## Procedure

In order to solve a linear system in two variables using elimination method, follow these steps:

- Decide on which of the two variables to eliminate (this choice can be made easy if one of the two variables has a coefficient of );
- Ensure that the chosen variable has the
same

coefficients in both equations (if not, carry out a preliminarymultiplication

to force equality); - Add/subtract the equations to eliminate the variable with the same coefficient;
- Solve the resulting one-variable linear equation and substitute back your answer to get the other variable;
- Check that your solution satisfies the original equations.

#### Example 1

Solve the linear system:

(1)

(2)

Either or can be easily eliminated because of the numerical value of their coefficients. Suppose we choose to eliminate , then we have to ** add ** the two equations (because then ). So:

Next, we substitute this value of into any of the original equations, say the first equation:

Thus

is the solution. Finally, we check that the solution satisfies the original equations:

That’s it! Easy; very easy.

#### Example 2

(3)

(4)

Here, neither of the two variables has the same coefficients, so we’ll need a preliminary multiplication, whichever variable is chosen for elimination. We don’t know why we often like to eliminate ; we’re doing it again. Multiply equation (3) by , and then multiply equation (4) by to make the coefficient of equal to in both equations:

Then ** subtract **:

Substitute into equation (3) and solve for :

Finally, we check that the solution () satisfies the original equations:

Done!

#### Example 3

(5)

(6)

The variable is on the right side in equation (5), so we first re-arrange the equation:

Multiply equation (6) by (since we want to eliminate ):

**Subtract:**

Substitute into any of the previous equations and obtain . We leave the verification of the solution to you.

Don’t always expect the solution to a linear system to be integers. Sometimes, it may not be.

#### Example 4

(7)

(8)

Substitute into any of the original equations (7) and (8) to get . Finally, verify the solutions.

#### Example 5 (an inconsistent system)

(9)

(10)

Divide both sides of equation (10) by to obtain . So, the refactored

equations are:

**Subtract:** , which is impossible. So there’s **no solution** to the original linear system; the equations simply represent two straight lines with the same slope but different -intercepts (that is, parallel lines).

#### Example 6 (a system with infinitely many solutions)

(11)

(12)

Observe that equation (12) equation (11), so the two equations actually represent the same straight line (that is, they are **coincident**). Such a system possesses **infinitely many** solutions.

#### Example 7 (a system with fractional coefficients)

(13)

(14)

We first **clear** fractions from each equation by multiplying by the appropriate **lowest common denominator** or LCD (for equation (13) it is ; while the LCD for equation (14) is ). This way we obtain the refactored equations:

Then we proceed as before:

The last step can be skipped because multiplication by doesn’t alter an equation. Now **add** to eliminate :

Substitute into equation (13) to find :

So and . We leave you to verify this solution.

#### Example 8 (a word problem involving integers)

The **sum** of two numbers is , while their **difference** is . Find the two numbers.

- Begin by assigning variables to represent the numbers; let’s choose and .
- Understand the highlighted terms (
**sum**and**difference**). - Formulate appropriate linear equations based on the given information.
- Solve the resulting equations.

Following the above procedure, we’re led to the linear system

(15)

(16)

**Add:** .

Substitute to get . Check that and .

Don’t be afraid of word problems.

#### Example 9 (another integer related word problem)

The **sum** of the digits of a **two-digit** number is . If the digits are **reversed**, the original number **decreases** by . Find the number.

Let the digits of the two-digit number be and ; for simplicity, let be the **tens** digit and let the **ones** digit. Then the actual number is ; but when the digits are reversed, the number becomes . This, together with the information given in the question, leads to the linear system:

(17)

(18)

Equation (18) can be simplified as , or . Our refactored equations are:

**Add:** . Substitute in equation (17) to find :

The original number is . Check that and that .

#### Example 10 (still on numbers)

Two numbers are in the ratio of . If the numbers add up to , find the numbers.

Sometimes, a word problem is so easy that the solution can be obtained by inspection. Just like in this case. However, we want to use linear system to solve it because that’s our topic for now. So, let the numbers be and (assume that is the smaller number). We have:

(or , or ) and .

(19)

(20)

So, . Substitute in equation (19) to get , whence .

The numbers are and . Check that , and , as required.

There are other class of word problems that generally lead to linear systems. For example, problems relating to speed/distance/time, problems relating to mixtures, and problems relating to money. And so on. You’ll find these in your textbook.