Solving equation x / 15-2 = x / 30 + 10 / 60

x/15 - 2 = x/30 + 10/60
Multiply 60 on both sides of the equation to get:
4x - 120 = 2x + 10
4x - 2x = 10 + 120
2x = 130
x = 60

Known about X, y binary linear equations!
Solve the binary linear equations about X, Y!
It is known that the solutions of a ′ x + B ′ y = C ′, a ′ x + B ′ y = C ′ are x = 3, y = 4. The solutions of 3A ′ x + 2B ′ y = 5C ′, 3a ″ x + 2B ″ y = 5C ″ are obtained
Three students put forward their own views on this problem. A said: "it seems that the conditions of this problem are not enough." B said: "their coefficients have certain rules, you can try them." C said: "can you divide both sides of the two equations of the two equations by 5, and solve them by substitution." referring to their discussion, what do you think the solution of this problem should be?
Details need to be clear! Five more points!

It can be solved. Divide both sides of the second system of equations by 5 to form a '(3x / 5) + B' (2Y / 5) = C ', a ″ (3x / 5) + B ″ (2Y / 5) = C ″. If 3x / 5 and 2Y / 5 are regarded as X and y of the first system of equations, they have the same form as the first system of equations. Then 3x / 5 can be obtained according to the solution of the first system of equations x = 3, y = 4

System of linear equations of two variables with x = 1 and y = 1

When x ≥ 1, y = X-1 + X + 1 = 2x ≥ 2
When - 1

The system of quadratic equations of two variables: Y / 3 - (x + 1) / 6 = 31, 2 (X-Y / 2) = 3 (x + Y / 18) 2

X = - 7, y = 6. Specifically, the equation is simplified, the two equations are added, subtracted and cancelled, and then only one unknown number is left to solve the equation

Solving equation x / 15-2 = x / 30 + 10 / 60