Решите способом подстановки пожалуйста!

Question 1

Question 2

$\begin{cases}5x+y=1\\11x+2y=8 \end{cases}\\\\\begin{cases} y=-5x+1\\11x+2 (-5x+1)=8 \end{cases}\\\\\begin{cases} y=-5x+1\\11x -10x+2=8 \end{cases}\\\\\begin{cases} y=-5x+1\\ n =8-2 \end{cases} \\\\\begin{cases} y=-5*6+1\\ x =6 \end{cases} \\\\\begin{cases} y=-29\\ x =6 \end{cases}$

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$\begin{cases}y=5x-1\\6x+2y=14 \end{cases}\\\\ \begin{cases}y=5x-1\\6x+2(5x-1)=14 \end{cases}\\\\ \begin{cases}y=5x-1\\6x+10x-2=14 \end{cases}\\\\ \begin{cases}y=5x-1\\16x =14+2 \end{cases}\\\\ \begin{cases}y=5x-1\\16x =16\ \ /:16 \end{cases}\\\\ \begin{cases}y=5*1-1\\ x =1\end{cases}\\\\\begin{cases}y=4\\ x =1 \end{cases}$

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$\begin{cases}3x-5y=-21\\7x+10y=16 \end{cases}\\\\ \begin{cases}3x=5y -21\ \ /:3\\7x+10y=16 \end{cases}\\\\ \begin{cases} x=\frac{5}{3}y-7\\7 (\frac{5}{3}y-7)+10y=16 \end{cases}\\\\ \begin{cases} x=\frac{5}{3}y-7\\ \frac{35}{3}y- 49+10y=16\ \ /*3 \end{cases}\\\\ \begin{cases} x=\frac{5}{3}y-7\\ 35y- 147+30y=48 \end{cases}\\\\ \begin{cases} x=\frac{5}{3}y-7\\ 65y=48+147 \end{cases}$

$\begin{cases} x=\frac{5}{3}y-7\\ 65y=195\ \ /:65 \end{cases} \\\\\begin{cases} x=\frac{5}{3} *3-7\\ y=3 \end{cases} \\\\\begin{cases} x=5-7\\ y=3 \end{cases} \\\\\begin{cases} x=-2\\ y=3 \end{cases}$

Question 3

Спасибо