First, let's solve the system:
[tex]\begin{cases}3x-2y=10 \\ x+y=5\end{cases}[/tex]
Clearing y from equation 2, substituting in equation 1 and solving for x :
[tex]\begin{gathered} x+y=5\rightarrow y=5-x \\ 3x-2y=10 \\ \rightarrow3x-2(5-x)=10 \\ \rightarrow3x-10+2x=10 \\ \rightarrow5x=20 \\ \Rightarrow x=4 \end{gathered}[/tex]
Substituting and solving for y :
[tex]\begin{gathered} y=5-x \\ \Rightarrow y=1 \end{gathered}[/tex]
We get that the solution for the system is:
[tex](4,1)[/tex]
In order for two lines to be parallel, they need to have the same slope. This means that we'll use a slope of 2.
Using this, the point calculated and the slope-point form:
[tex]\begin{gathered} y-1=2(x-4) \\ \rightarrow y-1=2x-8 \\ \rightarrow y=2x-7 \end{gathered}[/tex]
We get that the equation of the line that is parallel to y = 2x+1 and passes through (4 ,1) is:
[tex]y=2x-7[/tex]
