Solving Simultaneous Equations: Finding ( x ) and ( y ) in ( xy a ) and ( x - y b )

When faced with a pair of simultaneous equations, such as ( xy a ) and ( x - y b ), several methods can be used to find the values of ( x ) and ( y ). Here we discuss the step-by-step process, including methods of substitution and elimination, and provide examples for better clarity.

Method of Elimination

Let's consider the system of equations:

( xy 6 ) - Equation 1 ( x - y 2 ) - Equation 2

To solve for ( x ) and ( y ), we can use the method of elimination. First, let's add both equations together:

( xy (x - y) 6 2 ) ( xy x - y 8 ) ( x(y 1) 8 )

However, a simpler approach would be:

Add Equation 1 and Equation 2 directly: ( (xy) (x - y) 6 2 ) ( x y x - y 8 ) ( 2x 8 ) ( x 4 )

With ( x 4 ), substitute back into Equation 2 to find ( y ):

( 4 - y 2 ) ( y 4 - 2 ) ( y 2 )

So the values are:

( x 4 ) and ( y 2 ).

Using Algebraic Manipulation

Sometimes, we can manipulate the equations algebraically to simplify the process:

( xy 6 ) - Equation 1 ( x - y 2 ) - Equation 2

Add the equations to get:

( xy x - y 8 ) ( x(y 1) 8 ) ( x frac{8}{y 1} )

By substituting back, we can solve for ( x ) and ( y ):

( x - y 2 ) ( x y 2 ) ( (y 2)y 6 ) ( y^2 2y - 6 0 ) ( y frac{-2 pm sqrt{4 24}}{2} ) ( y frac{-2 pm 4.9}{2} ) ( y 2 ) or ( y -3.45 )

{Note: The simpler solution ( y 2 ) is used for this example, and it fits the context of the problem.}

Thus, ( x 4 ) and ( y 2 ).

In another example, consider:

( xy 8 ) ( x - y 2 )

Adding the equations:

( xy (x - y) 10 ) ( 2x 10 ) ( x 5 )

Substituting ( x 5 ) back into Equation 1:

( 5y 8 ) ( y frac{8}{5} 1.6 ) ( y 3 ) (as an approximation)

So, ( x 5 ) and ( y 3 ).

Conclusion

By utilizing the methods of elimination and substitution, we can effectively solve a system of simultaneous equations involving multiplication and subtraction. The step-by-step approach provides clarity and helps in systematically finding the values of the unknown variables.