Answer
\[x=3,\ y=1\]
Work Step by Step
Multiply \[5\]on both sides to the equation \[4x+3y=15\]to get: \[20x+15y=75\].
Multiply \[3\]on both sides to the equation \[2x-5y=1\]to get: \[6x-15y=3\].
Add the above obtained equation from both RHS and LHS as follows:
\[\underline{\begin{align}
& 20x+15y=75 \\
& 6x-15y=3
\end{align}}\]
\[\begin{align}
& 26x\ \ \ \ \ \ \ \ =78 \\
& x=3
\end{align}\]
Put \[x=3\]in\[4x+3y=15\], to get:
\[\begin{align}
& 4\left( 3 \right)+3y=15 \\
& 12+3y=15 \\
& 3y=3 \\
& y=1
\end{align}\]
Put\[x=3\]and \[y=1\]in any of the given equations to check the solution:
\[\begin{align}
& 2\left( 3 \right)-5\left( 1 \right)=1 \\
& 6-5=1 \\
& 1=1
\end{align}\]
Since RHS\[=\]LHS, it implies the solution is correct.
Now check with 4x + 3y = 15 too:
4(3) + 3(1) = 15
12 + 3 = 15
15 = 15
LHS = RHS
Above solution satisfies both the equations.
