GMAT Book 1: Linear Algebra

Linear equations | GMAT Study Material

Linear Equations is a topic from which you could expect 1 to 2 questions in the GMAT quant section. The concept is tested both in the problem solving format - mainly as a word problem - and in the data sufficiency format - mainly in testing your knowledge about unique solutions to a pair of linear equations. The topic is an important one.

Syllabus Covered in Wizako's GMAT Book | Linear Algebra

Simple linear Equations, equations of the first order with one, two or three variables are tested in GMAT Problem Solving and Data Sufficiency sections of the GMAT CAT Test. Wizako's GMAT Math Lesson Book in this chapter covers the following concepts:

  1. Definition of simple equation and degree of equation.
  2. Method to solve equations.
  3. Definition of linear equation.
  4. Illustrative examples to explain the above concepts.
  5. Definition of independent equation, dependent equation, and linear equation in two variables.
  6. Introduction to the concept of system of linear equations and the two methods to solve simultaneous linear equations algebraically, namely, solving by substitution and solving by elimination.
  7. Different types of solutions to a system of linear equations.
  8. Special cases in linear equations - equations reducible to linear equations in two variables.
  9. 3 illustrative examples and 13 solved examples that would help understand the concepts explained in the chapter. Includes word problems.
  10. 8 exercise problems with the answer key and also explanatory answers.
  11. 34 questions test on Linear and Quadratic equations as part of the work book.

Here is a typical solved example in Wizako's GMAT Book from this chapter

Sample Question

Solve the following system of equations
5x + 4y -z = 26; 3x + y + z = 5; -2x + 5y - 2z = 19

Explanatory Answer

5x + 4y - z = 26 ...... (1)
3x + y + z = 5 ....... (2)
-2x + 5y - 2z = 19 ...... (3)

Adding equations (1) and (2), we get   8x + 5y = 31 ...... (4)

Multiply equation (2) by 2 and add the result with equation (3)
6x + 2y + 2z = 10 ...... 2 * (2)
-2x + 5y - 2z = 19
----------------------
4x + 7y = 29 ...... (5)
----------------------

Multiply equation (5) by 2 and deduct equation (4) from the result
8x + 14y = 58 ...... 2 * (5)
8x + 5y = 31
--------------
9y = 27
--------------
Or y = 3

Substitute y = 3 in equation (5): x = 2
Substitute values of x and y in equation (1): z = -4


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