GRE Quantitative: Algebra - Solving Linear Equations

A system of equations is called as a set of linear equations, if it includes two or more linear equations containing two or more variables. The term “solving” in the language of the system of equations means to find the value of each variable that makes the equation true. Probably you can remember from your school days that there are many methods of solving a system of equations. However, if you can’t remember them then you should not be bothered as this article will assist you to prepare for this section in the GRE Quantitative Reasoning syllabus. But before refreshing these concepts we should first understand what a linear equation means.

Linear equation is a combination of variables and constants accompanied with the standard operations, in which no variable is raised to any power.

For example: 7 + x = 23

x + 5y = 34

p +3q = 2a

All these are examples of linear equations.

Let’s refresh these methods for solving the linear equations now.

  1. Add and Subtract: This is the least used method. In this method, you simply add and subtract to isolate a variable and then after getting the value of one of the variable put it in any one equation to get the value of another variable.
    Example: Solve for x and y in; x + 3y = 10 and x + 12y = 19
    Here, subtract both the equations, which will eliminate x
    Leaving us with, -9y = – 9
    y = 1
    So, x = 10-3 = 7


  1. Substitution method: Substitution is the most loved method by all students. In this method, you substitute the value of one variable from the other and put it in the second equation so that only one variable is left and hence solving it for a solution.
    For example; x +y = 23 and 9x + 5y = 18
    From first equation we can get, y = 23- x

Put it in second equation, which gives us
9x + 115 – 5x = 18
4x = -97
x = -24.25

So, y = 47.25


  1. Elimination method: It is just like addition and subtraction method but it needs more manipulation than the prior one.
    For example; Solve for x and y; 7x + 2y = 1 and 5x + 3y = 5
    multiply 1st equation with 5 and 2nd equation with 7
    This will make both the x components equal
    So, it becomes, 35x + 10y = 5 and 35x + 21y = 35
    Subtract these two, it gives
    -11y = -30
    So , y = 2.73
    Hence, x = -0.64


  1. Last method is for tricky problems. In this you probably won’t be asked to find the value of the variables but will be asked to perform some calculations.

For example;

Question: The cost of 7 mangoes, 8 apples and 3 peaches is Rs 20. The cost of 3 apples and 4 peaches and 5 guavas is Rs 21. The cost of 4 mangoes, 4 peaches and 6 guavas is Rs 25. What is the cost of 1 mango, 1 apple, 1 peach and 1 guava?

Solution: According to question;

7M + 8A +3 P = 20

3A + 4P + 5G = 21

4M + 4 P + 6 G = 25

Adding all the three equations;

11M + 11 A + 11 P + 11 G = 66

So, M + A + P + G = 6611 = 6

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