Giải hệ phương trình sau bằng phương pháp cộng đại số:
a) \(\left\{{}\begin{matrix}49x+7y=-1\\\dfrac{-4}{3}x-2y=\dfrac{4}{3}\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}4x+3y=13\\5x-3y=-31\end{matrix}\right.\)
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24 tháng 1 2021
a) Ta có: \(\left\{{}\begin{matrix}\sqrt{2}x-y=3\\x+\sqrt{2}y=\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2}x-y=3\\\sqrt{2}x+2y=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-3y=1\\x+\sqrt{2}y=\sqrt{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{3}\\x=\sqrt{2}-\sqrt{2}y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{3}\\x=\sqrt{2}-\sqrt{2}\cdot\dfrac{-1}{3}=\dfrac{4\sqrt{2}}{3}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{4\sqrt{2}}{3}\\y=-\dfrac{1}{3}\end{matrix}\right.\)
b) Ta có: \(\left\{{}\begin{matrix}\dfrac{x}{2}-2y=\dfrac{3}{4}\\2x+\dfrac{y}{3}=-\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-8y=3\\2x+\dfrac{1}{3}y=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{25}{3}y=\dfrac{10}{3}\\2x-8y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{2}{5}\\2x=3+8y=3+8\cdot\dfrac{-2}{5}=-\dfrac{1}{5}\end{matrix}\right.\)
hay \(\left\{{}\begin{matrix}x=-\dfrac{1}{10}\\y=-\dfrac{2}{5}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-\dfrac{1}{10}\\y=-\dfrac{2}{5}\end{matrix}\right.\)
c) Ta có: \(\left\{{}\begin{matrix}\dfrac{2x-3y}{4}-\dfrac{x+y-1}{5}=2x-y-1\\\dfrac{x+y-1}{3}+\dfrac{4x-y-2}{4}=\dfrac{2x-y-3}{6}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5\left(2x-3y\right)}{20}-\dfrac{4\left(x+y-1\right)}{20}=\dfrac{20\left(2x-y-1\right)}{20}\\\dfrac{4\left(x+y-1\right)}{12}+\dfrac{3\left(4x-y-2\right)}{12}=\dfrac{2\left(2x-y-3\right)}{12}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}10x-15y-4x-4y+4=40x-20y-20\\4x+4y-4+12x-3y-6=4x-2y-6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x-19y+4-40x+20y+20=0\\16x+y-10-4x+2y+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-34x+y=-24\\12x+3y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-102x+3y=-72\\12x+3y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-114x=-76\\12x+3y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\12\cdot\dfrac{2}{3}+3y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\3y=4-8=-4\end{matrix}\right.\)
hay \(\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-\dfrac{4}{3}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-\dfrac{4}{3}\end{matrix}\right.\)
7 tháng 11 2021
\(1,\Leftrightarrow\left\{{}\begin{matrix}x=2y+4\\-4y-8+5y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\cdot5+4=14\\y=5\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}5x-30+6x=3\\y=10-2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=4-2y\\6y-12+y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{10}{7}\\y=\dfrac{19}{7}\end{matrix}\right.\)
15 tháng 1 2023
a: =>8x+2y=4 và 8x+3y=5
=>y=1 và 4x=2-1=1
=>x=1/4 và y=1
b: 3x-2y=11 và 4x-5y=3
=>12x-8y=44 và 12x-15y=9
=>7y=35 và 3x-2y=11
=>y=5 và 3x=11+2*y=11+2*5=21
=>x=7 và y=5
c: 5x-4y=3 và 2x+y=4
=>5x-4y=3 và 8x+4y=16
=>13x=19 và 2x+y=4
=>x=19/13 và y=4-2x=4-38/13=52/13-38/13=14/13
d: 3x-y=5 và 5x+2y=28
=>6x-2y=10 và 5x+2y=28
=>11x=38 và 3x-y=5
=>x=38/11 và y=3x-5=104/11-5=104/11-55/11=49/11
24 tháng 1 2021
a) Ta có: \(\left\{{}\begin{matrix}-x+2y=3\\3x+y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-3x+6y=9\\3x+y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7y=8\\-x+2y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{8}{7}\\-x=3-2y=3-2\cdot\dfrac{8}{7}=\dfrac{5}{7}\end{matrix}\right.\)
hay \(\left\{{}\begin{matrix}x=-\dfrac{5}{7}\\y=\dfrac{8}{7}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-\dfrac{5}{7}\\y=\dfrac{8}{7}\end{matrix}\right.\)
