Solving simultaneous equations by elimination

1 Solving simultaneous equations by elimination

One way of solving simultaneous equations is by elimination . As the name implies, elimination, involves removing one or more of the unknowns. Note that if both sides of an equation are multiplied or divided by a non-zero number an exactly equivalent equation results. For example, if we are given the equation

$x + 4 y = 5$

then by multiplying both sides by 7 we find

$7 x + 28 y = 35$

and this modified equation is equivalent to the original one.

Given two simultaneous equations, elimination of one unknown can be achieved by modifying the equations so that the coefficients of that unknown in each equation are the same and then subtracting one modified equation from the other. Consider the following example.

Example 27

Solve the simultaneous equations

$3 x + 5 y = 31$ (1)

$2 x + 3 y = 20$ (2)

Solution

We first try to modify each equation so that the coefficient of $x$ is the same in both equations. This can be achieved if Equation (1) is multiplied by 2 and Equation (2) is multiplied by 3. This gives

$6 x$	$+$	$10 y$	=	62
$6 x$	$+$	$9 y$	=	60

Now the unknown $x$ can be eliminated if the second equation is subtracted from the first:

$6 x$	$+$	$10 y$	=	62
subtract $6 x$	$+$	$9 y$	=	60
$0 x$	$+$	$1 y$	=	2

The result implies that $1 y = 2$ and we see immediately that $y$ must equal 2. To find $x$ we substitute the value found for $y$ into either of the given Equations (1) or (2). For example, using Equation (1),

$\begin{array}{rcl} 3 x + 5 (2) & = & 31 \\ 3 x & = & 21 \\ x & = & 7 \end{array}$

Thus the solution of the simultaneous equations is $x = 7$ , $y = 2$ .

N.B. You should always check your solution by substituting back into both of the given equations.

Example 28

Solve the equations

$- 3 x + y = 18$ (3)

$7 x - 3 y = - 44$ (4)

Solution

We modify the equations so that $x$ can be eliminated. For example, by multiplying Equation (3) by 7 and Equation (4) by 3 we find

$- 21 x$	$+$	$7 y$	=	126
$21 x$	$-$	$9 y$	=	$- 132$

If these equations are now added we can eliminate $x$ . Therefore

$- 21 x$	$+$	$7 y$	=	126
add $21 x$	$-$	$9 y$	=	$- 132$
$0 x$	$-$	$2 y$	=	$- 6$

from which $- 2 y = - 6$ , so that $y = 3$ . Substituting this value of $y$ into Equation (3) we obtain:

$- 3 x + 3 = 18 so that - 3 x = 15 so x = - 5.$

The solution is $x = - 5, y = 3.$

Example 29

Solve the equations

$5 x + 3 y = - 74$ (5)

$- 2 x - 3 y = 26$ (6)

Solution

Note that the coefficients of $y$ differ here only in sign.

By adding Equation (5) and Equation (6) we find $3 x = - 48$ so that $x = - 16$ .

It then follows that $y = 2$ , and the solution is $x = - 16, y = 2$ .

Task!

Solve the equations

$5 x - 7 y = - 80$

$2 x + 11 y = 106$

The first step is to modify the equations so that the coefficient of $x$ is the same in both.

If the first is multiplied by 2 then the second equation must be multiplied by what?

5 Write down the resulting equations:

$10 x - 14 y = - 160$ , $10 x + 55 y = 530$

Subtract one equation from the other to eliminate $x$ and hence find $y$ :

$55 y - (- 14 y) = 530 - (- 160)$ so $69 y = 690$ so $y = 10$ .

Now substitute back to find $x$ :

$x = - 2$

1 Solving simultaneous equations by elimination

Example 27

Solution

Example 28

Solution

Example 29

Solution

Task!

Answer

Answer

Answer

Answer