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4. Algebraic Solutions of Linear Systems

a. Solving Systems of Equations Using Substitution

This method involves subsituting y (or `x` if it is easier) from one equation into the other equation. This simplifies the second equation and we can solve it easily.

Example 1

Solve the system

x + y = 3 [1]

3x − 2y = 14 [2]

using substitution.

(The numbers in square brackets, [1] and [2], are used to name each equation. This makes it easier when referring to them in the solution.)


From line [1], we subtract x from both sides and get `y = −x + 3`.

We substitute this in the place of y in line [2]:

3x − 2(−x + 3) = 14

This gives us: `3x + 2x − 6 = 14`

Therefore `x = 4`.

Now, using line [1] we get ` y = −1`.

If we have the right numbers, they should also work in the other equation.

Checking in line [2]:

`3(4) − 2(−1) = 14` [OK]

So our solution `(4, −1)` is correct.

b. Solving Systems of Equations Using Elimination

Our aim here is to eliminate one of the variables. It doesn't matter which one - we usually just do the easiest one.

Example 2

Solve the system using elimination.

3x + y = 10 [1]

x − 2y = 1 [2]


If we subtract one row from the other row, we don't eliminate anything. However, if we multiply one of the rows, we can eliminate one of the variables by adding the rows.

Row [1] `× 2` gives us

6x + 2y = 20 [3]

x − 2y = 1 [2] (no change)

If we add lines [3] and [2], we eliminate y.

`7x = 21`

So ` x = 3` and using line [1], `y = 1`.

Check in line [2]: `3 − 2(1) = 1` [OK]

So our solution is `(3, 1)`.

In a later chapter we will see how to solve systems of equations using determinants (okay for paper-based solutions) and matrices (very powerful and the best way to do it on computers).

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