In this tutorial, we will learn how to solve systems of linear equations. A system of linear equations means two or more unknowns need to be calculated from two or more linear equations. The number of equations needed is equal to the number of unknowns.

Consider the following equation:

7x = 70

Here we have one unknown (x), so we can calculate the value of x from one equation which is     

x = \frac{70}{7} = 10

But, if we are given two unknowns (x, y) as given below, we cannot determine the value of x and y from this equation. We need two equations to solve for x and y.

7x + 3y = 20

Let’s solve the following system of linear equations:

7x + 3y = 20 \ldots \ldots (1) \\

10x + 7y = 30 \ldots \ldots (2) \\

Method 1: Solve by variable substitution

From equation (1), we get the value of y as

3y = 20 - 7x \\

or, y = \frac{20-7x}{3} \ldots \ldots (3)

Substitute the value of y in equation (2) [Note: don’t substitute in the same equation]

10x + 7\times \frac{20-7x}{3} = 30 \\

Multiply both sides of the equation with 3 we get,

30x + 140 - 49x = 90

or, -19x = -50 \\

or, x = \frac{50}{19}

Substitute the value of x in equation (3) we get the value of y as

y = \frac{20 - 7\times \frac{50}{19}}{3}

or, y = \frac{30}{57} = \frac{10}{19}

Method 2: Solve by variable elimination

Let’s take the above equations as

7x + 3y = 20 \ldots \ldots (1) \\

10x + 7y = 30 \ldots \ldots (2) \\

In variable elimination method, we eliminate one variable by multiplying both equation with some numbers such that the coefficients of the variable to be eliminated becomes same. Then subtract one equation from the other.

Multiplying equation (1) with 10 and equation (2) with 7, we get

70x + 30y = 200 \ldots \ldots (3) \\

70x + 49y = 210 \ldots \ldots (4) \\

Subtract equation (4) from equation (3),

-19y = -10 \\

or, y = \frac{10}{19}

Substitute the value of y in any of the given equation [here equation (1) ] we get

7x + 3\times \frac{10}{19} = 20 \\

or, 7x = 20 - \frac{30}{19} = \frac{350}{19} \\

or, x = \frac{50}{19}

So, the solution is (x, y) = (\frac{50}{19}, \frac{10}{19} )

Method 3 : Using Cramer’s rule

For the system of linear equations:

a_1x + b_1y = c_1 \\

a_2x + b_2y = c_2 \\

Cramer’s rule for determining x and y is as follows:

den = \begin{vmatrix}a_1&b_1\\a_2&b_2\end{vmatrix} \\

\text{numx} = \begin{vmatrix}c_1&b_1\\c_2&b_2\end{vmatrix} \\

\text{numy} = \begin{vmatrix}a_1&c_1\\a_2&c_2\end{vmatrix} \\

x = \frac{\text{numx}}{\text{den}} = \frac{\begin{vmatrix}c_1&b_1\\c_2&b_2\end{vmatrix}}{\begin{vmatrix}a_1&b_1\\a_2&b_2\end{vmatrix}} \\

or, x = \frac{c_1b_2 - b_1c_2}{a_1b_2-a_2b_1}

y = \frac{\text{numy}}{\text{den}} = \frac{\begin{vmatrix}a_1&c_1\\a_2&c_2 \end{vmatrix}}{\begin{vmatrix}a_1&b_1\\a_2&b_2\end{vmatrix}} \\

or, y = \frac{a_1c_2 - a_2c_1}{a_1b_2 - a_2b_1}

So, for our given equations applying Cramer’s rule gives,

x = \frac{\begin{vmatrix}20&3\\30&7\end{vmatrix}}{\begin{vmatrix}7&3\\10&7\end{vmatrix}} = \frac{140-90}{49-30} = \frac{50}{19} \\

y = \frac{\begin{vmatrix}7&20\\10&30\end{vmatrix}}{\begin{vmatrix}7&3\\10&7\end{vmatrix}} = \frac{210-200}{49-30} = \frac{10}{19}

Practical Example:

Suppose that you have x dollar and you spent all the money buying 3 books where unit price of the book is y dollar. If you bought one book only, you could have saved 60 dollars. Determine how much money you had (here x) and what is the unit price of the book?

Unit price  = y

So, price of 3 books = 3 y

Since all the money were spent buying 3 books,  x = 3 y    … … (1)

If one book was bought, remaining money = x – y

So, x – y = 60   … … (2)

Substitute x = 3y in equation (2)

3 y – y = 60

So, 2 y =  60

Y = 30

Substitute the value of y in equation (1)

X = 3×30 = 90 (dollar)