Solve for x and y
3x²+5=2y²
3x-2y-1=0
To solve for x and y, we will use the substitution method.
Given equations:
1) 3x² + 5 = 2y²
2) 3x - 2y - 1 = 0
From equation 2, we can isolate x in terms of y:
3x = 2y + 1
x = (2y + 1) / 3
Now, substitute x in equation 1 with (2y + 1) / 3:
3((2y + 1) / 3)² + 5 = 2y²
(2y + 1)² + 5 = 2y²
(4y² + 4y + 1) + 5 = 2y²
4y² + 4y + 6 = 2y²
4y² + 4y + 6 - 2y² = 0
2y² + 4y + 6 = 0
Now, solve the quadratic equation for y. We can use the quadratic formula for this:
y = [-b ± √(b² - 4ac)] / 2a
a = 2, b = 4, c = 6
y = [-4 ± √(4² - 4(2)(6))] / (2*2)
y = [-4 ± √(16 - 48)] / 4
y = [-4 ± √(-32)] / 4
y = (-4 ± 4√2i) / 4
y = -1 ± √2i
So, there are two possible values for y:
1) y = -1 + √2i
2) y = -1 - √2i
Now, substitute the values of y back into the equation from step 2 to find the corresponding values of x:
For y = -1 + √2i:
x = (2(-1 + √2i) + 1) / 3
x = (-2 + 2√2i + 1) / 3
x = (-1 + 2√2i) / 3
x = -1/3 + 2√2i/3
For y = -1 - √2i:
x = (2(-1 - √2i) + 1) / 3
x = (-2 - 2√2i + 1) / 3
x = (-1 - 2√2i) / 3
x = -1/3 - 2√2i/3
Therefore, the solutions for x and y are:
1) x = -1/3 + 2√2i/3, y = -1 + √2i
2) x = -1/3 - 2√2i/3, y = -1 - √2i