An arbitrary system of m linear equations in n unknowns has the following general form:

where x_{1}, x_{2}, …, x_{n} are the unknowns and the subscripted a’s and b’s denote constants.

As with the systems of linear equations in general, the system of linear equation in 2 unknowns can have a unique solution, no solution, or even infinitely many solutions. In this post we will restrict ourselves to the case where m = n = 2 and the system of linear equations has a unique solution.

Below is an example of a system of linear equations.

By rewriting the system as , it can be easily seen that it is a system of 2 linear equations in 2 unknowns.

The systems of linear equations in 2 unknowns can be solved by elimination or substitution techniques.

**ELIMINATION TECHNIQUE**

In essence, this technique reduces the number of unknowns step by step. However, its effectiveness decreases as the number of unknowns increases.

**Example 1**

Determine the solution to the following system of equations.

**Answer**

The given equations are as follows.

3x – 5y = 1 ……………………………………………………………………………………………………………………………………………… (1)

2x + 3y = 7 ……………………………………………………………………………………………………………………………………………… (2)

Choose an unknown to eliminate. Suppose that x is selected. To eliminate x, multiply (1) by 2 and multiply (2) by 3. Then subtract the result of the latter from the result of the former.

2(1) …………………………… 6x – 10y = 2

3(2) …………………………… 6x + 9y = 21 –

-19y = -19

y = 1

To obtain x, eliminate y. To do this, multiply (1) by 3 and multiply (2) by -5. Then subtract the result of the latter from the result of the former.

3(1) …………………………… 9x – 15y = 3

-5(2) ………………………….. -10x – 15y = -35 –

19x = 38

x = 2

Thus, the solution to the system of linear equations is x = 2 and y = 1.

**Finding solution(s) to a system of 2 linear equations in 2 unknowns can be viewed as finding the coordinates of the point of intersection of two lines lying on a Cartesian plane.** In Example 1, the lines with equations 3x – 5y = 1 and 2x + 3y = 7 intersect at (2,1). (See the figure below.)

**SUBSTITUTION TECHNIQUE**

In fact, there is no substantial difference between this technique and the previous technique. When eliminating y, we substracted -5(2) from 3(1). It can be viewed as substituting -15y in 3(1) into -5(2). By substitution technique, we require that at the first stage, we select one of the equations and solve it for one of its unknowns. Suppose that the selected unknown is y. Then substitute the solution just-obtained for y in the other equation. This will produce the value of x. Having found this value, substitute it for x in the solution to the first equation, which is the equation chosen at the first stage. This will produce the value of y. (Of course we could have selected x, not y, at the first stage.)

**Example 2**

By applying the substitution technique, find the solution to the following system of equations.

**Answer**

Arbitrarily choose 3x – 5y = 1. The solution to the equation is . Substitute the value just-obtained into the other equation, that is 2x + 3y = 7. This results in:

Multiply both sides of the equation by 5. This gives:

10x + 3(3x – 1) = 35

19x – 3 = 35

19x = 38

x = 2

Substituting x = 2 into , we have:

Thus, the solution is x = 2 and y = 1.

Indeed there are other techniques to find the solution to such systems, but they will be discussed in some other articles on this website.