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

In our previous article ** Basic and rich Algebra concepts for elegant solutions of SSC CGL problems** we have explained the basic and rich algebra concepts along with powerful problem solving strategies and techniques to solve tough problems quickly and surely in a few simple steps. In addition to the basic concepts,

**ten numbers of rich concepts**for efficient algebraic manipulation have been covered.

In this article **more such rich algebraic concepts** are explained with problem solving examples to bring out the power of each concept and technique in simplifying the solution of complex algebra problems.

### More Rich algebra concepts and techniques

**11. Simplification technique**

Though practically all concepts and techniques in Algebra are meant for simplification of complex algebraic expressions, by this generic name we state a really simple but often ignored step in any simplification process. It states simply,

If you find an opportunity to simplify a complex expression,

simplify it immediatelywithout any delay.

This seems obvious, but it simplifies the expression preparing the ground for applying more sophisticated techniques. We will highlight the usefulness of the concept through an example problem. Our recommendation, never miss a step of simplification by **using the most basic algebraic concepts.**

**Example Problem 1.**

If $x + \displaystyle\frac{1}{x} = 5$, then the value of $\displaystyle\frac{x^4 + 3x^3 + 5x^2 + 3x + 1}{x^4 + 1}$ is,

- $\frac{47}{21}$
- $\frac{41}{23}$
- $\frac{43}{23}$
- $\frac{45}{21}$

**Solution: first stage objective**

Though the given expression is an inverse sum, looking at the **assymmetric fraction of large target expressions** we decide not to evaluate inverse sums in other suitable powers of $x$.

Rather, we focus our attention on the large fraction in the target expression for simplifying. This is our first objective.

**Note:** It is important to note that we have reached this first objective by examining the end state or the target expression comparing it with the beginning state or the given expression by applying the most frequently used **End State analysis approach, a powerful general problem solving concept.**

When we compare the denominator with the numerator, we identify that the numerator has an expression same as the denominator, that is, $x^4 +1$, though a bit hidden away, the first term in the beginning and the second term at the end of the expression. But we have detected its presence by using our * pattern recognition skill*, an essential skill in any type of problem solving.

We take this opportunity to immediately simplify by using the **most basic concept of dividing part of the numerator by the denominator.** This is * simplification technique* at work.

Thus we have the target expression,

$E = 1 + \displaystyle\frac{3x^3 + 5x^2 + 3x}{x^4 + 1} = 1 + \displaystyle\frac{E_1}{E_2}$, the fractional parts are given names.

This is a significant improvement and will make further simplification much easier.

From a complex fraction of $\displaystyle\frac{x^4 + 3x^3 + 5x^2 + 3x + 1}{x^4 + 1}$, the fraction to be evaluated has been simplified to $\displaystyle\frac{3x^3 + 5x^2 + 3x}{x^4 + 1}$ in which **the critical improvement has been elimination of the $x^4$ term and the numeric digit $1$ from the numerator** with a reduction of number of terms from 5 to 3.

We'll see shortly in the same example how use of this simple step **prepared the ground for application of more sophisticated algebraic techniques** towards the final solution.

**Recommendation**

Never miss an opportunity to simplify an expression using the most basic mathematical concepts. That step will enable you to take more powerful steps towards solution later. This is as true in Algebra as in real life problem solving.

#### 12. Many ways technique

This is a general problem solving and learning technique and we will use this technique in Algebra now with great positive results. The problem example used is the same and we will start from where we left in explaining the previous technique.

Here we will use the technique stated as,

If the most obvious approach does not seem to produce immediate results, explore any other approach available and use the approach to see if it breaks the immediate bottleneck.

In short, don't be locked in the habit of using the most obvious approach, explore to see if any other approach is available and use it.

Again, this concept seems to be obvious, but when applied its value would be clear to you.

