To solve by elimination, look at the coefficients of x and y in each equation. To eliminate a variable, you need to multiply one of the equations by a constant that makes the coefficient of either x or y the same absolute value as its coefficient in the other equation. In this case, the coefficient of x is 1 in the first equation and 3 in the second equation. You can multiply the first equation by 3 to get 3[x - 5y = -3]. Distribute the 3 to get [3x - 15y = -9]. Now you can subtract this from the second equation as follows: [3x - 5y = 11] - [3x - 15y = -9], so [3x - 3x - 5y + 15y = 11 + 9]. Simplify to 10y = 20, and you can divide both sides by 10 to get y = 2. Plug this back into the first equation to get x - 5(2) = -3, so x - 10 = -3 and x = 10 - 3 = 7.
To solve by substitution, isolate one of the variables in an equation. In the first equation, you can get x by itself by adding 5y to both sides to make x = 5y - 3. Now plug the expression for x into the second equation to get 3[5y - 3] - 5y = 11. Distribute the 3 to get 15y - 9 - 5y = 11. Combine like terms to get 10y - 9 = 11. Add 9 to both sides to get 10y = 20, and you can divide both sides by 10 to get y = 2. You can now solve for x as with the other method.
To solve the equations graphically, find two points on the first line. It's convenient to let y = 0, so x - 5(0) = -3, so x = -3. Now let x = 0, so 0 - 5y = -3, so y = 3/5. You now have two points: (-3,0) and (0,3/5). Connect the dots and use a straightedge to draw the line further out on your graph. Now find two points on the other line: let x = 0, so 3(0) - 5y = 11, so -5y = 11, so y = -11/5. Let y = 0, so 3x - 5(0) = 11, so 3x = 11, so x = 11/3. You now have two points: (0,-11/5) and (11/3, 0). Connect the dots and extend the line with a straightedge. This line will intersect the line you drew earlier in either 1) one point (one solution) , 2) nowhere (parallel lines), or 3) everywhere (overlapping lines). In this case, they will intersect in one point, and that's your solution.
