix}\right.\).
giải hệ phương trình
\(\left\{{}\begin{matrix}\sqrt{x-2}+\sqrt{y-3}=3\\2\sqrt{x-2}-3\sqrt{y-3}=-4\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{3x}{x+1}+\dfrac{2}{y+4}=4\\\dfrac{2x}{x+1}-\dfrac{5}{y+4}=4\end{matrix}\right.\)
a.
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge2\\y\ge3\end{matrix}\right.\)
\(\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\2\sqrt{x-2}-3\sqrt{y-3}=-4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\5\sqrt{x-2}=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\\sqrt{x-2}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}=1\\\sqrt{y-3}=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=7\end{matrix}\right.\)
b.
ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\y\ne-4\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{15x}{x+1}+\dfrac{10}{y+4}=20\\\dfrac{4x}{x+1}-\dfrac{10}{y+4}=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{15x}{x+1}+\dfrac{10}{y+4}=20\\\dfrac{19x}{x+1}=28\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{x+1}=\dfrac{28}{19}\\\dfrac{1}{y+4}=-\dfrac{4}{19}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}19x=28x+28\\4y+16=-19\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{28}{9}\\y=-\dfrac{35}{4}\end{matrix}\right.\)
câu 3: giải hệ phương trình
a) \(\left\{{}\begin{matrix}5a+b=5\\b-10a=-19\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\dfrac{5x}{6}-y=\dfrac{-5}{6}\\\dfrac{2x}{2x+y}+3y=\dfrac{-2}{3}\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}x\sqrt{3}+3y=1\\2x-y\sqrt{3}=\sqrt{3}\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{6}{y}\\\dfrac{5}{x}+\dfrac{6}{y}=13\end{matrix}\right.=17\)
giúp mk vs ạ mk cần gấp
a) \(\left\{{}\begin{matrix}5a+b=5\\b-10a=-19\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}5a+b=5\\15a=24\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{8}{5}\\b=-3\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{6}{y}=17\\\dfrac{5}{x}+\dfrac{6}{y}=13\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{6}{y}=17\\\dfrac{6}{x}=30\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=-\dfrac{1}{2}\end{matrix}\right.\)
