What are solutions of simultaneous equations 1/x=(y-1)/y 3/x=(2y-1)/y
2 Answers
justaguide | College Teacher | (Level 2) Distinguished Educator
We have to solve the equations 1/x=(y-1)/y and 3/x=(2y-1)/y
1/x=(y-1)/y
=> x = y/(y - 1)
Substitute x = y/ (y - 1) in 3/x=(2y-1)/y
=> 3/x=(2y-1)/y
=> 3 /[y/ (y - 1)] = (2y-1)/y
=> 3(y - 1)/y = (2y - 1)/y
=> 3y - 3 = 2y - 1
=> y = 2
x = 2/(2 - 1) = 2
Therefore we get x = 2 and y = 2
giorgiana1976 | College Teacher | (Level 3) Valedictorian
We'll re-write the given equations by cross multiplying:
y = x(y-1)
We'll remove the brackets:
y = xy - x
We'll add x both sides:
x + y = xy (1)
We'll change the 2nd equation in:
3y = x(2y-1)
3y = 2xy - x
We'll add x both sides:
3y + x = 2xy (2)
We'll substitute (1) in (2):
3y + x = 2(x+y)
3y + x = 2x + 2y
3y - 2y = 2x - x
y = x (3)
We'll substitute (3) in (2)
3x + x = 2x^2
4x = 2x^2
We'll divide by 2 and we'll use the symmetric property:
x^2 - 2x = 0
We'll factorize by x:
x(x-2) = 0
x = 0
x-2 = 0
x = 2
So, because x = y => y = 2
The solution of the system is {2 ; 2}.