How to solve the following linear equations by substitution method: 1) x^2+y^2=13 and 2x-y=4 2) 4x^2+9y^2=72 and 2x-y=4 3)4x^2+9y^2=72 and x-y^2=-1?

At least one of the equations in each set is not a linear equation.

Substitution is the replacement of a variable or expression by its equivalent.

1) Solve the second equation for y, then substitute for y in the first equation.
2x - y = 4
2x = 4 + y    (add y)
2x - 4 = y    (now we have an equivalent for y that we can use in the first equation)
x^2 + (2x - 4)^2 = 13    (make the substitution)
x^2 + 4x^2 - 16x + 16 = 13    (carry out the square operation)
5x^2 - 16x + 3 = 0    (subtract 13)
(5x - 1)(x - 3) = 0    (factor)
x = 1/5, 3    (find solutions to make the factors zero)
y = -3 3/5, 2    (find corresponding values for y)
The solutions are (x, y) = (1/5, -3 3/5) and (3, 2).

2) Use the same substitution as before.
4x^2 + 9(4x^2 - 16x + 16) = 72
40x^2 -144x + 72 = 0    (rewrite to standard form)
5x^2 - 18x + 9 = 0    (divide by 8)
(5x-3)(x-3) = 0    (factor)
x = (3/5, 3); y = -2 4/5, 2
The solutions are (x, y) = (3/5, -2 4/5), (3, 2).

3) You can solve the second equation for y^2 and make a substitution for that.   x - y^2 = -1
x+1 = y^2
4x^2 + 9(x+1) = 72    (make the substitution for y^2)
4x^2 + 9x - 63 = 0    (put in standard form)
(4x+21)(x-3) = 0    (factor it)
x = -21/4, 3    (only the positive solution is useful here)
3+1 = y^2, so y = +/- 2
The solutions are (x, y) = (3, 2), (3, -2).
thanked the writer.