## Use basic and rich Algebra concepts to solve tough problems of SSC CGL in a few simple steps

The problems involving Algebra are taught in schools and also form an important part of many of the competitive job tests SSC CGL and the likes.

Algebra is one of the most useful disciplines in Maths as a whole. It is used in Number system problems, Trigonometric problems and other topic areas. Its use is not limited to real number domain, it is used as well in dealing with problems involving imaginary numbers.

Its widespread use originates from its **inherent abstraction** in defining and dealing with its component elements.

### Component elements in Algebra

The smallest component elements in algebra are the **variables** and the **constant coefficients.** These two components combine to form **algebraic terms **and the terms combine to form the **algebraic expressions.** The combining process happens through the use of **usual mathematical operators.**

*When an expression has an equality operator we get an equation.*

Let us present an example.

$ax^2 + bx + c$ is an algebraic expression whereas, $ax^2 + bx + c = 0$ is a frequently encountered algebraic equation.

In this equation, we define $x$ to be the variable and $a$ and $b$ to be the constant coefficients of the terms $ax^2$ and $bx$, the third $c$ being an independent coefficient and term.

When we define $x$ to be the variable, it means, we don't know its exact value, its value can vary within a range or sometimes the variable can take up any range of values, or for that matter may not have any definable value even. A variable in an expression **may have real values** or **imaginary values.**

The **nature of the expression** is determined by the maximum power of its variables. In this specific example as above, the **expression is a quadratic** one in **single variable** $x$. If power of all variables is unity, the equation is classified as a **linear equation.**

### All arithmetic rules apply to algebraic expression manipulation

This is inherent in the definition of the discipline and its component elements.

The basic concepts of addition, subtraction, multiplication and division along with the concepts of fraction manipulation, factorization, LCM, HCF, Indices are all applicable to algebraic expressions and equations.

Similarly the other rules such as associative rules and the such that are elaborated in our article on * Number System* are also applicable to Algebraic manipulations.

### Mathematical modeling

Introduction of the concept of variables, value of which are not known in the beginning of a large scale real life analysis, further leads to formation of relationships between important and relevant variables expressed in the form of expressions, equalities and inequalities and then on to analysis of these to finally arrive at useful results.

This whole process is done by forming first the **mathematical model representing close to real life situation** and then analyzing the model using plethora of tools, techniques and strategies that are well known and not so well known.

**Mathematical Modeling of most of the important real life problems though is infeasible. **

### An example of a simple model

#### Problem example 1

Khokon bought 5 oranges and 5 lemons at a total cost of Rs. 80, and 7 days later bought 3 oranges and 4 lemons at a total cost Rs. 50. If the prices remained unchanged in the two purchases, what was the per piece price of orange in Rs.?

**Standard Solution**

Let us assume per piece price of organge be $a$ and that of lemon be $b$.

These assumptions enable us to express the first purchase transaction in the form of the equation,

$5\times{a} + 5\times{b} = 80$.

And the second transaction as,

$3\times{a} + 4\times{b} = 50$.

This is the **simple mathematical model** *representing the two transactions.*

The process of **solution for values of two variables in two linear equations** is well known and easily understood. We need to eliminate one of the variables by **first transforming** one equation and **then subtracting the transformed one from the other.**

We will transform, let us say the first equation of our problem, with the target to eliminate variable $b$ (as we need to know the value of the other variable $a$ to get the answer). To do this * we use the useful property of an equation* that,

You can multiply, divide, add, and subtract any value to (or from) the two expressions on both sides of an equation without changing the nature of the equation.

Here, as we want to make the coefficient of variable $b$ equal to $4$ (so that $b$ cancels out when we subtract one equation from the other after transformation), we multiply the first equation (effectively each term of the first expression) by $\displaystyle\frac{4}{5}$. This transforms the first equation to,

$4a + 4b = 64$.

Subtracting the second equation from this transformed equation we have the desired answer of the per piece price of oranges in Rs. as,

$a = 14$.

#### General rule of solvability of a system of linear equations

Linear equations means all variables in a set of equations have only power of unity.

The **solvability rule** says,

You can solve and get values of all of the $n$ variables if you have

at least $n$number ofunique equationsin $n$ variables.