b) Ta có: \(\left\{{}\begin{matrix}2x+2\sqrt{3}\cdot y=1\\\sqrt{3}x+2y=-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{3}x+6y=\sqrt{3}\\2\sqrt{3}x+4y=-10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2y=\sqrt{3}+10\\\sqrt{3}x+2y=-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{\sqrt{3}+10}{2}\\x\sqrt{3}+2\cdot\dfrac{\sqrt{3}+10}{2}=-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{\sqrt{3}+10}{2}\\x\sqrt{3}=-5-\sqrt{3}-10=-15-\sqrt{3}\end{matrix}\right.\)
hay \(\left\{{}\begin{matrix}x=-1-5\sqrt{3}\\y=\dfrac{\sqrt{3}+10}{2}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-1-5\sqrt{3}\\y=\dfrac{\sqrt{3}+10}{2}\end{matrix}\right.\)
29 tháng 11 2023
a: \(\left\{{}\begin{matrix}\dfrac{x}{2}-\dfrac{y}{3}=1\\5x-8y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}+1\\5x-8y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}y+2\\5\cdot\left(\dfrac{2}{3}y+2\right)-8y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y+2\\\dfrac{10}{3}y+10-8y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{14}{3}y=-7\\x=\dfrac{2}{3}y+2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=7:\dfrac{14}{3}=7\cdot\dfrac{3}{14}=\dfrac{3}{2}\\x=\dfrac{2}{3}\cdot\dfrac{3}{2}+2=1+2=3\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}3x+2y=2\\6x-3y=18\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x=2-2y\\2\cdot3x-3y=18\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3x=2-2y\\2\left(2-2y\right)-3y=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4-7y=18\\3x=2-2y\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}7y=-14\\3x=2-2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-2\\3x=2-2\cdot\left(-2\right)=6\end{matrix}\right.\)
=>x=2 và y=-2
4 tháng 1
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: \(\left\{{}\begin{matrix}\dfrac{x}{y}=\dfrac{2}{3}\\x+y-10=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y\\x+y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{3}y+y=10\\x=\dfrac{2}{3}y\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{5}{3}y=10\\x=\dfrac{2}{3}y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=6\\x=\dfrac{2}{3}\cdot6=4\end{matrix}\right.\)
14 tháng 6 2022
a: \(\left\{{}\begin{matrix}3x-2y=1\\2x+4y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x-4y=2\\2x+4y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}8x=5\\3x-2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{8}\\2y=3x-1=\dfrac{15}{8}-1=\dfrac{7}{8}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{8}\\y=\dfrac{7}{16}\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}4x-3y=1\\-x+2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x-3y=1\\-4x+8y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=-1+2y=-1+2=1\end{matrix}\right.\)
c: \(\left\{{}\begin{matrix}\dfrac{2}{3}x+\dfrac{4}{3}y=1\\\dfrac{1}{2}x-\dfrac{3}{4}y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+4y=3\\2x-3y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{41}{14}\\y=-\dfrac{5}{7}\end{matrix}\right.\)
a) Ta có: \(\left\{{}\begin{matrix}49x+7y=-1\\-\dfrac{4}{3}x-2y=\dfrac{4}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}98x+14y=-2\\-\dfrac{28}{3}x-14y=\dfrac{28}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{266}{3}x=\dfrac{22}{3}\\49x+7y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{133}\\49\cdot\dfrac{11}{133}+7y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{133}\\7y=-1-\dfrac{77}{19}=-\dfrac{96}{19}\end{matrix}\right.\)
hay \(\left\{{}\begin{matrix}x=\dfrac{11}{133}\\y=-\dfrac{96}{133}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{11}{133}\\y=-\dfrac{96}{133}\end{matrix}\right.\)
b) Ta có: \(\left\{{}\begin{matrix}4x+3y=13\\5x-3y=-31\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}9x=-18\\4x+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\3y=13-4x=13-4\cdot\left(-2\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\3y=21\end{matrix}\right.\)
hay \(\left\{{}\begin{matrix}x=-2\\y=7\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-2\\y=7\end{matrix}\right.\)