If x = - 3, y = 5, is the solution of the system of equations 5x + ay = 0, BX + 7Y + 1 = 0, then a = () B = ()

If x = - 3, y = 5, is the solution of the system of equations 5x + ay = 0, BX + 7Y + 1 = 0, then a = () B = ()

5x+ay=0
bx+7y+1=0
If x = - 3, y = 5, then
-15+5a=0 a=3
-3b+35+1=0
b=12
therefore
a=3 b=12

It is known that the solutions of the equations of X and Y: 5x-7y = - 2.5ax + 7by = 31 and ax by = - 1.x + 5Y = 6 are the same, so the value of a and B can be obtained

The solutions of the equations 5x-7y = - 2,5ax + 7by = 31 and ax-by = - 1, x + 5Y = 6 are the same. The solutions of the equations must make 5x-7y = - 2 and X + 5Y = 6 become immediate. The solutions of the equations 5x-7y = - 2, x + 5Y = 6 are the solutions of the original two equations. The solutions of the equations 5x-7y = - 2, x + 5Y = 6 get x = 1, y = 1, then the solutions must make 5AX + 7by = 31

Find the positive integer solution of the system of Diophantine equations 5x + 7Y + 2Z = 243x-y-4z = 4

5x + 7Y + 2Z = 24 ① 3x-y-4z = 4 ②, ① × 2, 10x + 14y + 4Z = 48 ③ We can get 13X + 13y = 52, that is, x + y = 4, ∵ x, y, Z are positive integers, ∵ x = 1, y = 3 or x = 2, y = 2 or x = 3, y = 1, substituting x = 1, y = 3 into ②, 3-3-4z = 0, z = 0, which is not suitable; substituting x = 2, y = 2 into ②, 6 -

If | x + 12y-3 | and | 2x-4y-144 | are opposite to each other, calculate the value of 10x + 5yx − 2Y

∵| x + 12y-3 | and | 2x-4y-144 | are opposite to each other, ∵| x + 12y-3 | + | 2x-4y-144 | = 0, ∵ x + 12y-3 = 0, 2x-4y-144 = 0. The solution is x = 845, y = - 1385, ∵ 10x + 5yx − 2Y = 10 × 845 + 5 × (− 1385) 845 − 2 × (− 1385) = 512

If the two roots of two quadratic equations of one variable are equal, are a, B and C exactly the same?

Not necessarily
For example, the two solutions of (x-1) (X-2) = 0 and 2 (x-1) (X-2) = 0 are the same, but not necessarily equal

Given the quadratic equation of one variable about X, we can find two unequal real roots

If the square of B minus 4ac is less than zero, then the equation has no solution
If the square of B minus 4ac equals zero, then the equation has two identical real solutions
The square of B minus 4ac is greater than zero
X1 is equal to {negative B plus [root (square of B minus 4ac)]}
X2 is equal to {negative B minus [root (square of B minus 4ac)]}
Because it's not easy to use symbols to express the root sign, I hope you can understand it

When t takes what value, the univariate quadratic equation x & # / 4 + (1 / 2x + T) = 1 has two equal roots
Wrong question... Should be "X & # 178 / 4 + (1 / 2x + T) &# 178; = 1"

By sorting out the equation, we get: X & # / 4 + 1 / 2x + T-1 = 0
Because: the univariate quadratic equation x & # / 4 + (1 / 2x + T) = 1 of X has two equal roots
So: ⊿ = (1 / 2) &# 178; - 4 * (1 / 4) * (t-1) = 0
1/4-t+1=0
t=5/4

When t takes what value, there are two equal roots for the univariate quadratic equation x & # 178 / 4 + (1 / 2x + T) &# 178; = 1 of X

x²/4+（1/2x+t）²=1
It is reduced to (1 / 2) x & # 178; + TX + T & # 178; - 1 = 0
If the equation has two equal roots, the discriminant is
Δ=t²-4(1/2)(t²-1)=-t²+2=0
t=±√2

When k takes what value, the two roots of the quadratic equation kx2 + 2kx + K-3 = 0 with respect to X are negative?

kx^2+2kx+k-2=0
k(x^2+2x+1)-2=0
k(x+1)^2=2
(x+1)^2=2/k
Because (x + 1) ^ 2 & gt; 0, so 2 / K & gt; 0
x1*x2=-2/k<0,
So the two equations have different numbers

A formula for finding imaginary roots of quadratic equation of one variable

ax^2+bx+c=0
Δ=b^2-4ac
When Δ