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How to solve difficult SSC CGL algebra problems in a few steps 14

When no clear solution is visible, Reduce the number of variables by component expression substitution

How to solve difficult SSC CGL algebra problems in a few steps 14

Usually we solve difficult Algebra problems in a few steps using basic and rich Algebra concepts together with powerful general problem solving techniques visualizing the solution in the beginning itself. Occasionally for a problem the solution path is not visible till the end. For instance, the chosen problem in this session required more of deductive reasoning, mathematical reasoning, and exploratory application of core algebraic concepts such as symmetry in algebraic expressions, variable number reduction by component expression substitution and secondary term value sharing.

The highly effective general algebraic technique of variable number reduction by component expression substitution says,

If number of variables in an expression is large, reduce the number by appropriate substitution of a component expression involving more than one variable by a single dummy variable, simplify the resulting simpler expression and substitute back the expression substituted at the final stage (by reverse substitution).

Often new possible paths open up when the original expression is simplified by this component expression substitution, and its clutter reduced.

In this session we will present three ways to solve the problem.

In the first solution, we will highlight the reasoning and the concepts that went into solving the chosen problem by reducing the number of variables with substitution of a multi-variable component expression by a single variable and then breaking up and sharing the secondary resource term among the symmetric primary terms of the main expression.

In the second solution which is an advanced form of the first, instead of combining just the single $p$ with each primary term, we further breakup the remaining $p=a+b+c$ into three individual variables and combine $(p+a)$, $(p+b)$ and $(p+c)$ with the first, second and the third primary terms respectively on the LHS. We consider this as the most elegant solution.

Lastly, in the third solution, the rich problem will again be solved in a completely different approach using deductive reasoning along with the objective of denominator elimination. The reasons behind the decisions for specific actions will enhance the explanation of the solution process.

Overall, in all three solutions concepts and strategies will play prominent roles.

You may watch the video below,

Let us go through the problem solving processes in this interesting problem.

Chosen Problem.

If $\displaystyle\frac{x-a^2}{b+c} +\displaystyle\frac{x-b^2}{c+a} +\displaystyle\frac{x-c^2}{a+b}=4(a+b+c)$, with $a$ $b$, and $c$ positive real variables, value of $x$ is,

  1. $a^2+b^2+c^2$
  2. $ab +bc +ca$
  3. $a^2+b^2+c^2 - ab - bc - ca$
  4. $(a+b+c)^2$

Before going ahead further try to solve the problem yourself.

Problem analysis - first observation

The three primary terms on the LHS are perfectly symmetric with variable group $a$, $b$, $c$ present in the same manner in each of the terms. This is the hallmark of a symmetric expression.

In such a symmetric expression if one of the three variables is interchanged with another, the three term expression remains unchanged.

To simplify such a three term expression we look towards the RHS for a term (that is secondary in nature) which can be split into three equal parts and shared among the three terms for additive combining. Usually when we are able to take such a step, solution is visible immediately. But this problem is more complex with more number of hidden layers. With secondary term value sharing at this stage, instead of simplification the situation would turn out to be far too complex.

Identifying one of the main difficulties we are facing in this problem as large number of variables for clean immediate solution, we decide to substitute the expression $a+b+c$ by $p$, a single dummy variable. This step is Reduction in number of variables by component expression substitution.

If we are able to make such a substitution, the number of variables reduces, the target expression becomes simpler to deal with and new promising avenues become visible.

This is an application of Principle of representative variable where a dummy variable represents a component expression involving a group of variables.

As this action will never increase the complexity of the expression, rather will make both sides of the equation simpler, with no other promising path being available, we go ahead with the component expression substitution applying the principle of representative variable,


Or, $b+c=p-a$,

Or, $c+a=p-b$, and


Problem solving execution first step

With the substitution the given equation is transformed as,

$\displaystyle\frac{x-a^2}{p-a} +\displaystyle\frac{x-b^2}{p-b} +\displaystyle\frac{x-c^2}{p-c}=4p$.

Problem solving execution second step

With this simplified form of the equation now, we take the exploratory but often effective technique of splitting the secondary term $4p$ into four $p$s, and combine three of the $p$s with the three terms on the LHS, leaving only one $p$ on the RHS. As a result we get the transformed equation as,

$\left[\displaystyle\frac{x-a^2}{p-a} -p\right]+\left[\displaystyle\frac{x-b^2}{p-b}-p\right] +\left[\displaystyle\frac{x-c^2}{p-c}-p\right]=p$,

Let us simplify the first term on the LHS (as the terms are similar in the symmetric expression),

$\left[\displaystyle\frac{x-a^2}{p-a} -p\right]$



$=\displaystyle\frac{x-p^2}{p-a} +a$.