** Application of the technique**

We have simplified the target expression in the previous problem as,

$E = 1 + \displaystyle\frac{3x^3 + 5x^2 + 3x}{x^4 + 1}$ with the given expression as,

$x + \displaystyle\frac{1}{x} = 5$.

In most cases, we would use the given inverse sum along the lines of * principle of inverses* and will consider evaluating the required inverse sums of squares or cubes. But in this case

**examining the large fraction of the target expression**we do not find any immediate use of the more frequently used form of

**principle of inverses.**

Being aware of the many ways technique, we choose instead to **transform the given expression in its second form,**

$x + \displaystyle\frac{1}{x} = 5$,

Or, $x^2 - 5x + 1 = 0$.

We will use this second form of input expression to simplify the target expression invoking Many ways technique.

Just as you try to solve a problem in many ways to improve your problem solving skills and learn to solve problems efficiently, you may need to use a given resource also in more than one way.

Here instead of using an inverse sum in its usual manner **we are now using it in expanded second form.**

Now we will invoke another powerful algebraic concept where we will use this second form of the given expression to simplify the target expression greatly.

#### 13. Continued factor extraction technique for simplification

When there is no other way, we apply this powerful algebraic technique for simplifying quite intimidating expressions **with the help of a simpler input expression of zero value.**

Essentially, we extract the input expression as a factor stage by stage from the higher order target expression and at each stage, after extraction as a factor in a part of the target expression, **we substitute zero value** for the input expression **to eliminate that part altogether.**

Then again we take up second stage extraction of the input expression as a factor in the remaining portion of the target expression.

At every stage the target expression thus gets simplified in big steps.

Let's see how this is done with the numerator of the fraction in the target expression,

$E = 1 + \displaystyle\frac{3x^3 + 5x^2 + 3x}{x^4 + 1} = 1 + \displaystyle\frac{E_1}{E_2}$, where $E_1$ and $E_2$ are the numerator and denominator respectively,

and the given expression expanded in the second form of a zero valued expression,

$x^2 - 5x + 1 = 0$.

We will extract the LHS of the equation as a factor and eliminate all terms involved in the factored part of the expression.

Applying the technique on the numerator,

$E_1 = 3x^3 + 5x^2 + 3x = 3x(x^2 - 5x + 1) + 15x^2 + 5x^2$

$= 20x^2$.

In the first step (or in any stage) we have extracted or formed the LHS of the given expression as a factor, **absorbing the highest order term of the given expression** with the other terms automatically formed.

In this case, when we formed the LHS $(x^2 - 5x + 1)$ as a factor from the target expression absorbing the highest order term $3x^3$, we find the third term 1 of the factor equivalent to $3x$ already existing in the target expression. Thus **we needed to compensate** only for the second term of $-5x$ equivalent to $+15x^2$ outside the brackets in the main part of the rest of the expression.

At any stage, absorbing the highest order (with highest power) term in the factor by the first term of the factor, writing down the factored expression and compensating for the rest of the terms with opposite signs outside the factored expression are the main steps to carry out in this technique.

As the value of the factor is zero, the whole lot of terms involved and absorbed in the factored expression are reduced to zero and are eliminated at one stroke. The result becomes a simple $15x^2 + 5x^2 = 20x^2$, a very significant simplification.

This is **one stage factor extraction only** but it shows the power of simplification in no uncertain terms.

Just to satisfy academic curiosity, we will show the final steps to the solution below **which you may skip.**

**Final steps to the solution:**

So, the target expression is transformed to,

$E = 1 + \displaystyle\frac{20x^2}{x^4 + 1}$

$=1 + \displaystyle\frac{20}{x^2 + \displaystyle\frac{1}{x^2}}$

*At last we have got our inverse expression in the target.*

We evaluate the inverse in squares in no time,

$x + \displaystyle\frac{1}{x} = 5$,

Or, $x^2 + \displaystyle\frac{1}{x^2} = 25 -2 = 23$.

