Tips and Tricks and Shortcuts To Solve Algebra

March 7, 2024

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.

Algebra Tips

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 = (x2−x1)2+(y2−y1)2\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 = (y1−y2)(x1−x2)\frac{(y_{1}- y_{2})}{(x_{1}- x_{2})}
  • Slope of line when linear equation is given ax + by = c => –ab – \frac{a}{b}
  • Midpoint =x1+ x22,y1+ y22\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 = mx2+nx1m+n\frac{ m x_{2} + nx_{1}}{m + n}

y = my2+ny1m+n\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 =mx2–nx1m–n\frac{ m x_{2} – nx_{1}}{m – n}

 y = my2–ny1m–n\frac{my_{2} – ny_{1} }{m – n}

  • Centroid of a triangle with its vertices (x1,y1), (x2,y2), (x3,y3)
    C = x1+x2+x33\frac{x_{1}+ x_{2} + x_{3}}{3}, y1+y2+y33\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 =12×[(x1(y2−y3)+(x2(y3−y1) )+(x3(y1−y2) )]\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= mx2+nx1m+n\frac{ m x_{2} + nx_{1}}{m + n} and y= my2+ny1m+n\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: a(b−c)2(c−a)(a−b)+b(c−a)2(a−b)(b−c)+c(a−b)2(b−c)(c−a)=?\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. a2+b2+c2a^{2} + b ^{2} + c ^{2}
C. 3
D. abc

Correct Answer: A
Explanation:

Degree of a(b−c)2(c−a)(a−b)\frac{a(b-c)^{2}}{(c-a)(a-b)} = 1+21+1\frac{1+2}{1+1} = 32\frac{3}{2} = 3−23 -2 =1

Degree of b(c−a)2(a−b)(b−c)\frac{b(c-a)^{2}}{(a-b)(b-c)} = 1+21+1\frac{1+2}{1+1} = 32\frac{3}{2} = 3−23 -2 = 1

Degree of c(a−b)2(b−c)(c−a)\frac{c(a-b)^{2}}{(b -c)(c-a)} = 1+21+1\frac{1+2}{1+1} = 32\frac{3}{2} = 3−23 -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 (1y−1x)=2(\frac{1}{y}-\frac{1}{x})=2, then find the value of x3−y3x^{3}-y^{3}.

Options

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

Correct Answer: 14

Explanation:

Given, 

x-y=2 …(i)

(1y−1x)=2(\frac{1}{y}-\frac{1}{x})=2

=> x−yxy=2\frac{x-y}{xy}=2

=> xy = 1 ….(ii)

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

(x−y)2=</span><spanstyle="font−weight:400;"> 22</span><spanstyle="font−weight:400;">(x-y)^{2} =</span><span style="font-weight: 400;">  2^{2}</span><span style="font-weight: 400;">

=> x2+y2=6x^{2}+y^{2}=6 …(iii)

We know that, a3−b3=(a−b)(a2+b2+ab)a^{3}-b^{3}=(a-b)(a^{2}+b^{2}+ab)

After putting all the values, we get:

= 14.

Question 5: If x+1x=3x+\frac{1}{x}=3, then find the value of:  x2+20x+1x2+10x+1\frac{x^{2}+20x+1}{x^{2}+10x+1}.

Options

  1. 2313\frac{23}{13}
  2. 1323\frac{13}{23}
  3. 2211\frac{22}{11}
  4. 1122\frac{11}{22}



Correct Answer: 2313\frac{23}{13}

Explanation:

Given,

x2+20x+1x2+10x+1\frac{x^{2}+20x+1}{x^{2}+10x+1}

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

= x2+20x+1xx2+10x+1x\frac{\frac{x^{2}+20x+1}{x}}{\frac{x^{2}+10x+1}{x}} 

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

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

So, equation (i) becomes:

= 3+203+10\frac{3+20}{3+10} 

= 2313\frac{23}{13}.

 

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