Question

Solve for x and y: $ 6(ax + by) = 3a + 2b; $ $ 6(bx - ay) = 3b - 2a $

Answer 1

Hint: In the given two equations, first of all convert the coefficient any of the variable common in both the equations and then use addition or subtraction as applicable to remove one of the variables by simplification of the equations and find the values of the variables one by one.

Complete step-by-step answer:
Take the given expressions:
 $ 6(ax + by) = 3a + 2b $
Open the brackets multiplying the term outside the bracket with the terms inside the bracket and also multiply “a” on both the sides of the equation –
 $ 6{a^2}x + 6aby = 3{a^2} + 2ab $ ….. (A)
 $ 6(bx - ay) = 3b - 2a $
Open the brackets multiplying the term outside the bracket with the terms inside the bracket –
 $ 6bx - 6ay = 3b - 2a $
Multiply the above expression with “b”
 $ 6{b^2}x - 6aby = 3{b^2} - 2ab $ ….. (B)
Add equation (A) and the equation (B) –
 $ 6{a^2}x + 6aby + 6{b^2}x - 6aby = 3{b^2} - 2ab + 3{a^2} + 2ab $
Combine the like terms in the above expression –
\[6{a^2}x + 6{b^2}x - 6aby + 6aby = 3{b^2} + 3{a^2} + 2ab - 2ab\]
Like terms with the same value and the opposite sign cancels each other.
\[6{a^2}x + 6{b^2}x = 3{b^2} + 3{a^2}\]
Take common multiple common on the right hand side of the equation –
\[6x({a^2} + {b^2}) = 3({b^2} + {a^2})\]
Make the required term, “x” the subject - term multiplicative on one side if moved to the opposite side then it goes to the denominator.
\[x = \dfrac{{3({b^2} + {a^2})}}{{6({a^2} + {b^2})}}\]
Common multiples from the numerator and the denominator cancel each other.
 $ \Rightarrow x = \dfrac{1}{2} $ ….. (C)
Place above value in the equation (A)
 $ 6{a^2}\dfrac{1}{2} + 6aby = 3{a^2} + 2ab $
Simplify the above equation considering that common multiples from the numerator and the denominator cancel each other.
 $ 3{a^2} + 6aby = 3{a^2} + 2ab $
Same term with the same value and sign cancels each other on both the sides of the equation.
 $ 6aby = + 2ab $
Make “y” the subject –
 $ y = \dfrac{{2ab}}{{6ab}} $
Common factors from the numerator and the denominator cancel each other.
 $ y = \dfrac{1}{3} $ …. (D)
Hence, the values are $ x = \dfrac{1}{2} $ and $ y = \dfrac{1}{3} $
So, the correct answer is “ $ x = \dfrac{1}{2} $ and $ y = \dfrac{1}{3} $ ”.

Note: Be careful about the sign convention while simplifying the equations. When the terms are on one side of the equation then like terms with the opposite sign and same values cancel each other and if the terms are in the opposite direction, then the terms with the same sign cancel each other.