Giải các hệ phương trình sau bằng phương pháp cộng

Giải SBT Toán 9 Bài 4: Giải hệ phương trình bằng phương pháp cộng đại số

Bài 25 trang 11 SBT Toán 9 Tập 2: 

a)$\begin{array}{l}2x-11y=-7\\ 10x+11y=31\end{array}$

b)$\begin{array}{l}4x+7y=16\\ 4x-3y=-24\end{array}$

c)$\begin{array}{l}0,35x+4y=-2,6\\ 0,75x-6y=9\end{array}$

d)$\begin{array}{l}\sqrt{2}x+2\sqrt{3}y=5\\ 3\sqrt{2}x-\sqrt{3}y=\frac{9}{2}\end{array}$

e)$\begin{array}{l}10x-9y=8\\ 15x+21y=0,5\end{array}$

f)$\begin{array}{l}3,3x+4,2y=1\\ 9x+14y=4\end{array}$

Lời giải:

a)$\begin{array}{l}2x-11y=-7\\ 10x+11y=31\end{array}$$\iff \begin{array}{l}\left(2x-11y\right)+\left(10x+11y\right)=-7+31\\ 2x-11y=-7\end{array}$

$\iff \begin{array}{l}12x=24\\ 2x-11y=-7\end{array}\iff \begin{array}{l}x=2\\ 2.2-11y=-7\end{array}$

$\iff \begin{array}{l}x=2\\ 4-11y=-7\end{array}\iff \begin{array}{l}x=2\\ -11y=-11\end{array}\iff \begin{array}{l}x=2\\ y=1\end{array}$

Vậy hệ phương trình đã cho có nghiệm (x; y) = (2; 1)

b)$\begin{array}{l}4x+7y=16\\ 4x-3y=-24\end{array}$$\iff \begin{array}{l}4x+7y=16\\ \left(4x+7y\right)-\left(4x-3y\right)=16-\left(-24\right)\end{array}$

$\iff \begin{array}{l}4x+7y=16\\ 10y=40\end{array}\iff \begin{array}{l}4x+7y=16\\ y=4\end{array}$

$\iff \begin{array}{l}4.x+7.4=16\\ y=4\end{array}\iff \begin{array}{l}4x+28=16\\ y=4\end{array}$

$\iff \begin{array}{l}4x=-12\\ y=4\end{array}\iff \begin{array}{l}x=-3\\ y=4\end{array}$

Vậy hệ phương trình đã cho có nghiệm (x; y) = (-3; 4).

c)$\begin{array}{l}0,35x+4y=-2,6\\ 0,75x-6y=9\end{array}$$\iff \begin{array}{l}3.\left(0,35x+4y\right)=3.\left(-2,6\right)\\ 2.\left(0,75x-6y\right)=2.9\end{array}$

$\iff \begin{array}{l}1,05x+12y=-7,8\\ 1,5x-12y=18\end{array}\iff \begin{array}{l}2,55x=10,2\\ 1,5x-12y=18\end{array}$

$\iff \begin{array}{l}x=4\\ 1,5.4-12y=18\end{array}\iff \begin{array}{l}x=4\\ 6-12y=18\end{array}$

$\iff \begin{array}{l}x=4\\ -12y=12\end{array}\iff \begin{array}{l}x=4\\ y=-1\end{array}$

Vậy hệ phương trình đã cho có nghiệm (x; y) = (4; -1).

d)$\begin{array}{l}\sqrt{2}x+2\sqrt{3}y=5\\ 3\sqrt{2}x-\sqrt{3}y=\frac{9}{2}\end{array}$$\iff \begin{array}{l}3\left(\sqrt{2}x+2\sqrt{3}y\right)=5.3\\ 3\sqrt{2}x-\sqrt{3}y=\frac{9}{2}\end{array}$

$\iff \begin{array}{l}3\sqrt{2}x+6\sqrt{3}y=15\\ 3\sqrt{2}x-\sqrt{3}y=\frac{9}{2}\end{array}\iff \begin{array}{l}\sqrt{2}x+2\sqrt{3}y=5\\ \left(3\sqrt{2}x+6\sqrt{3}y\right)-\left(3\sqrt{2}x-\sqrt{3}y\right)=15-\frac{9}{2}\end{array}$

$\iff \begin{array}{l}\sqrt{2}x+2\sqrt{3}y=5\\ 7\sqrt{3}y=10,5\end{array}\iff \begin{array}{l}\sqrt{2}x=5-2\sqrt{3}y\\ y=\frac{\sqrt{3}}{2}\end{array}$

$\iff \begin{array}{l}\sqrt{2}x=5-2\sqrt{3}.\frac{\sqrt{3}}{2}\\ y=\frac{\sqrt{3}}{2}\end{array}\iff \begin{array}{l}\sqrt{2}x=2\\ y=\frac{\sqrt{3}}{2}\end{array}\iff \begin{array}{l}x=\sqrt{2}\\ y=\frac{\sqrt{3}}{2}\end{array}$Vậy hệ phương trình đã cho có nghiệm $\left(x;y\right)=\left(\sqrt{2};\frac{\sqrt{3}}{2}\right)$

e)$\begin{array}{l}10x-9y=8\\ 15x+21y=0,5\end{array}$

$\iff \begin{array}{l}30x-27y=24\\ 30x+42y=1\end{array}\iff \begin{array}{l}30x-27y=24\\ \left(30x+42y\right)-\left(30x-27y\right)=1-24\end{array}$

$\iff \begin{array}{l}30x-27y=24\\ y=-23:69\end{array}$$\iff \begin{array}{l}30x-27y=24\\ y=-23:69\end{array}$

$\iff \begin{array}{l}30x-27.\left(\frac{-1}{3}\right)=24\\ y=\frac{-1}{3}\end{array}$$\iff \begin{array}{l}30x=24-9\\ y=\frac{-1}{3}\end{array}$

$\iff \begin{array}{l}30x=15\\ y=\frac{-1}{3}\end{array}\iff \begin{array}{l}x=\frac{1}{2}\\ y=\frac{-1}{3}\end{array}$

Vậy hệ phương trình đã cho có nghiệm (x; y) = $\left(\frac{1}{2};\frac{-1}{3}\right)$.

f)$\begin{array}{l}3,3x+4,2y=1\\ 9x+14y=4\end{array}\iff \begin{array}{l}\frac{30}{11}\left(3,3x+4,2y\right)=\frac{30}{11}\\ 9x+14y=4\end{array}$

$\iff \begin{array}{l}9x+\frac{126}{11}y=\frac{30}{11}\\ 9x+14y=4\end{array}\iff \begin{array}{l}9x+14y=4\\ \left(9x+14y\right)-\left(9x+\frac{126}{11}y\right)=4-\frac{30}{11}\end{array}$

$\iff \begin{array}{l}9x+14y=4\\ \frac{28}{11}y=\frac{14}{11}\end{array}\iff \begin{array}{l}9x+14y=4\\ y=\frac{1}{2}\end{array}$

$\iff \begin{array}{l}9x+14.\frac{1}{2}=4\\ y=\frac{1}{2}\end{array}\iff \begin{array}{l}9x+7=4\\ y=\frac{1}{2}\end{array}$

$\iff \begin{array}{l}9x=-3\\ y=\frac{1}{2}\end{array}\iff \begin{array}{l}x=-\frac{1}{3}\\ y=\frac{1}{2}\end{array}$

Vậy hệ phương trình đã cho có nghiệm (x; y) = $\left(-\frac{1}{3};\frac{1}{2}\right)$.