Answer
\[\left\{ \left( -1,3 \right) \right\}\]
Work Step by Step
To find the solution of provided system of equation by substitution method and using the following steps:
Step1:
The first equation, \[2x-y=-5\], solve it for the y in terms of x.
\[\begin{align}
& 2x-y=-5 \\
& y=2x+5
\end{align}\]
Step2:
Substitute the value of \[y=2x+5\] into the other equation \[x+5y=14\]. The equation became in one variable x. Solve the equation for x:
\[\begin{align}
& x+5y=14 \\
& x+5\left( 2x+5 \right)=14 \\
& x+10x+25=14 \\
& 11x=14-25
\end{align}\]
Simplify the above equation:
\[\begin{align}
& 11x=14-25 \\
& 11x=-11 \\
& x=-1
\end{align}\]
Step3:
Now, substitute the value of x obtained in step2 in \[y=2x+5\]:
\[\begin{align}
& y=2x+5 \\
& y=2\cdot \left( -1 \right)+5 \\
& y=-2+5 \\
& y=3 \\
\end{align}\]
Step4:
The value of x and y obtained in step2 and step3, is the solution of the provided system of equations.
Hence \[\left\{ \left( -1,3 \right) \right\}\]is the required solution.
Step5:
Now, to verify that the obtained solution is correct, substitute the values of x and y in both equations.
Put \[x=-1\text{ and }y=3\]
\[\begin{align}
& 2x-y=-5 \\
& 2\cdot \left( -1 \right)-3=-5 \\
& -2-3=-5 \\
& -5=-5
\end{align}\]
\[\begin{align}
& x+5y=14 \\
& -1+5\cdot 3=14 \\
& -1+15=14 \\
& 14=14
\end{align}\]
Since, \[\left( -1,3 \right)\]satisfies both equations, the set \[\left( -1,3 \right)\]is the solution of the provided system of equations.
