Answer
$x = 2$
Work Step by Step
Square both sides of the equation to get rid of the radicals:
$(\sqrt{5x-3})^2=(\sqrt{2x+3})^2\\
5x - 3 = 2x + 3$
Subtract $2x$ from each side of the equation to move variables to the left side of the equation:
$5x-3-2x=2x+3-2x\\
3x - 3 = 3$
Add $3$ to each side of the equation to move constants to the right side of the equation:
$3x-3+3 =3+3\\
3x= 6$
Divide both sides of the equation by $3$ to solve for $x$:
$\frac{3x}{3}=\frac{6}{3}\\
x = 2$
To check for extraneous solutions, substitute $2$ for $x$ into the original equation:
$\sqrt {5(2) - 3} = \sqrt {2(2) + 3}$
$\sqrt {10 - 3} = \sqrt {4 + 3}$
$\sqrt {7} = \sqrt {7}$
Both sides of the equation equal one another; therefore, the answer is correct. There are no extraneous solutions.
