Given that x = 10, y = 3A square-2b, and 2x-3y + 5 = 0, when a = (- 0.5), find the value of B and y I'll get higher marks

Given that x = 10, y = 3A square-2b, and 2x-3y + 5 = 0, when a = (- 0.5), find the value of B and y I'll get higher marks

2x-3y + 5 = 0, x = 10, y = 25 / 3
Y = the square of 3a-2b, that is 25 / 3 = the square of 3a-2b, a = (- 0.5), the square of 3A = 3 / 4
So 2B = 3 / 4-25 / 3 B = - 91 / 24

Given 2x + 3Y = 5, when x > 3A + 2B / 2, the value range of Y is obtained

2x>3a+2b
2x+3y>3a+2b+3y
That is, 5 > 3A + 2B + 3Y
3y

If the equation 2x2a + B-4 + 4y3a-2b-3 = 1 is a quadratic equation with respect to X and y, then the values of a and B are______ ．

According to the meaning of the question, 2A + B − 4 = 13a − 2B − 3 = 1, the solution is a = 2B = 1