**This time we have invoked the Many ways technique in its most used form by *** transforming* the given expression by the

**techniques embodied in principle of inverses.**Thus we get,

$E = 1 + \displaystyle\frac{20}{23}$

$=\displaystyle\frac{43}{23}$

**Answer:** Option c: $\displaystyle\frac{43}{23}$.

This is an example of one stage factoring out. Occasionally longer expressions require two or more stages of factor extraction, each stage being simplified greatly because of zero value of the factor.

You may refer to the full treatment of these three concepts in our session on **How to solve difficult Algebra problems in a few simple steps 6.**

**14. Reuse technique**

It is a simple but effective technique that saves time for solution. It states,

When you are at a specific evaluation stage, instead of starting from the beginning point, use all the evaluated results that can be used for saving maximum time.

We will highlight the use of this simple but useful technique through a fragnent of a problem.

**Example problem 2.**

We are given, $a \gt 1$ and $a + \displaystyle\frac{1}{a} = 2\frac{1}{12}$, and we are to evaluate the value of $a^4 - \displaystyle\frac{1}{a^4}$.

Getting the inverse sum in square is easy,

$a + \displaystyle\frac{1}{a} = 2\frac{1}{12} = \frac{25}{12}$,

Or, $a^2 + \displaystyle\frac{1}{a^2} = \frac{25^2}{12^2} - 2 = \frac{337}{144}$.

At this point we take stock and understand clearly that if we get the value of $a^2 - \displaystyle\frac{1}{a^2}$, we can easily calculate the value of the target expression. And to evaluate, $a^2 - \displaystyle\frac{1}{a^2}$, we need to get the value of $a - \displaystyle\frac{1}{a}$, as value of $a + \displaystyle\frac{1}{a}$ is already available.

Thus we form a subtraction of square of inverses directly from the last available result of $a^2 + \displaystyle\frac{1}{a^2} = \frac{337}{144}$,

$a^2 - 2 + \displaystyle\frac{1}{a^2} = \displaystyle\frac{337}{144} - 2$

Or, $\left(a - \displaystyle\frac{1}{a}\right)^2 = \displaystyle\frac{49}{144}$,

Or, $\left(a - \displaystyle\frac{1}{a}\right) = \displaystyle\frac{7}{12}$, as $a \gt 1$.

We have reused the value of already evaluated expression, $a^2 + \displaystyle\frac{1}{a^2}$.

Instead, we could have started from the beginning with the given expression, $a + \displaystyle\frac{1}{a} = 2\frac{1}{12}$ and squaring and subtracting 4, and then taking the square root we would have had our desired value of $\left(a - \displaystyle\frac{1}{a}\right)$.

The difference between the two approaches would be,

In the first approach of using

Reuse technique, we have calculated $\displaystyle\frac{337}{144} - 2$ which took little time, while in the second approach we had to first form the square and then calculate $\displaystyle\frac{625}{144} - 4$.

The second step would have taken 5 to 10 seconds extra which in itself may seem to be small, but in competitive tests may turn out to be the difference between success and failure.

Apart from success in tests,

Our main philosophy being

Efficient Problem Solving, we won't like to miss any opportunity in finding a shorter way to the solution.

That's where lies the importance of this apparently obvious technique to be practiced as a habit in all situations.

In fact,

All innovations resuse.

It is a foundational technique.

You may refer to detailed use of this technique in our session on **How to solve difficult Algebra problems in a few simple steps 5.**

#### 15. Surd rationalization property

Though surds form a part of number system, it is closely interwined with Algebra in forming Surd Algebra problems that are usually a bit more complex than usual. That's why we include the basic surd concepts as rich algebraic concepts here.

We have already seen how Surd rationalization works in our first part of the basic and rich algebra concepts. Now we will go through two very important concepts. The first is the Surd rationalization property.

