A problem about quadratic equation of one variable When k takes what value, the system of equations: x-y-k = 0 and X * x-8y = 0, has a real solution? And the solution of the system of equations is obtained. (the second equation is the square of x minus 8y)

x-y-k=0
y=x-k
Substituting X & sup2; - 8y = 0
x²-8x+8k=0
If there is a solution, the discriminant is equal to 0
64-32k=0
k=2
x²-8x+16=0
(x-4)²=0
x=4
y=x-k=2

A simple exercise on quadratic equation of one variable
It is known that the value of the discriminant of the root of the quadratic equation MX & sup2; - (3m-1) x + 2m-1 = 0 is 1. Find the value of M and the root of the equation
PS: please write down the process of solving the problem

b²-4ac=（3m-1）²-4m（2m-1）=1
m²-2m=0
M1 = 2, M2 = 0 (rounding off)
When m = 2, the original equation is: 2x & sup2; - 5x + 3 = 0
（2x-3）（x-1）=0
x1=3/2 x2=1

A and B are solving the equations 3x + by = 5
When AX = by = 1, a misinterprets A and gets x = 3, y = 2. B writes B in one of the equations as an opposite number and gets x = 1, y = - 1. Find the values of a and B
The equations are 3x + by = 5
Ax + by = 1, wrong number on it

A only misread a, so x = 3, y = 2 conform to: 3x + by = 5, that is, 3 * 3 + b * 2 = 5, B = - 2
Substituting B = - 2, x = 1, y = - 1 into the equation 3x + by = 5, it is found that the left side = 3 * 1-2 * (- 1) = 5 = the right side, so B misread the equation: ax + by = 5, and the misread equation becomes: ax + 2Y = 5
a*1+2*(-1)=1,a=3
So a = 3, B = - 2

Solving quadratic equations of one variable
To solve the system of quadratic equations of one variable - x + X + Half x + quarter x = 2Y + 3
Two Y + y + Half y + quarter y = 4x-12
I'm tired of solving this system of equations
To solve the system of quadratic equations of one variable x + X + Half x + quarter x = 2Y + 3
Y + y + Half y + quarter y = 4x-12
I'm tired of solving this system of equations

I don't understand, is this: ① x + X + X / 2 + X / 4 = 2Y + 3 ② y + y + Y / 2 + Y / 4 = 4x-12 = = > ① both sides with * 4 get ③ 4x + 4x + 2x + x = 8y + 1211x = 8y + 12 ② both sides with * 4 get ④ 11y = 16x-48 substitute ③ get: x = (8y + 12) / 11 ④ get 11y = 16 * (8y + 12) / 11-48 both sides with * 11 get: 121y = 128y + 192-528

A problem about quadratic equation of one variable When k takes what value, the system of equations: x-y-k = 0 and X * x-8y = 0, has a real solution? And the solution of the system of equations is obtained. (the second equation is the square of x minus 8y)