[tex] \bf \begin{cases}
\boxed{y}=5x+2\\
3x=-y+10
\end{cases}~\hspace{4em}\stackrel{\textit{substituting \boxed{y} in the second equation}}{3x=-\left( \boxed{5x+2} \right)+10}
\\\\\\
3x=-5x-2+10\implies 8x=8\implies x=\cfrac{8}{8}\implies \blacktriangleright x=1 \blacktriangleleft
\\\\\\
\stackrel{\textit{substituting the \underline{x} in the first equation}}{y=5(1)+2}\implies \blacktriangleright y = 7\blacktriangleleft \\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
~\hspace{14em}(1,7) [/tex]
Answer: (1,7)
Step-by-step explanation:
The given system of linear equations :-
[tex]y=5x+2...................(1)\\\\3x=-y+10..................(2)[/tex]
Substituting the value of y from (1) in equation (2), we get
[tex]3x=-(5x+2)+10\\\\\Rightarrow 3x=-5x-2+10\\\\\Rightarrow\ 8x=8\\\\\Rightarrow\ x=1[/tex]
Substituting the value of x in (1), we get
[tex]y=5(1)+2=7[/tex]
Hence, the solution to the given system of equations : (x,y)=(1,7)