**Input transformation by Surd Rationalization Property**

In dealing with surds, we examine a two term surd expression always to detect if it satisfies the * Surd Rationalization Property* which states,

If the difference of the square of the two terms in the surd expression is 1, when we inverse the expression and rationalize, its denominator will become 1, and the denominator will thus be eliminated very conveniently, leaving us with a second complementary surd expression.

**For example,** if the starting expression were $2 + \sqrt{3}$ that satisfies Surd rationalization property, after inverting and rationalizing we get as a result, its complementary expression, $2 - \sqrt{3}$.

As an example we will take up a third example problem.

**Example problem 3.**

If $x = 3 + 2\sqrt{2}$, then the value of $\displaystyle\frac{x^6 + x^4 + x^2 + 1}{x^3}$ is,

- 192
- 216
- 204
- 198

**Problem analysis and solution**

Without waiting, we divide the numerator of the target expression by the denominator, because we notice the possibility of generating inverse expressions that we are comfortable with. This also is a case of using * Simplification technique* that we discussed earlier.

$\displaystyle\frac{x^6 + x^4 + x^2 + 1}{x^3}$

$= x^3 + x + \displaystyle\frac{1}{x} + \displaystyle\frac{1}{x^3}$

Our intention is to find a useful value of the sum of inverse of $x$ and applying the techniques of principle of inverses evaluate the target expression with ease.

Observing the surd rationalizing property present in the given expression $x = 3 + 2\sqrt{2}$, immediately we take the steps to invert $x$ and rationalize using rationalization technique,

$\displaystyle\frac{1}{x} = \displaystyle\frac{3 - 2\sqrt{2}}{3^2 - (2\sqrt{2})^2} = 3 - 2\sqrt{2}$, a very convenient expression.

With this result we get the value of sum of inverses,

$x + \displaystyle\frac{1}{x} = (3 + 2\sqrt{2}) + (3 - 2\sqrt{2}) = 6$

With this sum of inverses we would easily be able to evaluate sum of inverse squares and then the sum of inverse cubes to evaluate finally the target expression.

After simplifying the target expression in the form of sum of inverses, we needed a convenient value of such a sum of inverses as input. The given input expression though was in a different from, it had the Surd rationalization property.

Detecting this property, evaluating first the inverse by inverting and rationalizing and then summing the inverse of $x$ with the direct $x$ produced the desired sum of inverses in convenient form.

**Recommendation:** Wherever you detect presence of surd rationalization property in a two term surd expression you may take up inversion and then rationalization.

You may refer to the detailed use of this useful concept in our sessions on * How to solve difficult Algebra problems in a few simple steps 5 *and

**How to solve difficult Surd Algebra problems in a few simple steps 4.**#### 16. Transformation to square sum technique

This forms two of the most important techniques in surd expression simplification.

The objective that these pair of techniques fulfill is to transform a two term surd expression into a square of two term surd expression.

Wherever the original expression is under square roots, you have to apply one of these two techniques to bring the expression out of the square root for further simplification steps. In all such problems involving square root of surd expressions, application of one of these two techniques is indispensable.

**First technique of transformation of a two term surd expression as a square of a second two term surd expression:**

We will highlight the technique using a problem example.

**Example problem 4.**

Given $x = 3 + 2\sqrt{2}$, evaluate, $\sqrt{x} - \displaystyle\frac{1}{\sqrt{x}}$.

**First stage pattern identification**

Examining the given expression we notice that the surd term $2\sqrt{2}$ has a coefficient of 2. This is the first essential property for expressing the surd expression as a square of sum. This will form the second term $2ab$ in the general square expression $(a + b)^2 = a^2 + 2ab + b^2$.

The second essential property is, the sum of squares of two target terms in the square of sum must be equal to the first term in integer form. In our case it is 3 and it satisfies this second condition as, $3 = (\sqrt{2})^2 + 1^2$.

Unless the given two term surd expression satisfies these two conditions, transforming it into a square of two term surd expression won't be possible.

