The two sides of an equation contain both variable (unknown quantity) and constants (numerals). In such cases, we first simplify two sides in their simplest forms and then transpose (shift) terms containing variable on R.H.S. to L.H.S and constant :erms on L.H.S to R.H.S. By transposing a term from one side to the other side, we mean changing its sign and carrying it to the other side. In transposition the plus sign of the :er-m changes into minus sign on the other side and vice-versa.
The transposition method involves the following steps:
Step I Obtain the linear equation.
Step II Identify the variable (unknown quantity) and constants (numerals).
Step – III Simplify the L.H.S. and R.H.S. to their simplest forms by removing brackets.
Step – IV Transpose all terms containing variable on L.H.S. and constant terms on R.H.S. Note that the sign of the terms will change in shifting them from L.H.S. to R.H.S. and vice-versa.
Step V Simplify L.H.S. and R.H.S. in the simplest form so that each side contains just one term.
Step VI Solve the equation obtained in step V by dividing both sides by the coefficient of the variable on L.H.S.
Example : \( (2x+3)^2 +(2x-3)^2 =(8x+6)(x-1)+ 22 \)
Solution : \( (2x + 3)^2 + (2x- 3)^2 = (8x +6) (x-1) + 22 \)
= \( 2{(2x)^2 +3^2}=x(8x+6)-(8x+6)+ 22 \)
= \( 2(4x^2+9) 8x^2+6x-8x – 6+22 \)
= \( 8x^2+18 = 8x^2-2x + 16 \)
= \( 8x^2-8x^2+2x= 16-18 \)
= \( 2x = -2 \)
= \( x = – 1 \)