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How to solve a difficult Algebra problem mentally in quick time 15

In a symmetric balanced three-term target expression concentrate on simplifying one term

How to solve a difficult SSC-CGL algebra problem mentally in quick time 15

In the problem we have chosen this time, the given and target expression with three terms both are perfectly symmetric and balanced between three variables $a$, $b$ and $c$. This property exists when interchanging any two variables in both the target and given expressions, the problem does not change.

In this situation, strategically we take first step as concentrating on one single term of the target expression with the goal of simplifying it in as few steps as possible.

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Let us take up process of solving the difficult algebra problem to substantiate the reasoning of problem solving.

Chosen Problem.

If $bc+ca+ab=abc$, then the value of $\displaystyle\frac{b+c}{bc(a-1)} + \displaystyle\frac{c+a}{ca(b-1)}+\displaystyle\frac{a+b}{ab(c-1)}$ is,

  1. $0$
  2. $1$
  3. $-\frac{1}{2}$
  4. $-\frac{3}{2}$

Before going ahead further, try to solve the problem yourself.

Problem analysis

The given expression as well as the target expression are fully symmetric and balanced between the three variables $a$, $b$ and $c$. Following is the property of a balanced symmetric expression,

In a symmetric expression involving three variables, if any two of the three variables are interchanged, the expression remains essentially unchanged.

In such a problem, especially when we find no way to combine the three terms of the target expression, the strategy to be followed is to take first step as,

Concentrate on a single term of the three term target expression and simplify it in as few steps as possible.

This is a pattern-method pair for moving towards the solution with assurance.

Thus, in the second stage, we examine the first term of the target expression, and with the help of the given expression, try to simplify it as quickly as possible. The first term here is,


In this algebraic fraction, the only way simplification can be achieved is to create a common factor between the numerator and denominator. It is deduction based on mathematical reasoning.

This objective can be achieved by transforming either the numerator or the denominator.

Now comes the second strategic action in play, that is effective in general,

While simplifying an algebraic fraction, simplify the denominator first, especially so if it is the more complex expression compared to the numerator.

The reason is obvious—the denominator in this case is more complex than the numerator, and stands in the way of combining the three terms towards a simple final result as given in the choice values. 

Examining the denominator, we ask, can we expand the denominator to,


and use the given expression on it with positive results?

At this point we see the lightly hidden useful pattern immediately,


Or, $abc-bc=ca+ab=a(b+c)$.

The factor $(b+c)$ cancels out leaving only $a$ in the denominator. Simplifying the other two terms in the same way, we get a much simplified target expression as,


Till now we carried out the manipulations in mind. Now also we combine the three simple inverse terms in mind and reach the result in quick time,



$=1$, using the given expression again.

Answer: Option b: $1$.

The analysis and strategies followed are automatic and quick. Following such pattern based strategies usually would give you quickest results, though we should mention an additional pattern based strategy that would have been equally effective here,

When a term of the target expression is to be simplified by using the given expression that involves three balanced two factor terms on the LHS and a three factor term on the RHS, we have to use simplifying expression in the form of, $abc-ca=ab+bc$.

We could have proceeded with this strategy instead, and reached the result in same number of steps.

Any other approach would be more time consuming.

Key concepts used: Problem analysis -- Key pattern identification -- Symmetric expression -- Strategic problem solving -- Deductive reasoning -- Denominator simplification -- Rich algebraic techniques.

The list of Difficult algebra problem solving in a few steps quickly is available at, Quick algebra.

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