In our problem then we can directly have our transformation,

$x = 3 + 2\sqrt{2} = (\sqrt{2})^2 + 2\times{1}\times{\sqrt{2}} + 1^2$

$= (\sqrt{2} + 1)^2$.

So, $\sqrt{x} = \sqrt{2} + 1$.

Now only we can think of evaluating the target expression.

We need to form a sum of inverses and we detect presence of * surd rationalization property* in $\sqrt{x}$. Thus we invert, rationalize and then subtract,

$\sqrt{x} - \displaystyle\frac{1}{\sqrt{x}}$

$=\sqrt{2} + 1 - \displaystyle\frac{1}{\sqrt{2} + 1}$

$=\sqrt{2} + 1 - (\sqrt{2} - 1) = 2$.

This first form appears in surd expressions frequently.

**Second technique of transformation of a two term surd expression as a square of a second two term surd expression:**

This second form appears less frequently and so is harder to detect. It needs first a minor transformation of the given expression to convert the surd term to a term with a coefficient of 2 and then this transformed expression can be treated with the first technique as detailed above.

Let us highlight the use of this technique through a problem example.

** Example problem 5.**

If $x = \displaystyle\frac{\sqrt{3}}{2}$, then the value of $\displaystyle\frac{\sqrt{1 + x}}{1 + \sqrt{1 + x}} + \displaystyle\frac{\sqrt{1 - x}}{1 - \sqrt{1 - x}}$ is,

- $2$
- $2 - \sqrt{3}$
- $\frac{2}{\sqrt{3}}$
- $1$

**Solution:**

**First stage Problem analysis:**

Looking at the target expression we decide, unless we express $(1 + x)$ and $(1 - x)$ as* square of sums*, we won't be able to come out of the square root on these two main expressions and can't reach the solution for a long time to come.

By substituting the value of $x$ we get,

$\sqrt{1 + x} = \sqrt{1 + \displaystyle\frac{\sqrt{3}}{2}}$

$=\sqrt{\displaystyle\frac{2 + \sqrt{3}}{2}}$.

**First stage input transformation:**

The target is more defined now. * We need to express the numerator of the surd sum under the root as a square* which at first glance seems to be not possible.

We remember our experience of dealing with such two-term surd expression where, multiplying the expression by 2 we made the expression a square in our **Algebra Solution Set 13.**

Taking the cue, we multiply the numerator and denominator both by 2,

$\sqrt{1 + x} =\sqrt{\displaystyle\frac{4 + 2\sqrt{3}}{4}}$

$=\sqrt{\displaystyle\frac{(\sqrt{3} + 1)^2}{4}}$

$=\displaystyle\frac{\sqrt{3} + 1}{2}$.

This is our **main breakthough. **And it is a case of applying** Transformation to square of sum technique **in surds.

Similarly we get,

$\sqrt{1 - x} = =\displaystyle\frac{\sqrt{3} - 1}{2}$.

With this main breakthough the rest of the steps don't pose much of difficulty.

For detailed treatment, you may refer to our session on * Algebra Solution Set 13 *and

**How to solve difficult Surd Algebra problems in a few simple steps 4.**#### 17. Numerator simplification technique

Most times our objective in simplifying algebraic expressions is to simplify either numerator or denominator or both of the expressions. By Numerator simplification technique we address a specific case of simplifying the numerator with great positive results.

The two conditions for application of this very useful technique occurs when in a fraction of two term algebraic expressions,

- Both the numerator and the denominator have one common term, and
- The other terms left out are formed by same variable or surd but with different coefficients.

When we detect such a pattern we simply subtract a 1 from the fraction expression and compensate it by addition of another 1. The subtraction operation cancels out the common terms and transforms the numerator to a single term expression, a great improvement.

We will highlight the use of this technique through a problem fragment.

