2.05 Simultaneous equations

Solving a single linear equation is when we solve for one variable, such as $x$x. For example, rearranging $x+2=5$x+2=5 will create a unique answer of $3$3 for $x$x. So what happens when there is more than one variable in an equation?

 

Exploration

Consider $x+y=6$x+y=6. This could have the solution $x=2$x=2 and $y=4$y=4, or another solution of $x=40$x=40 and $y=-34$y=−34. In fact, there are infinitely many solutions to this equation. Just pick any value for $x$x, and then use the equation to find the corresponding value of $y$y required to make the equation hold true.

If we have two equations with the same two variables in them (such as $x$x and $y$y), then this is known as a system of equations. They are also commonly referred to as simultaneous equations.

To solve a system of linear equations is to find a common pair of $x$x and $y$y values that satisfies both of these equations simultaneously. If there is a single pair of values of $x$x and $y$y that successfully does this, then the system of equations has a unique solution.

 

The graphical method

One way to solve a system of linear equations is to graph both linear equations on the same coordinate plane. The solution to a system of equations is represented by the point of intersection of the two equations' graphs (where the two graphs cross over). So the $x$x- and $y$y-values of the solution will take the form of coordinates of the intersection point $\left(x,y\right)$(x,y).

 

 

Practice question

Question 1

 

The substitution method

One way to solve simultaneous equations is called the substitution method. In this chapter, the simultaneous equations that will be mostly studied involve $2$2 variables and $2$2 equations. In these cases, the substitution method works by solving one variable first through 'substituting' one equation into the other.

 

Worked example

Example 1

To solve something like $3y+4=x$3y+4=x and $y=2+x$y=2+x we can combine the two equations by substituting $y=2+x$y=2+x into $3y+4=x$3y+4=x. This gives us:

$3\left(2+x\right)+4$3(2+x)+4	$=$=	$x$x
$6+3x+4$6+3x+4	$=$=	$x$x
$3x-x$3x−x	$=$=	$-4-6$−4−6
$2x$2x	$=$=	$-10$−10
$x$x	$=$=	$-5$−5

 

Now that we know the value of $x$x we can easily solve for $y$y by using the equation we already have that has $y$y as the subject: $y=2+x$y=2+x. Therefore $y=2+\left(-5\right)$y=2+(−5) which is equal to $-3$−3. So our answer is the pair $\left(-5,-3\right)$(−5,−3).

 

Practice questions

Question 2

 

The elimination method

This is another method of solving simultaneous equations. It works by adding or subtracting equations from one another to eliminate one variable. For simultaneous equations with $2$2 variables and $2$2 unknowns that have a unique solution, the elimination method will create $1$1 linear equation with $1$1 unknown variable, which can then be solved.

 

Worked example

Example 2

Find the unique solution to the following simultaneous equations

$2x-y=1$2x−y=1		(1)
$5x+y=2$5x+y=2		(2)

 

One way to use the elimination method is to add equations (1) and (2) together:

$2x-y+5x+y=1+\left(-2\right)$2x−y+5x+y=1+(−2) (3)

While collecting the like terms, we can see that $-y+y=0$−y+y=0, and so the $y$y terms have been eliminated from equation (3). This equation simplifies to $7x=3$7x=3, which doesn't have any $y$y terms. This linear equation can now be solved, so that $x=\frac{3}{7}$x=37​.

This solution can now be substituted into either equation (1) or (2) to then find the corresponding $y$y-value. Using equation (1), $2\left(\frac{3}{7}\right)-y=1$2(37​)−y=1, and so $y=2\left(\frac{3}{7}\right)-1=-\frac{1}{7}$y=2(37​)−1=−17​.

So the unique $\left(x,y\right)$(x,y) solution to the above system of equations is $\left(\frac{3}{7},-\frac{1}{7}\right)$(37​,−17​). Remember this is equivalent to the point of intersection of the graphs of these two lines on a number plane.

 

Using multiplication & division

Sometimes two simultaneous equations will not have the same value coefficients for any of the $2$2 variables that need to be eliminated. In this case we can multiply or divide the whole equation by a constant until both equations have $1$1 variable with matching coefficients.

 

Worked example

Example 3

Find the unique solution to the following simultaneous equations

$x+3y=5$x+3y=5		(1)
$2y+2x=1$2y+2x=1		(2)

 

Notice that the sum or difference of both these equations will not eliminate either variable. However, the coefficient of $x$x is $1$1 in equation (1) and $2$2 in equation (2). The idea then is to multiply equation (1) by $2$2 so that $x$x has a coefficient of $2$2 for both equations. Then the same logic can be applied as before to then add or subtract the equations. Let's see this in action:

Multiplying equation (1) by $2$2:

$2\left(x+3y\right)$2(x+3y)	$=$=	$2\times5$2×5
$2x+6y$2x+6y	$=$=	$10$10	call this new equation, equation (3)

Now let's find the difference between equation (3) and equation (2):

$2x+6y-\left(2y+2x\right)$2x+6y−(2y+2x)	$=$=	$10-1$10−1
$2x+6y-2y-2x$2x+6y−2y−2x	$=$=	$9$9
$4y$4y	$=$=	$9$9
$y$y	$=$=	$\frac{9}{4}$94​

It is now a simple case of substituting $y=\frac{9}{4}$y=94​ back in any of the equations (1, 2 or 3) to get the simultaneous value for $x$x:

$x+3y$x+3y	$=$=	$5$5
$x+3\times\frac{9}{4}$x+3×94​	$=$=	$5$5
$x+\frac{27}{4}$x+274​	$=$=	$5$5
$x$x	$=$=	$5-\frac{27}{4}$5−274​
$x$x	$=$=	$-\frac{7}{4}$−74​

So the unique $\left(x,y\right)$(x,y) solution to the above system of equations is $\left(-\frac{7}{4},\frac{9}{4}\right)$(−74​,94​)

 

Practice questions

Question 3

Question 4

 

Solving simultaneous equations using a CAS calculator

A CAS calculator can be utilised in two different ways to solve a linear system of equations. A graphical approach would be to use a CAS calculator to graph both linear equations on the same set of axis, then use the CAS calculator to find the point of intersection.

Alternatively, the solve function of the CAS calculator can be used to find the point of intersection immediately with no graphing involved.

Select your brand of calculator below to view instructions for these two approaches.

Casio Classpad

TI Nspire

 

Practice questions

Question 5