Ai giúp em giải câu b,c bằng phương pháp cộng đại số
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23 tháng 10 2021
b: \(\left\{{}\begin{matrix}3x-2y=4\\2x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-2y=4\\4x+2y=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-x=-6\\2x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\y=5-2x=5-12=-7\end{matrix}\right.\)
PH
1
1 tháng 12 2021
\(\left\{{}\begin{matrix}3x+2y=-2\\3x-2y=-3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3x+2y=-2\\3x+2y-3x+2y=-2+3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3x+2y=-2\\4y=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3x+2.\dfrac{1}{4}=-2\\y=\dfrac{1}{4}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{6}\\x=\dfrac{1}{4}\end{matrix}\right.\)
PH
2
1 tháng 12 2021
\(\left\{{}\begin{matrix}-5x+2y=4\\6x-3y=-7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}15x-6y=-12\\12x-6y=-14\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}15x-6y=-12\\15x-6y-12x+6y=-12-\left(-14\right)\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}15x-6y=-12\\3x=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}15.\dfrac{2}{3}-6y=-12\\x=\dfrac{2}{3}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=\dfrac{11}{3}\end{matrix}\right.\)
1 tháng 12 2021
\(\Leftrightarrow\left\{{}\begin{matrix}15x-6y=-12\\12x-6y=-14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=2\\-5x+2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\2y-\dfrac{10}{3}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=\dfrac{11}{3}\end{matrix}\right.\)
PH
0
30 tháng 12 2021
4: \(\Leftrightarrow\left\{{}\begin{matrix}-3y=9\\x+y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-3\\x=-1-y=-1-\left(-3\right)=2\end{matrix}\right.\)
30 tháng 12 2021
\(\Leftrightarrow\left\{{}\begin{matrix}6x-2y=10\\5x+2y=23\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=3x-5=4\end{matrix}\right.\)
KV
19 tháng 10 2023
3x - 2y = 5 ⇔ 6x - 4y = 10
và 2x + 3y = 12 ⇔ 6x + 9y = 36
⇔ 13y = 26
và 2x + 3y = 12
⇔ y = 2
và 2x + 3.2 = 12
⇔ y = 2
và 2x = 6
⇔ y = 2
và x = 3
Vậy S = {(3; 2)}
30 tháng 12 2021
\(\Leftrightarrow\left\{{}\begin{matrix}x-2y=-1\\x-2y=-1\end{matrix}\right.\Leftrightarrow\left(x,y\right)\in R\)
4 tháng 10 2021
\(1,3x+2y=7\\ \Leftrightarrow2y=7-3x\left(1\right)\)
Vì \(2y⋮2\)
\(\Leftrightarrow3x-7⋮2\\ \Leftrightarrow3x-9⋮2\\ \Leftrightarrow3\left(x-3\right)⋮2\\ \Leftrightarrow x-3⋮2\\ \Leftrightarrow x.lẻ\)
Đặt \(x=2k+1\left(k\in Z\right)\)
Thay vào (1), ta được :
\(\left(1\right)\Leftrightarrow2y=3\left(2k+1\right)-7\\ \Leftrightarrow2y=6k+3-7\\ \Leftrightarrow2y=6k-4\\ \Leftrightarrow y=3k-2\)
Vậy \(x=2k+1;y=3k-2\left(k\in Z\right)\)
\(2,C_1:\left\{{}\begin{matrix}-2x+y=1\\4x+5y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-4x+2y=2\\4x+5y=3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4x+5y=2\\7y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{7}\\y=\dfrac{5}{7}\end{matrix}\right.\\ C_2:\left\{{}\begin{matrix}-2x+y=1\\4x+5y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1+2x\\4x+5y=3\end{matrix}\right.\Leftrightarrow4x+5+10x=3\\ \Leftrightarrow x=-\dfrac{1}{7}\Leftrightarrow y=1-\dfrac{2}{7}=\dfrac{5}{7}\)
\(b,\left\{{}\begin{matrix}2x+y=3\\x-y=7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3x=10\\x-y=7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{10}{3}\\\dfrac{10}{3}-y=7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{10}{3}\\y=-\dfrac{11}{3}\end{matrix}\right.\)
\(c,\left\{{}\begin{matrix}2x-3y=6\\2x+y=3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}-4y=3\\2x+y=3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-3}{4}\\2x-\dfrac{3}{4}=3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-3}{4}\\x=\dfrac{15}{8}\end{matrix}\right.\)
b.
\(\left\{{}\begin{matrix}2x+y=3\\x-y=7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3x=10\\x-y=7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{10}{3}\\\dfrac{10}{3}-y=7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{10}{3}\\y=-\dfrac{11}{3}\end{matrix}\right.\)
Vậy...
c.
\(\left\{{}\begin{matrix}2x-3y=6\\2x+y=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-2x+3y=-6\\2x+y=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}4y=-3\\2x+y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{3}{4}\\2x-\dfrac{3}{4}=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{3}{4}\\x=\dfrac{15}{8}\end{matrix}\right.\)