**Example problem 6.**

Simplify, $\displaystyle\frac{5\sqrt{3} - 2\sqrt{5}}{5\sqrt{3} - 4\sqrt{5}} + \displaystyle\frac{6\sqrt{3} - 3\sqrt{5}}{4\sqrt{3} - 3\sqrt{5}}$.

**Useful Pattern recognition**

In the first term we notice the term $5\sqrt{3}$ common between the numerator and denominator and in the second term also we find similar pattern of the term $3\sqrt{5}$ common between the numerator and denominator. Thus the first condition is satisfied.

Furthermore, the other terms also are in same surds $\sqrt{5}$ and $\sqrt{3}$ respectively, but with different coefficients and thus the situation satisfies the second condition for applying the numerator simplification technique.

When we find such a pattern, the common terms are eliminated by applying the ** numerator simplification technique, **more specifically, by subtracting 1 from the fraction term (or adding 1 to the fraction term when the common terms are of opposite signs between the numerator and denominator) and compensating the subtraction by addition of another 1.

As the second terms also are in terms of the same surd (or variable), the numerator is simplified to a single term expression.

Applying the **numerator simplification technique** on the target expression we get,

$E=\displaystyle\frac{5\sqrt{3} - 2\sqrt{5}}{5\sqrt{3} - 4\sqrt{5}} - 1 + 1$

$\hspace{10mm} + \displaystyle\frac{6\sqrt{3} - 3\sqrt{5}}{4\sqrt{3} - 3\sqrt{5}} - 1 + 1$

$=\displaystyle\frac{2\sqrt{5}}{5\sqrt{3} - 4\sqrt{5}} + \displaystyle\frac{2\sqrt{3}}{4\sqrt{3} - 3\sqrt{5}} + 2$.

This is significant simplification, but the last barrier remains to be crossed.

For detailed use of this technique you may refer to our session on **How to solve difficult Surd Algebra problems in a few simple steps 4.**

#### 18. Simplification by factoring out technique

This is similar but not the same as the Continued extraction of factor technique. When we apply this technique, we detect a factor common to all terms of an expression and take it out of the expression thus simplifying it.

This again is a naturally obvious technique which should invariably be applied whenever opportunity arises. **The important point to emphasize is,**

In general, the common factor is not so easily visible.

This is the reason for our inclusion of this obvious simplification approach as an important rich algebraic concept. You have to use your pattern identification skills consciously to find out the hidden common factor.

We will highlight the use of this technique through a fragment of a problem. This is taken from the point where we left in highlighting the numerator simplification technique.

**Example problem 7.**

Simplify, $\displaystyle\frac{2\sqrt{5}}{5\sqrt{3} - 4\sqrt{5}} + \displaystyle\frac{2\sqrt{3}}{4\sqrt{3} - 3\sqrt{5}} + 2$

**Identifying hidden pattern**

Examining the two denominators closely we detect the possibility of taking out the term $\sqrt{5}$ as a factor from the first denominator and $\sqrt{3}$ as a factor from the second denominator,

$E=\displaystyle\frac{2\sqrt{5}}{\sqrt{5}(\sqrt{15} - 4)} + \displaystyle\frac{2\sqrt{3}}{\sqrt{3}(4 - \sqrt{15})} + 2$

$=\displaystyle\frac{2}{\sqrt{15} - 4} + \displaystyle\frac{2}{4 - \sqrt{15}} + 2$

$=\displaystyle\frac{2}{4 - \sqrt{15}} -\displaystyle\frac{2}{4 - \sqrt{15}} + 2$

$=2$, a really simple result.

For detailed use you may refer to our session on **How to solve difficult Surd Algebra problems in a few simple steps 4.**

In itself these concepts and techniques are practical ways to simplify otherwise complex algebraic expressions quickly in a few steps, but one must recognize the fact that these powerful concepts and techniques can't be used in isolation.

