22034中２・連立方程式・計算問題・応用問題4

(1)答$\begin{array}{r}\{\begin{array}{l}x=2\\ y=-3\end{array}\end{array}$

$\begin{array}{r}x+y=7x+5y=-1\end{array}$

連立方程式の形にして解きます。

$\begin{array}{r}\{\begin{array}{lll}x+y=-1& \cdots \mathrm{①}& \\ 7x+5y=-1& \cdots \mathrm{②}& \end{array}\end{array}$

$\begin{array}{rcl}\mathrm{①}\times 77x+7y& =& -7\\ \mathrm{③}-)7x+5y& =& -1\\ 2y& =& -6\\ y& =& -3\end{array}$

$y=-3$を ① に代入して，

$\begin{array}{rcl}x-3& =& -1\\ x& =& 2\end{array}$

(2)答$\begin{array}{r}\{\begin{array}{l}x=-2\\ y=-9\end{array}\end{array}$

$\begin{array}{r}8x-2y-2=5x-y+1=0\end{array}$

連立方程式の形にして解きます。

$\begin{array}{r}\{\begin{array}{lll}8x-2y-2=0& \cdots \mathrm{①}& \\ 5x-y+1=0& \cdots \mathrm{②}& \end{array}\end{array}$

$\begin{array}{rcl}\mathrm{①}8x-2y-2& =& 0\\ \mathrm{②}\times 2-)10x-2y+2& =& 0\\ -2x-4& =& 0\\ x& =& -2\end{array}$

$x=-2$を ① に代入して，

$\begin{array}{rcl}8\times (-2)-2y-2& =& 0\\ y& =& -9\end{array}$

(3)答$\begin{array}{r}\{\begin{array}{l}x=-4\\ y=1\end{array}\end{array}$

$\begin{array}{r}4x-9y=6x-1=2x-5y-12\end{array}$

連立方程式の形にして解きます。

$\begin{array}{r}\{\begin{array}{lll}4x-9y=6x-1& \cdots \mathrm{①}& \\ 6x-1=2x-5y-12& \cdots \mathrm{②}& \end{array}\end{array}$

① の式を変形します。

$\begin{array}{rcl}4x-9y& =& 6x-1\\ 2x+9y& =& 1\cdots \mathrm{③}\end{array}$

② の式を変形します。

$\begin{array}{rcl}6x-1& =& 2x-5y-12\\ 4x+5y& =& -11\cdots \mathrm{④}\end{array}$

$\begin{array}{rcl}\mathrm{③}\times 24x+18y& =& 2\\ \mathrm{④}-)4x+5y& =& -11\\ 13y& =& 13\\ y& =& 1\end{array}$

$y=1$を ③ に代入して，

$\begin{array}{rcl}2x+9\times 1& =& 1\\ x& =& -4\end{array}$

(4)答$\begin{array}{r}\{\begin{array}{l}x={\displaystyle \frac{2}{3}}\\ y=1\end{array}\end{array}$

$\begin{array}{r}0.6x+0.1y={\displaystyle \frac{12x-y}{14}}={\displaystyle \frac{1}{2}}\end{array}$

連立方程式の形にして解きます。

$\begin{array}{r}\{\begin{array}{lll}0.6x+0.1y={\displaystyle \frac{1}{2}}& \cdots \mathrm{①}& \\ {\displaystyle \frac{12x-y}{14}}={\displaystyle \frac{1}{2}}& \cdots \mathrm{②}& \end{array}\end{array}$

① の両辺に $10$ をかけます。

$\begin{array}{rcl}(0.6x+0.1y)\times 10& =& {\displaystyle \frac{1}{2}}\times 10\\ 6x+y& =& 5\cdots \mathrm{③}\end{array}$

② の両辺に $14$ をかけます。

$\begin{array}{rcl}({\displaystyle \frac{12x-y}{14}})\times 14& =& {\displaystyle \frac{1}{2}}\times 14\\ 12x-y& =& 7\cdots \mathrm{④}\end{array}$

$\begin{array}{rcl}\mathrm{③}\times 212x+2y& =& 10\\ \mathrm{④}-)12x-y& =& 7\\ 3y& =& 3\\ y& =& 1\end{array}$

$y=1$を ③ に代入して，

$\begin{array}{rcl}6x+1& =& 5\\ x& =& {\displaystyle \frac{2}{3}}\end{array}$