Converting the other two terms in the same way we will have the equation then as,

$\displaystyle\frac{x-p^2}{p-a}+\displaystyle\frac{x-p^2}{p-b}+\displaystyle\frac{x-p^2}{p-c}+(a+b+c)=(a+b+c)$, we make reverse substitution of $p$.

Or, $(x-p^2)\left[\displaystyle\frac{1}{b+c}+\displaystyle\frac{1}{c+a}+\displaystyle\frac{1}{a+b}\right]=0$,

This is a very promising result.

As the second factor cannot be zero,


Or, $x=(a+b+c)^2$.

Even after the second action of additive sharing of $p$ with each primary term on the LHS, final solution was not immediately visible. We needed to rearrange, pair up the terms in the numerator suitably and form a product of sums, that too for a part of the numerator.

Answer: Option d: $(a+b+c)^2$.

This is a mathematically sound solution based on problem analysis and application of promising techniques.

Key concepts used: Problem analysis -- Primary barrier identification -- Symmetric expression -- Deductive reasoning -- Variable reduction technique -- Component expression substitution -- Dummy variable -- Principle of representative variable -- Secondary resource sharing -- Denominator elimination -- Factorization -- Reverse substitution -- Basic algebraic concepts -- Rich algebraic techniques.

Note: Denominator elimination as an objective comes in much later in the solution process, and we share only a part of the secondary term resource, not the full.

Second alternate solution - improving the first solution

In this second solution, instead of additive combining of one $p$ with each primary term, we further break up the remaining $p=a+b+c$ on the RHS and combine additively $(p+a)$, $(p+b)$ and $(p+c)$ with the first, second and the third primary terms.This is two level secondary resource sharing. In this approach as the secondary resource is fully shared among the three primary terms leaving a 0 on the RHS, the positive effect will naturally be much more.

Let us show the result,

$\displaystyle\frac{x-a^2}{b+c} +\displaystyle\frac{x-b^2}{c+a} +\displaystyle\frac{x-c^2}{a+b}=4(a+b+c)$,

Or, $\displaystyle\frac{x-a^2}{p-a} -p+\displaystyle\frac{x-b^2}{p-b}-p +\displaystyle\frac{x-c^2}{p-c}-p=p$,

Or, $\displaystyle\frac{x-a^2}{p-a} -(p+a)+\displaystyle\frac{x-b^2}{p-b}-(p+b) +\displaystyle\frac{x-c^2}{p-c}-(p+c)=0$,

Or, $\displaystyle\frac{x-a^2-(p^2-a^2)}{p-a}+\displaystyle\frac{x-b^2-(p^2-b^2)}{p-b} +\displaystyle\frac{x-c^2-(p^2-c^2)}{p-c}=0$,

Or, $\displaystyle\frac{x-p^2}{p-a}+\displaystyle\frac{x-p^2}{p-b}+\displaystyle\frac{x-p^2}{p-c}=0$,

Or, $(x-p^2)\left[\displaystyle\frac{1}{b+c}+\displaystyle\frac{1}{c+a}+\displaystyle\frac{1}{a+b}\right]=0$.

As the second factor cannot be zero,


Or, $x=p^2=(a+b+c)^2$.

Key concepts used: Problem analysis -- Primary barrier identification -- Symmetric expression -- Deductive reasoning -- Variable reduction technique -- Component expression substitution -- Dummy variable -- Principle of representative variable -- Two level Secondary resource sharing -- Numerator equalization -- Factorization -- Reverse subsitution -- Basic algebraic concepts -- Rich algebraic techniques -- Many ways technique.

Note: This solution is naturally a more elegant and faster solution than the first.

Third alternate solution - problem analysis and solving

In this solution we take up the analysis with the generally effective objective of how to eliminate the denominators, that is, denominator elimination at the very beginning only.

To achieve this goal, say for the first term, $\displaystyle\frac{x-a^2}{b+c}$, we need a factor $(b+c)$ in the numerator, which we can imagine to be in the form $(a+b+c)-a$.

Continuing the deductive reasoning, we observe that with such a factor on the numerator, for generating the term $-a^2$ after multiplying with the second factor, a very convenient second factor will be, $(a+b+c)+a$, resulting in $(a+b+c)^2 -a^2$ after multiplication. This is mathematical reasoning based scenario analysis.

Thus if $x=(a+b+c)^2$, the denominators are eliminated conveniently generating he RHS term in LHS.

In this solution, we do not try out each choice value directly. That would have been far too time taking.

Note: We classify this beautiful problem to be comparatively more difficult, that required exploratory steps towards the solution without seeing clearly the solution at the start itself which happens for most of the other difficult problems. We understand the reason to be the solution hidden behind multiple opaque barriers.

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

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