) 2times 1+x_(2)=1 ytimes (+5x_(2)+3x_(2)+x_(4)=y 3times 1-14x_(2)+3x_(3)+5xy=1 6x_(4)+4times 2+3y_(2)=30

The given system of equations is:<br /><br />$\begin{cases} 2\times 1+x_{2}=1\\ y\times (+5x_{2}+3x_{2}+x_{4}=y\\ 3\times 1-14x_{2}+3x_{3}+5xy=1\\ 6x_{4}+4\times 2+3y_{2}=30\end{cases}$<br /><br />Let's solve this system step by step.<br /><br />1. Simplify the first equation:<br /> $2 + x_{2} = 1$<br /> $x_{2} = -1$<br /><br />2. Simplify the second equation:<br /> $y(5x_{2} + 3x_{2} + x_{4}) = y$<br /> $y(8x_{2} + x_{4}) = y$<br /> $8x_{2} + x_{4} = 1$<br /><br />3. Simplify the third equation:<br /> $3 - 14x_{2} + 3x_{3} + 5xy = 1$<br /> $-14x_{2} + 3x_{3} + 5xy = -2$<br /><br />4. Simplify the fourth equation:<br /> $6x_{4} + 4 \times 2 + 3y_{2} = 30$<br /> $6x_{4} + 8 + 3y_{2} = 30$<br /> $6x_{4} + 3y_{2} = 22$<br /><br />Now we have the following system of equations:<br /><br />$\begin{cases} x_{2} = -1\\ 8x_{2} + x_{4} = 1\\ -14x_{2} + 3x_{3} + 5xy = -2\\ 6x_{4} + 3y_{2} = 22\end{cases}$<br /><br />We can solve this system by substitution or elimination method. Let's use the substitution method.<br /><br />From the first equation, we have $x_{2} = -1$. Substitute this value into the other equations:<br /><br />$\begin{cases} 8(-1) + x_{4} = 1\\ -14(-1) + 3x_{3} + 5xy = -2\\ 6x_{4} + 3y_{2} = 22\end{cases}$<br /><br />Simplifying the first equation, we get $x_{4} = 9$.<br /><br />Substituting $x_{2} = -1$ and $x_{4} = 9$ into the second equation, we have:<br /><br />$-14(-1) + 3x_{3} + 5xy = -2$<br />$14 + 3x_{3} + 5xy = -2$<br />$3x_{3} + 5xy = -16$<br /><br />Substituting $x_{2} = -1$ and $x_{4} = 9$ into the third equation, we have:<br /><br />$6(9) + 3y_{2} = 22$<br />$54 + 3y_{2} = 22$<br />$3y_{2} = -32$<br />$y_{2} = -\frac{32}{3}$<br /><br />Now we have the following system of equations:<br /><br />$\begin{cases} x_{2} = -1\\ x_{4} = 9\\ 3x_{3} + 5xy = -16\\ y_{2} = -\frac{32}{3}\end{cases}$<br /><br />We can solve this system by substitution or elimination method. Let's use the substitution method.<br /><br />From the first equation, we have $x_{2} = -1$. Substitute this value into the other equations:<br /><br />$\begin{cases} 3x_{3} + 5xy = -16\\ y_{2} = -\frac{32}{3}\end{cases}$<br /><br />We cannot solve this system further without additional information. Therefore, the solution to the given system of equations is:<br /><br />$\begin{cases} x_{2} = -1\\ x_{4} = 9frac{32}{3}\end{cases}$