The quotient of number a divided by number B is 14, the remainder is 2, the sum of number a, number B, quotient and remainder is 63, what's number B? We don't need to use the method of setting x ` y The child is still young

The quotient of number a divided by number B is 14, the remainder is 2, the sum of number a, number B, quotient and remainder is 63, what's number B? We don't need to use the method of setting x ` y The child is still young

Analysis: a number △ B number = 14.2 a number + B number + quotient + remainder = 63 quotient + remainder = 16 a number + B number = 63-16 = 47, then a number = B number × 14 + 2, so B number = (a number - 2) △ 14, because: a number includes 14 b, plus 2, then: 47-2 = 45 = 14 b + 1 B (itself) according to the above analysis formula: [63 - (14 + 2) - 2]

If the number a is divided by the number B, and the number B is divided by the number C, the quotient is equal, and the remainder is 2, and the remainder and quotient cannot be divisible. The sum of the two numbers a and B is 478

Let a = x, B = y, C = Z
x+z=478 z=478-x
Because the remainder is 2
(x-2)/y=(y-2)/(428-x)
So a = 359, B = 119, C = 39

If the number a is divided by the number B, the quotient of 3 is more than 5. If the number a is increased by 4 times, the quotient is exactly 14. What is the number a

Let number B be X.4 (3x + 5) = 14x, x = 10 10 * 3 + 5 = 35 A: number a is 35

If x = - 1 and y = 2 are solutions of the equation ax-4y = 3,3x + by = 5, then a =?, B =?

-A-8=3；
A=-11；
-3+2B=5；
B=4；

(2x-1) (3x + 4) = 2x-1. What method is used to solve this equation?
Especially "(2x-1) (3x + 4-1) = O" can be solved from there? (2x-1) (3x + 4) = 2x-1 (2x-1) (3x + 4) - (2x-1) = 0 (2x-1) (3x + 4-1) = 0 (2x-1) (3x + 3) = 0 x1 = 1 / 2 x2 = - 1

(2x-1) (3x + 4) = 2x-1 (2x-1) (3x + 4) - (2x-1) = 0 (shift) (2x-1) (3x + 4-1) = 0 (extract common factor) (2x-1) (3x + 3) = 0 (merge) X1 = 1 / 2 x2 = - 1

If x = 1y = − 2 is a solution of the equation AX by = 1 about X and y, and a + B = - 3, then 5a-2b=______ ．

Substituting x = 1y = − 2 into the equation AX by = 1, we get a + 2B = 1. Because a + B = - 3, we get the binary linear equation system a + 2B = 1A + B = − 3 about a and B. solving this equation system, we get b = 4, a = - 7, so 5a-2b = 5 × (- 7) - 2 × 4 = - 35-8 = - 43

If x = 1, y = 2 is a solution of the equation ax-by = - 3, and a + B = 3, then 5a-2b=

If x = 1, y = 2 is a solution of the equation ax-by = - 3 about X and y
a-2b=-3
∵a+b=3
∴ a=1,b=2
5a-2b=5-4=1

If x = 1y = - 2, is a solution of the equation AX by = 1 about XY, and a + B = - 3, then 5a-2b =?

Substituting x = 1, y = - 2 into the equation AX by = 1, we get the following result:
a+2b=1 （1）
And a + B = - 3 (2)
(1) 2
b=4
Substituting B = 4 into (1) yields:
a=-7
So 5a-2b
=-35-8
=-43

If x = 1y = − 2 is a solution of the equation AX by = 1 about X and y, and a + B = - 3, then 5a-2b=______ ．

Substituting x = 1y = − 2 into the equation AX by = 1, we get a + 2B = 1. Because a + B = - 3, we get the binary linear equation system a + 2B = 1A + B = − 3 about a and B. solving this equation system, we get b = 4, a = - 7, so 5a-2b = 5 × (- 7) - 2 × 4 = - 35-8 = - 43

When a is a value, the binary equation x ^ 2-ax + A ^ 2-4 = 0, ① two positive roots, ② two negative roots, ③ one positive root

Let y = x ^ 2-ax + A ^ 2-4, the coefficient of quadratic term > 0, and the opening of parabola upward
① Two positive roots
△≥0
When x = 0, Y > 0
Axis of symmetry > 0
② Two negative roots
△≥0
When x = 0, Y > 0
Axis of symmetry < 0
③ A root
1) When △ = 0, the axis of symmetry is greater than 0
2) When △ 0, x = 0, y < 0
My calculation is a little poor
① 2 ＜ a ≤ 4 / 3 root sign 3
② Negative 4 / 3 root sign 3 ≤ a < - 2
③ A equals 16 / 3 or - 2 < a < 2