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PREPINSTA PRIME
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
Correct Answer: A
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
Correct Answer: C
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
Correct Answer: A
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
- 14
- 13
- 16
- 15
Correct Answer: 14
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} =</span><span style="font-weight: 400;"> 2^{2}</span><span style="font-weight: 400;">
=> 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
- \frac{23}{13}
- \frac{13}{23}
- \frac{22}{11}
- \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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