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### Simultaneous linear equations

Problems in which there are two (or more) unknowns linked via linear equations often lead to simultaneous linear equations. There is a general method for solving systems of linear equations with more than two unknowns, but this will not be discussed here. In this section we will only consider systems of the form

\begin{align*} a_1x+b_1y &= c_1\\ a_2x+b_2y &= c_2, \end{align*}

where $$a_1,a_2,b_1,b_2,c_1,c_2$$ are given numbers and we seek to find the values of $$x$$ and $$y$$.

There are two basic algebraic approaches to this problem. Students need to be familiar and comfortable with both methods.

#### Example

Solve

\begin{align*} 2x+y &= 1\\ x-y &= -4. \end{align*}

#### Solution

Number the equations:

\begin{alignat*}{2} 2x+y &=1 \qquad\qquad &&(1)\\ x-y &=-4. &&(2) \end{alignat*}

Make $$y$$ the subject of equation (1), so $$y= 1-2x$$. Now substitute this expression for $$y$$ into equation (2):

\begin{align*} x-(1-2x) &= -4\\ 3x-1 &= -4\\ x &= -1. \end{align*}

We can now find the value of $$y$$ by substituting into the expression for $$y$$:

$y=1-(2\times -1)= 3.$

Hence, the solution is $$x = -1$$, $$y = 3$$.

It is always wise to check that the solution actually satisfies both of the original equations.

This method is often extended to deal with simultaneous equations which are not linear.

##### Exercise 9

Solve simultaneously

\begin{align*} y &= x^2\\ y-x &= 2. \end{align*}

#### The elimination method

This method is often quicker and forms the basis of the more general method for dealing with more than two unknowns.

As the name suggests, this method involves eliminating one of the unknowns by adding or subtracting one equation from a multiple of the other.

#### Example

Use the elimination method to solve

\begin{align*} 3x+2y &= 7\\ 5x+y &= 7. \end{align*}

#### Solution

Number the equations:

\begin{alignat*}{2} 3x+2y &= 7 \qquad\qquad &&(1)\\ 5x+y &= 7. &&(2) \end{alignat*}

It is easiest here to eliminate $$y$$. We do this by multiplying equation (2) by 2. Number the resulting equation:

\begin{align*} 10x+2y &= 14. &&(3) \end{align*}

Now subtract equation (1) from equation (3) to produce

\begin{align*} 7x &= 7\\ x &= 1. \end{align*}

We can now find the value of $$y$$ by substituting into equation (1), thus

\begin{align*} 3+2y &= 7\\ y &= 2. \end{align*}

Hence, the solution is $$x = 1$$, $$y = 2$$.

Again, we should check that this solution satisfies the two original equations.

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