The Substitution Method is extremely useful when you have to solve algebra problems in competitive exams like SBI PO, IBPS, SSC CGL.
Innumerable algebra problems are asked in all competitive exams like SBI PO, SSC CGL, IBPS, which makes it a very important topic. Algebra problems can be both confusing and time taking. The substitution method comes in very handy here because not only is it quick but also simple.
So in the 22nd blog posts in the series of smart methods that will help you solve questions in 5- 10 seconds, this smart method will simplify algebra problems like never before!
We all know algebra is a part of mathematics where alphabets are used as variables to represent numbers and quantities in expressions, formulas and equations. Algebra problems can be a combination of any of these mentioned things and usually need you to determine or find the value of an expression or variable.
Example of Algebra ProblemsIf the value of a + b + c= 0, then find the value of a²⁄bc + b²⁄ca + c²⁄ab
Conventional method to Solve Algebra Problems
The regular method to solve a question like uses lots of algebraic formulas, substitution and many steps to get the right answer.
a + b + c= 0
a + b= - c
Cubing the expression (a + b)3 = -c3
a3 + b3 + 3ab (a+b) = -c3
a3 + b3 + c3 = 3abc
(a3 + b3 + c3)/abc = 3
(a3/abc) + (b3/abc) + (c3/abc) = 3
(a2/ bc) + (b2/ac) + (c2/ab) = 3
The correct answer but after tons of steps and unlike school no one cares about the steps but about the correct answer.
Smart Method to Solve Algebra Problems
Substitution Method is the key to all short cuts in any algebraic expression. In the above expression arrive at a number combination which after substitution will add upto zero.
Let a= 2, b= -1, c= -1, these values will satisfy the given expression, 2 -1 -1 = 0
Substituting the given values in the expression - a²⁄bc + b²⁄ca + c²⁄ab
[2²/ (-1) (1)] + [(-1)²/ (-1) (2)] + [(-1)²/ (-1) (2)]
(4/1) + (-1/2) + (-1/2) = 3
See our expert faculty explain the substitution method in the video.
Solve these algebra problems by using the substitution method-
Question 1: If a ⁄b = c ⁄d = e ⁄f = 3, then (2a² + 3c² + 4e²) ⁄ ( 2b² + 3d² + 4f ²)= ?
1) 2 2) 3 3) 4 4) 9 5) None of these
Question 2: If x= 4ab ⁄ (a+b) then (x+ 2a) ⁄ (x-2a) + (x+ 2b) ⁄ (x-2b) = ?1) a 2) b 3) 0 4) 2 5) None of these
Remember to leave your answer in the comments along with the time saved on using the substitution method.
For more smart methods, stay tuned for more and keep practicing.