In simpler terms, to solve for values of three variables you need at least three unique linear equations in terms of the three variables, for two you need at least two equations and so on.

**Unique equations** means, you must not be able to derive any equation from any other in the set of equations being evaluated. For example,

$4a + 12b = 24$ and $a + 3b = 6$ are **not two unique equations** as, just by multiplying the second equation by $4$ you get the first.

### Focal point is efficient Algebra problem solving

Though it starts simple, algebra is large and complex. We won't cover any more of equations, inequations, binomial theorems and such other concepts of algebra.

Our **main focus in this session** will be how to use the most **basic concepts** and the **rich derived concepts** along with powerful general problem solving strategies and techniques to solve algebraic problems involving expression manipulations in least number of steps and time.

In other words, we will focus on **elegant solutions of awkward looking algebraic problems that are relevant mainly to competitive tests.**

Nevertheless the concepts that follow *should be highly effective i n nurturing the problem solving abilities in the student's mind by focusing on reaching the solution quickly using analytical abilities in contrast to going through long series of manipulative routine steps wasting valuable time and energy. This applies to school students as well.*

### Causes behind difficulties with Algebra

**First :** Though the topic of Algebra is based on a small set of concepts, dealing with **abstract symbolic variables poses first level of difficulty** in forming the problem definition in comparison to problems involving numbers.

**Second:** Furthermore, based on the small number of building blocks of concepts, **Algebraic expressions can be made to appear very complex** posing **the second level of difficulty** to the student in understanding and solving such problems.

**Third:** *Identification of*** useful patterns in complex **** algebraic expressions is an essential skill** in solving Algebra problems, and, not everyone is comfortable in recognizing useful patterns which is not easily visible. This forms

**the third level of difficulty**with problems on Algebra in general.

Some of these **important and useful patterns** may systematically be compiled into a **set of rich concepts** on top of the basic concepts, for solving such problems easily in future.

**Fourth:** Knowledge of basic and rich concepts is though not enough - one needs to have **basic problem solving skills** using powerful general **problem solving strategies and techniques.**

**Fifth:** Lastly, **it is a fact of life and Algebra** that even after intensive skill enhancing practices, you **may** **face a totally new type of problem** in the final performing stage. You need to **be able to bridge the final gap**, as we say, to be in full control of any domain, in this case, Algebra.

**Concept layers**

Like any subject or activity area, the important and useful concepts fall in two layers - **the basic concept layer** that everyone knows, and the **rich concept layer** that is derived from the basic concepts and **actual problem solving experience.**

When a new rich concept is discovered in the process of solving a new type of problem in a few steps elegantly, it is **added to the rich concept layer incrementally and continually. ***Thus this rich concept layer is an extensible open ended layer of concepts.*

The **problem solving concept **and** process layer ***uses these subject concept layers to solve any problem* in the domain efficiently.

Remember, **it is not problem solving any which way** to us - *it is efficient problem solving **at low cost (in this case time) *.

**Basic algebra concepts**

The **basic operations** involved in Algebra are none other than all the **basic arithmetic operations**, but on abstract symbolic variables and expressions, not on numbers.

The more important **relationships in Algebra forming the basic concept layer of Algebra** are the following.

$(a + b)^2 = a^2 + 2ab + b^2$, in the form of square of sum of two variables.

$(a - b)^2 = a^2 - 2ab + b^2$, negative counterpart of square of sum of two variables.

$a^2 - b^2 = (a + b)(a - b)$, this is one of the most useful algebraic relationships.

$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$, in the form of cube of sum of two variables.

$(a + b)^3 = a^3 + b^3 + 3ab(a + b)$, this cube form is used frequently and it is better to remember it as a basic concept.

$(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$, negative counterpart of cube of sum of two variables.

$(a - b)^3 = a^3 - b^3 - 3ab(a - b)$, which again is useful.

**Derived from the cubes of sums** we get the **next set of basic relationships that are frequently used**.

$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$, a very useful relationship to be remembered and used in the right places.

$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$, negative counterpart and equally useful.

Similarly it pays to know the **similar expressions in three variables**.

