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Solve: $2x - 5\left[ {7 - \left( {x - 6} \right) + 3x} \right] - 28 = 39$

Answer
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Hint:The value of x in $2x - 5\left[ {7 - \left( {x - 6} \right) + 3x} \right] - 28 = 39$ can be found by using the method of transposition. We will first open the brackets and simplify the expression using the BODMAS rule. Then, we will shift the terms in the equation and do the required calculations in order to find the value of x.

Complete step by step answer:
So, we have, $2x - 5\left[ {7 - \left( {x - 6} \right) + 3x} \right] - 28 = 39$
Firstly, we simplify the equation using the BODMAS rule and simplify the expression by carrying out the operations in the order.
So, we get,
$ \Rightarrow 2x - 5\left[ {7 - x + 6 + 3x} \right] - 28 = 39$
Adding up the like terms, we get,
$ \Rightarrow 2x - 5\left[ {13 + 2x} \right] - 28 = 39$
Opening the brackets, we get,
$ \Rightarrow 2x - 65 - 10x - 28 = 39$

Keeping the x terms in left side of equation and shifting all the constant terms to right side of equation, we get,
$ \Rightarrow 2x - 10x = 39 + 65 + 28$
Remember to reverse the sign of the terms while shifting the terms in the equation.
Now, we add up the like terms.
$ \Rightarrow - 8x = 132$
Dividing both the sides of equation by $ - 8$, we get,
$ \Rightarrow x = \dfrac{{132}}{{ - 8}}$
Cancelling the common factors in numerator and denominator, we get,
$ \Rightarrow x = - \dfrac{{33}}{2}$
Expressing the answer in decimal, we get,
$ \therefore x = - 16.5$

Hence, the value of x in $2x - 5\left[ {7 - \left( {x - 6} \right) + 3x} \right] - 28 = 39$ is $ - 16.5$.

Note:Method of transposition involves doing the exact same mathematical thing on both sides of an equation with the aim of simplification in mind. This method can be used to solve various algebraic equations like the one given in question with ease. If we add, subtract, multiply or divide by the same number on both sides of a given algebraic equation, then both sides will remain equal. We must give the final answer in the simplest form by cancelling the common factor in numerator and denominator.
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