Giải các hệ phương trình Câu hỏi 27 trang 11 SBT Toán 9 Tập 2

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

Bài 27 trang 11 SBT Toán 9 Tập 2: Giải các hệ phương trình:

a) $\begin{array}{l}5\left(x+2y\right)=3x-1\\ 2x+4=3\left(x-5y\right)-12\end{array}$

b) $\begin{array}{l}4x^2-5\left(y+1\right)={\left(2x-3\right)}^2\\ 3\left(7x+2\right)=5\left(2y-1\right)-3x\end{array}$

c) $\begin{array}{l}\frac{2x+1}{4}-\frac{y-2}{3}=\frac{1}{12}\\ \frac{x+5}{2}=\frac{y+7}{3}-4\end{array}$

d) $\begin{array}{l}\frac{3s-2t}{5}+\frac{5s-3t}{3}=s+1\\ \frac{2s-3t}{3}+\frac{4s-3t}{2}=t+1\end{array}$

Lời giải:

a) $\begin{array}{l}5\left(x+2y\right)=3x-1\\ 2x+4=3\left(x-5y\right)-12\end{array}$

$\iff \begin{array}{l}5x+10y=3x-1\\ 2x+4=3x-15y-12\end{array}$

$\iff \begin{array}{l}5x+10y-3x=-1\\ 2x-3x+15y=-12-4\end{array}$

$\iff \begin{array}{l}2x+10y=-1\\ -x+15y=-16\end{array}$$\iff \begin{array}{l}2x+10y=-1\\ 2.\left(-x+15y\right)=\left(-16\right).2\end{array}$

$\iff \begin{array}{l}2x+10y=-1\\ -2x+30y=-32\end{array}$

$\iff \begin{array}{l}2x+10y=-1\\ \left(2x+10y\right)+\left(-2x+30y\right)=\left(-32\right)+\left(-1\right)\end{array}$

$\iff \begin{array}{l}2x+10y=-1\\ 40y=-33\end{array}\iff \begin{array}{l}2x+10.\frac{-33}{40}=-1\\ y=\frac{-33}{40}\end{array}$

$\iff \begin{array}{l}2x=\frac{29}{4}\\ y=\frac{-33}{40}\end{array}\iff \begin{array}{l}x=\frac{29}{8}\\ y=\frac{-33}{40}\end{array}$

Vậy hệ phương trình đã cho có nghiệm (x; y) = $\left(\frac{29}{8};\frac{-33}{40}\right)$

b) $\begin{array}{l}4x^2-5\left(y+1\right)={\left(2x-3\right)}^2\\ 3\left(7x+2\right)=5\left(2y-1\right)-3x\end{array}$

$\iff \begin{array}{l}4x^2-5y-5=4x^2-12x+9\\ 21x+6=10y-5-3x\end{array}$

$\iff \begin{array}{l}12x-5y=14\\ 24x-10y=-11\end{array}\iff \begin{array}{l}24x-10y=28\\ 24x-10y=-11\end{array}$

$\iff \begin{array}{l}\left(24x+10y\right)-\left(24x+10y\right)=28+11\\ 24x-10y=-11\end{array}$

$\iff \begin{array}{l}0x-0y=39\\ 24x-10y=-11\end{array}$(vô lí)

Vậy hệ phương trình đã cho vô nghiệm.

c) $\begin{array}{l}\frac{2x+1}{4}-\frac{y-2}{3}=\frac{1}{12}\\ \frac{x+5}{2}=\frac{y+7}{3}-4\end{array}$

$\iff \begin{array}{l}3\left(2x+1\right)-4\left(y-2\right)=1\\ 3\left(x+5\right)=2\left(y+7\right)-24\end{array}$

$\iff \begin{array}{l}6x+3-4y+8=1\\ 3x+15=2y+14-24\end{array}$

$\iff \begin{array}{l}6x-4y=-10\\ 3x-2y=-25\end{array}$

$\iff \begin{array}{l}3x-2y=-5\\ 3x-2y=-25\end{array}$

$\iff \begin{array}{l}\left(3x-2y\right)-\left(3x-2y\right)=-5+25\\ 3x-2y=-5\end{array}$

$\iff \begin{array}{l}0x+0y=20\\ 3x-2y=-5\end{array}$(vô lí)

Vậy hệ phương trình đã cho vô nghiệm.

d) $\begin{array}{l}\frac{3s-2t}{5}+\frac{5s-3t}{3}=s+1\\ \frac{2s-3t}{3}+\frac{4s-3t}{2}=t+1\end{array}$

$\iff \begin{array}{l}3\left(3s-2t\right)+5\left(5s-3t\right)=15\left(s+1\right)\\ 2\left(2s-3t\right)+3\left(4s-3t\right)=6\left(t+1\right)\end{array}$

$\iff \begin{array}{l}9s-6t+25s-15t=15s+15\\ 4s-6t+12s-9t=6t+6\end{array}$

$\iff \begin{array}{l}19s-21t=15\\ 16s-21t=6\end{array}$

$\iff \begin{array}{l}\left(19s-21t\right)-\left(16s-21t\right)=15-6\\ 16s-21t=6\end{array}$

$\iff \begin{array}{l}3s=9\\ 16s-21t=6\end{array}\iff \begin{array}{l}s=3\\ 16.3-21t=6\end{array}$

$\iff \begin{array}{l}s=3\\ -21t=6-48\end{array}\iff \begin{array}{l}s=3\\ -21t=-42\end{array}\iff \begin{array}{l}s=3\\ t=2\end{array}$

Vậy hệ phương trình đã cho có nghiệm (s; t) = (3; 2).