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

*,*

**End state analysis**

**Many ways technique,***,*

**Pattern recognition***and the likes along with the subject concepts.*

**Problem breakdown technique**### Problem examples and exercises on Algebra

We have not really included any serious problem examples or exercises for your understanding and honing your skill of using the basic and rich concepts on algebra. But you will find enough of problem exercises and solutions using these concepts explained in details in most of our sessions on Algebra. The following is the list of the sessions.

#### Exercise sessions on Algebra

**SSC CGL Question set 1 on Algebra 1**

**SSC CGL Question set 8 on Algebra 2**

**SSC CGL Question set 9 on Algebra 3**

**SSC CGL Question set 10 on Algebra 4**

**SSC CGL Question set 11 on Algebra 5**

**SSC CGL Question set 13 on Algebra 6**

**SSC CGL Question set 22 on Algebra 7**

**SSC CGL Question set 23 on Algebra 8**

**SSC CGL Question set 33 on Algebra 9**

**SSC CGL Question set 35 on Algebra 10**

**SSC CGL Question set 45 on Algebra 11**

**SSC CGL Question set 51 on Algebra 12**

**SSC CGL Question set 57 on Algebra 13**

**SSC CGL Question set 58 on Algebra 14**

**SSC CGL Question set 64 on Algebra 15**

#### Elegant solutions and explanations on Algebra question sets

**SSC CGL Solution set 1 on Algebra 1**

**SSC CGL Solution set 8 on Algebra 2**

**SSC CGL Solution set 9 on Algebra 3**

**SSC CGL Solution set 10 on Algebra 4**

**SSC CGL Solution set 11 on Algebra 5**

**SSC CGL Solution set 13 on Algebra 6**

**SSC CGL Solution set 22 on Algebra 7**

**SSC CGL Solution set 23 on Algebra 8**

**SSC CGL Solution set 33 on Algebra 9**

**SSC CGL Solution set 35 on Algebra 10**

**SSC CGL Solution set 45 on Algebra 11**

**SSC CGL Solution set 51 on Algebra 12**

**SSC CGL Solution set 57 on Algebra 13**

**SSC CGL Solution set 58 on Algebra 14**

**SSC CGL Solution set 64 on Algebra 15**

#### Elegant solutions of tough Algebra problems in a few steps

**How to solve complex SSC CGL level Algebra problem lightning fast**

**How to solve difficult Algebra problems in a few simple steps 1**

**How to solve difficult Algebra problems in a few simple steps 2**

**How to solve difficult Algebra problems in a few simple steps 3**

**How to solve difficult Surd Algebra problems in a few simple steps 4**

**How to solve difficult Algebra problems in a few simple steps 5**

**How to solve difficult Algebra problems in a few simple steps 6**

**How to solve difficult SSC CGL Algebra problems in a few steps 7**

**How to solve difficult SSC CGL Algebra problems in a few steps 8**

**How to solve difficult SSC CGL Algebra problems in a few steps 9**

**How to solve difficult SSC CGL level Algebra problems in a few assured steps 10**

**How to solve difficult SSC CGL level Algebra problems in a few steps 11**

**How to solve difficult SSC CGL level Algebra problems in a few steps 12**

**How to solve difficult SSC CGL level Algebra problems in a few steps 13**

**How to solve difficult SSC CGL level Algebra problems in a few steps 14**

#### SSC CGL Tier II level Question and Solution Sets

** SSC CGL Tier II level Question Set 1 Algebra 1**

**SSC CGL Tier II level Solution Set 1 Algebra 1**

**SSC CGL Tier II level Question Set 2 Algebra 2**

**SSC CGL Tier II level Solution Set 2 Algebra 2**

**SSC CGL Tier II level Question Set 3 Algebra 3**

**SSC CGL Tier II level Solution Set 3 Algebra 3**

**SSC CGL Tier II level Question Set 9 Algebra 4**

**SSC CGL Tier II level Solution Set 9 Algebra 4**