# Tips and Tricks and Shortcuts To Solve Algebra

## Tips and Tricks To Solve Algebra Questions

Algebra is a very wide topic which consists numerous problems. This page is all about Tips and Tricks To Solve Algebra. It may seem a problem to many but there’s a trump card for everything. Using conventional methods to solve algebra may take 5-10 minutes per question. Therefore, Tips and Tricks To Solve Algebra are very helpful. They will even help you to master the questions based on algebra quickly and easily.

## Algebraic Expressions:-

Here we have listed some of the most important Formulas For Algebra:-

• a2 – b2 = (a – b)(a + b)
• (a+b)2 = a2 + 2ab + b2
• a2 + b2 = (a – b)2 + 2ab
• (a – b)2 = a2 – 2ab + b2
• (a + b + c)2 = a2 + b2 + c2 + 2ab + 2ac + 2bc
• (a – b – c)2 = a2 + b2 + c2 – 2ab – 2ac + 2bc
• (a + b)3 = a3 + 3a2b + 3ab2 + b3 ,
which can be also denoted as – (a + b)3 = a3 + b3 + 3ab(a + b)
• (a – b)3 = a3 – 3a2b + 3ab2 – b3
• a3 – b3 = (a – b)(a2 + ab + b2)
• a3 + b3 = (a + b)(a2 – ab + b2)
• (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4)
• (a – b)4 = a4 – 4a3b + 6a2b2 – 4ab3 + b4)
• a4 – b4 = (a – b)(a + b)(a2 + b2)
• a5 – b5 = (a – b)(a4 + a3b + a2b2 + ab3 + b4)

## Formulas Required  for Solving Coordinate Geometry Questions:

• Distance between two points A(x1, y1) and B(x2, y2)
AB = $\sqrt{(x_{2}- x_{1})^2 + (y_{2}- y_{1})^2}$
• Slope of line when two points are given (x1, y1) and (x2, y2)
m = $\frac{(y_{1}- y_{2})}{(x_{1}- x_{2})}$
• Slope of line when linear equation is given ax + by = c => $– \frac{a}{b}$
• Midpoint =$\frac{x_{1} + x_{2}}{2},\frac{y_{1} + y_{2}}{2}$
• The co-ordinates of a point R(x,y) that divides a line segment joining two points A(x1, y1) and B(x2, y2) internally in the ratio m:n is given by
x = $\frac{ m x_{2} + nx_{1}}{m + n}$

y = $\frac{my_{2} + ny_{1} }{m + n}$

• The co-ordinates of a point R(x,y) that divides a line segment joining two points A(x1, y1) and B(x2, y2) externally in the ratio m:n is given by
x =$\frac{ m x_{2} – nx_{1}}{m – n}$

y = $\frac{my_{2} – ny_{1} }{m – n}$

• Centroid of a triangle with its vertices (x1,y1), (x2,y2), (x3,y3)
C = $\frac{x_{1}+ x_{2} + x_{3}}{3}$, $\frac{y_{1}+ y_{2} + y_{3}}{3}$
• Area of a Triangle with its vertices A(x1,y1), B(x2,y2), C(x3,y3)
A =$\frac{1}{2} ×[ (x_{1}(y_{2}-y_{3}) + (x_{2}(y_{3}-y_{1}) ) + (x_{3}(y_{1}-y_{2}) )]$
• Division of a line segment by a point
If a point p(x, y) divides the join of A(x1, y1) and B(x2, y2), in the ratio m: n, then
x= $\frac{ m x_{2} + nx_{1}}{m + n}$ and y= $\frac{my_{2} + ny_{1} }{m + n}$
• The equation of a line in slope intercept form is Y= mx+ c, where m is its slope.
The equation of a line which has gradient m and which passes through the point (x1, y1) is =
y – y1 = m(x – x1).

## Type 1: Evaluation or finding the value of a variable

Question 1: Find the value of x in 5x + 2 – 3x = 10.

Options

A. 4
B. 3
C. 1
D. None of the above

Explanation:

5x + 2 – 3x = 10
5x -3x +2 = 10
2x = 10-2
2x= 8
X= 4

## Type 2: Tips and Tricks and Shortcuts for Algebra by Substitution Methods

Question 2: What is the value of 2 (3a- 5)+ 2b (2- 6b) =10 when a= 2 and b =3?

Options

A. 89
B. 23
C. 18
D. None of the above

Explanation:

2 (3a- 5)+ 2(2a- 6b) =10
6a – 10a + 12b = 10
Substituting the values, we get
6 * 2 -10 *2 + 12 *3 = 10
12 – 20 + 36 =10
12 – 20 + 36 -10 = 0
18

Question 3: $\frac{a(b-c)^{2}}{(c-a)(a-b)} + \frac{b(c-a)^{2}}{(a-b)(b-c)} + \frac{c(a-b)^{2}}{(b -c)(c-a)} =?$

Options

A. a +b +c
B. $a^{2} + b ^{2} + c ^{2}$
C. 3
D. abc

Explanation:

Degree of $\frac{a(b-c)^{2}}{(c-a)(a-b)}$ = $\frac{1+2}{1+1}$ = $\frac{3}{2}$ = $3 -2$ =1

Degree of $\frac{b(c-a)^{2}}{(a-b)(b-c)}$ = $\frac{1+2}{1+1}$ = $\frac{3}{2}$ = $3 -2$ = 1

Degree of $\frac{c(a-b)^{2}}{(b -c)(c-a)}$ = $\frac{1+2}{1+1}$ = $\frac{3}{2}$ = $3 -2$ =1

Therefore, Degree of overall expression is 1.

Therefore, according to the given options :-

Degree Of Option A – 1
Degree Of Option B – 2
Degree Of Option C – 0
Degree Of Option D – 3

Hence, Option a +b +c is the correct option

Question 4: If x-y=2 and $(\frac{1}{y}-\frac{1}{x})=2$, then find the value of $x^{3}-y^{3}$.

Options

1. 14
2. 13
3. 16
4. 15

Explanation:

Given,

x-y=2 …(i)

$(\frac{1}{y}-\frac{1}{x})=2$

=> $\frac{x-y}{xy}=2$

=> xy = 1 ….(ii)

Now, squaring both side of eq. (i):

$(x-y)^{2} = 2^{2}$

=> $x^{2}+y^{2}=6$ …(iii)

We know that, $a^{3}-b^{3}=(a-b)(a^{2}+b^{2}+ab)$

After putting all the values, we get:

= 14.

Question 5: If $x+\frac{1}{x}=3$, then find the value of:  $\frac{x^{2}+20x+1}{x^{2}+10x+1}$.

Options

1. $\frac{23}{13}$
2. $\frac{13}{23}$
3. $\frac{22}{11}$
4. $\frac{11}{22}$

Correct Answer: $\frac{23}{13}$

Explanation:

Given,

$\frac{x^{2}+20x+1}{x^{2}+10x+1}$

Divide by ‘x’ in numerator and denominator, we get:

= $\frac{\frac{x^{2}+20x+1}{x}}{\frac{x^{2}+10x+1}{x}}$

= $\frac{x+\frac{1}{x}+20}{x+\frac{1}{x}+10}$ …(i)

Given, $x+\frac{1}{x}=3$

So, equation (i) becomes:

= $\frac{3+20}{3+10}$

= $\frac{23}{13}$.

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