Cần gấp vote 5 sao
Giải bằng hệ phương trình bậc nhất 2 ẩn
a) 0,35x+4y=-2,6 và 0,75x-6y=9
b) 10x-9y=8 và 15x+21y=0,5
c) 3,3x+4,2y=1 và 9x+14y=4
Cần gấp vote 5 sao
Giải bằng hệ phương trình bậc nhất 2 ẩn
a) 0,35x+4y=-2,6 và 0,75x-6y=9
b) 10x-9y=8 và 15x+21y=0,5
c) 3,3x+4,2y=1 và 9x+14y=4
Đáp án:
a) \(\left\{ \begin{array}{l}
x = 4\\
y = – 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\left\{ \begin{array}{l}
\dfrac{7}{{20}}x + 4y = – \dfrac{{13}}{5}\\
\dfrac{3}{4}x – 6y = 9
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = \left( {\dfrac{3}{4}x – 9} \right):6\\
\dfrac{7}{{20}}x + 4.\dfrac{{\dfrac{3}{4}x – 9}}{6} = – \dfrac{{13}}{5}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = \left( {\dfrac{3}{4}x – 9} \right):6\\
\dfrac{7}{{20}}x + \dfrac{{3x – 36}}{6} = – \dfrac{{13}}{5}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = \left( {\dfrac{3}{4}x – 9} \right):6\\
\dfrac{7}{{20}}x + \dfrac{{x – 12}}{2} = – \dfrac{{13}}{5}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = \left( {\dfrac{3}{4}x – 9} \right):6\\
\dfrac{{7x + 10x – 120 + 13.4}}{{20}} = 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = \left( {\dfrac{3}{4}x – 9} \right):6\\
17x = 68
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 4\\
y = – 1
\end{array} \right.\\
b)\left\{ \begin{array}{l}
10x – 9y = 8\\
15x + 21y = \dfrac{1}{2}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = \dfrac{{10x – 8}}{9}\\
15x + 21.\dfrac{{10x – 8}}{9} = \dfrac{1}{2}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = \dfrac{{10x – 8}}{9}\\
15x + \dfrac{{70x – 56}}{3} = \dfrac{1}{2}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = \dfrac{{10x – 8}}{9}\\
\dfrac{{45x + 70x – 56}}{3} = \dfrac{1}{2}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = \dfrac{{10x – 8}}{9}\\
230x – 112 = 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{1}{2}\\
y = – \dfrac{1}{3}
\end{array} \right.\\
c)\left\{ \begin{array}{l}
\dfrac{{33}}{{10}}x + \dfrac{{42}}{{10}}y = 1\\
9x + 14y = 4
\end{array} \right.\\
\to \left\{ \begin{array}{l}
33x + 42y = 10\\
9x + 14y = 4
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = – \dfrac{1}{3}\\
y = \dfrac{1}{2}
\end{array} \right.
\end{array}\)