$(a + b + c)^2 $

$\hspace{1mm} = a^2 + b^2 + c^2 + 2(ab + bc + ca)$,

$(a + b + c)^3 $

$\hspace{1mm} = a^3 + b^3 + c^3 + 3(a + b)(b + c)(c +a)$,

$a^3 + b^3 + c^3 - 3abc $

$\hspace{1mm} = (a + b + c) \times{}$

$\hspace{10mm} (a^2 + b^2 + c^2 - ab - bc - ca)$.

At the end, we would present * Sreedhar Acharya's formula for finding the two roots* of the genral quadratic equation,

$ax^2 + bx + c = 0$, where $a$, $b$ and $c$ are the two coefficients and the constant and $x$ is the variable.

The two roots are governed by the formula,

$\displaystyle\frac{-b \pm{\sqrt{b^2 - 4ac}}}{2a}$.

If $b^2 \lt {4ac}$, then the roots of $x$ are imaginary.

### Rich algebra concepts

**1. Three variable zero sum principle**

The rich concept of **three variable zero sum principle is,**

If $a + b + c = 0$, then $a^3 + b^3 + c^3 = 3abc$ which is a very useful concept.

**Aside:**

*Proof of three variable zero sum principle:*

*$a + b + c = 0$,*

*$(a + b + c)^3 $*

*$ = (a + b)^3 + $*

*$\hspace{10mm} 3(a + b)c(a + b + c) + c^3$*

*$ = a^3 + 3ab(a + b) + b^3 + c^3$, $a + b + c = 0$ eliminates second term of the last expression.*

*$ = a^3 - 3abc + b^3 + c^3 = 0$, again from $a + b + c = 0$, we use, $a + b = -c$,*

*Or, $a^3 + b^3 + c^3 = 3abc$, a compact and highly useful result.*

*To identify this highly useful result we have given the concept a name of Three variable zero sum principle and included it in our rich concept layer.*

You may refer to details of application of this valuable rich concept in our **second algebra simplification session.**

#### Second: Chained equation treatment technique

When we meet a chained equation of the form, $a^x = b^y = c^z$, which we call as a chained equation, a strategy of adding another artificial equality proves generally to be very useful as it frees each expression in the chained equation and allows independent pairwise comparison and useful result derivation.

The new equality is introduced in the form,

$a^x = b^y = c^z = q$, from which we can form independent equations, $a^x = q$, $b^y = q$ and $c^z = q$, all in terms of $q$.

You may refer the effective use of this technique in our **first algebra simplification session***.*

**Third: Base equalization technique**

We have found application of this profound technique in widely different areas, of Maths and in real life problems as well.

You may explore its use in * indices problems,* in

*,*

**our first algebra simplification session***and in*

**fraction problems**

**diverse areas**.**Fourth: Principle of inverses**

This is another very useful concept set. You may refer to its detailed treatment in our **session on inverses.**

Briefly one of the useful results of principle of inverses in Algebra is,

If $x + \displaystyle\frac{1}{x} = n$, where $n$ usually is a suitable positive integer, we can always derive similar expressions in sum of inverses for powers 2, 3 and beyond. The basic advantage with this type of expressions of inverses results from the variable $x$ disappearing when the two inverses are multiplied together, $x\times{\displaystyle\frac{1}{x}} = 1$.

**Example problem:**

If $x + \displaystyle\frac{1}{x} = 2$, find $x^3 + \displaystyle\frac{1}{x^3}$.

**Solution:**

$x + \displaystyle\frac{1}{x} = 2$,

Squaring we get,

$x^2 + \displaystyle\frac{1}{x^2} + 2 = 4$,

Or, $x^2 + \displaystyle\frac{1}{x^2} = 2$,

So, $x^3 + \displaystyle\frac{1}{x^3}$

$\hspace{5mm}= \left(x + \displaystyle\frac{1}{x}\right)\left(x^2 - 1 + \displaystyle\frac{1}{x^2}\right)$

$\hspace{5mm}= 2$.

The *principle of inverses can also be applied in solving real life problems.*

#### Fifth: Substitution technique or principle of representative

When in complex expressions we observe **some of the component expressions appear unchanged throughout the problem**, our first action will always be to **replace these more complex component expressions by single variables**.

This immediately **simplifies the clutter** and** visual complexity** of the problem, and **brings into focus commonly used results** that were not visible earlier due to the clutter.

