solve the equation
3y square-XY=0 equation(1)
2X-5Y=1 eqn (2)
To solve the given equations, we can use the method of substitution or elimination. Let's use the method of substitution:
Step 1: Solve equation (2) for X in terms of Y.
2X - 5Y = 1
2X = 5Y + 1
X = (5Y + 1)/2
Step 2: Substitute the value of X from equation (2) into equation (1).
3y^2 - XY = 0
3y^2 - (5Y + 1)/2 * Y = 0
Multiply both sides by 2 to eliminate the fraction:
6y^2 - (5Y + 1) * Y = 0
6y^2 - 5Y^2 - Y = 0
Rearrange the equation:
6y^2 - 5Y^2 - Y = 0
Step 3: Rearrange the equation to a standard quadratic form:
6y^2 - 5Y^2 - Y = 0
6y^2 - 5Y^2 - Y + 0 = 0
Step 4: Combine like terms and set equal to zero:
y^2 - Y = 0
Step 5: Factorize the equation:
y(y - 1) = 0
Step 6: Solve for y:
y = 0 or y - 1 = 0
y = 0 or y = 1
Step 7: Substitute the values of y back into equation (2) to find the corresponding values of X:
For y = 0:
2X - 5(0) = 1
2X = 1
X = 1/2
For y = 1:
2X - 5(1) = 1
2X - 5 = 1
2X = 6
X = 3
Therefore, the solutions to the given system of equations are:
X = 1/2, y = 0
X = 3, y = 1