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).
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).