Since the square of any real number is non-negative, the expression can only be equal to 0 when both terms inside the absolute value and the square equal to 0.
So, first, we set: 3x - 2y - 4 = 0 3x = 2y + 4 x = (2/3)y + 4/3
Now, we set: 3x - 5y + 3 = 0 3x = 5y - 3 x = (5/3)y - 1
We have two equations for x, so we can equate them: (2/3)y + 4/3 = (5/3)y - 1 2y + 4 = 5y - 3 4 + 3 = 5y - 2y 7 = 3y y = 7/3
Now we can substitute y back in to find x: x = (2/3)(7/3) + 4/3 x = 14/9 + 4/3 x = (14 + 12) / 9 x = 26 / 9
Therefore, the solution to the equation is: x = 26/9, y = 7/3
(3x - 2y - 4)² + |3x - 5y + 3| = 0
Since the square of any real number is non-negative, the expression can only be equal to 0 when both terms inside the absolute value and the square equal to 0.
So, first, we set:
3x - 2y - 4 = 0
3x = 2y + 4
x = (2/3)y + 4/3
Now, we set:
3x - 5y + 3 = 0
3x = 5y - 3
x = (5/3)y - 1
We have two equations for x, so we can equate them:
(2/3)y + 4/3 = (5/3)y - 1
2y + 4 = 5y - 3
4 + 3 = 5y - 2y
7 = 3y
y = 7/3
Now we can substitute y back in to find x:
x = (2/3)(7/3) + 4/3
x = 14/9 + 4/3
x = (14 + 12) / 9
x = 26 / 9
Therefore, the solution to the equation is:
x = 26/9, y = 7/3