This again is a **general problem solving principle** and is applied unknowingly *in many diverse real life problem areas*.

You may refer to the use of this powerful technique in our * second algebra simplification session *as well as

**the third algebra simplification session.**#### Sixth: Principle of collection of friendly terms

This is a very useful technique applied especially in Algebra with great effectiveness. It simply says,

In a complex algebraic expression with many individual terms, collect together the terms in small groups, so that each group can take up a new meaningful existence, thus simplifying the whole expression in one glorious sweep.

We have already highlighted the use of this concept * in the first algebra simplification session *as well as the

**second algebra simplification session.****Seventh:** Input transformation technique

Though this sounds simple, many tricky problems can be elegantly solved **by transforming the given expression in well known symmetric forms** that bear **direct relations to the end state expression.**

You may refer to the effective use of this technique in our * second algebra simplification session *as well as the

**third algebra simplification session.**#### Eighth: Principle of zero sum of square terms

This principle is based on fundamental mathematical principles. It says,

If,

$a^2 + b^2 + c^2 = 0$, and $a$, $b$ and $c$ are real, then, $a=b=c=0$.

On many occasions, use of this basic principle is the only way you can solve the problem quickly.

We presented an example of effective use of this valuable rich concept in our * second Algebra simplification session*.

#### Ninth: Use of a new factor technique

By this technique, an existing expression is multiplied with a new factor thus simplifying the whole expression.

For example, when we encounter the expression, $a^2 - ab + b^2$ we might like to multiply it by $a + b$ to simplify the product to $a^3 + b^3$.

You may refer to the perfect use of this technique in our **third algebra simplification session.**

**Tenth: Use of delayed evaluation technique**

In simplification of algebraic expression a very useful technique to use is the delayed evaluation technique. It says,

Delay evaluation of especially long time-consuming products till the last stage possible.

More often than not, at the last stage, the time-consuming product itself may not be required to be done.

**Example problem:**

If $a \gt 1$ and $a + \displaystyle\frac{1}{a} = 2\frac{1}{12}$, then the value of $a^4 - \displaystyle\frac{1}{a^4}$ is,

- $\frac{57895}{20736}$
- $\frac{57985}{20736}$
- $\frac{59825}{20736}$
- $\frac{58975}{20736}$

#### Final hurdle

We know $\left(a + \displaystyle\frac{1}{a}\right) = \displaystyle\frac{25}{12}$ and $\left(a - \displaystyle\frac{1}{a}\right) = \displaystyle\frac{7}{12}$.

So multiplying the two,

$\left(a^2 - \displaystyle\frac{1}{a^2}\right)=\displaystyle\frac{25\times{7}}{12^2}$.

Again we know,

$\left(a^2 + \displaystyle\frac{1}{a^2}\right)=\displaystyle\frac{337}{12^2}$.

Multiplying the two we get finally,

$\left(a^4 - \displaystyle\frac{1}{a^4}\right) = \displaystyle\frac{337\times{25}\times{7}}{12^4}$

#### Applying free resource use principle

In the beginning itself we have looked at the large choice values and decided to use whatever technique we know to avoid large multiplication. That is why we kept the factors, 7, 25 and 337 undisturbed without multiplying them.

First we test the numerators of choice values for factor 25 and quickly eliminate option a and b.

Next we test the two values of numerators of options c and d, that is, 59825 and 58975 for divisibility by 7. Either you apply the 7 divisibility technique or direct mental division here. We preferred the second approach and found 59825 not a multiple of 7 but 58975 to be a perfect multiple of 7. So the choice is option d.

Notice that we have ignored calculating $12^4$ altogether because all the denominators of four choices are same. If you actually calculate $12^4$, it will be equal to $20736$.

**Answer:** Option d: $\displaystyle\frac{58975}{20736}$.

Application of powerful * free resource use principle* made this quick elegant solution possible.

You may refer to the detailed treatment in **How to solve difficult algebra problems in a few simple steps 5.**

For effectively using these concepts in solving any algebra problem elegantly, it is necessary to use the powerful * general problem solving strategies and techniques* such as

*,*

**End state analysis***,*

**Pattern recognition***,*

**Solving a simpler problem***and the likes along with the subject concepts.*

**Problem breakdown technique**### Guided help on Algebra in Suresolv

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