6.10 Rearrangement of formulas

Changing the subject of a formula

Changing the subject of a formula is an important skill to learn. It can come in handy when you know the value of one algebraic symbol but not another. The subject of an equation is the pronumeral that is by itself on one side on the equals sign and it usually is at the start of the formula.

For example, in the formula $A=pb+y$A=pb+y, $A$A is the subject because it is by itself on the left hand side of the equals sign.

We have already been changing the subject of equations when we learnt to solve equations because we took steps to get the pronumeral by itself, for example we made $x$x the subject of equations. We can make any term in an equation the subject, even if it starts off as the denominator of a fraction. When we're changing the subject of a formula, we often have more than one pronumeral but we still use a similar process. 

 

• Group any like terms
• Simplify using the inverse of addition or subtraction
• Simplify further by using the inverse of multiplication or division
• Other inverse operations might be needed like square and square root

 

Formulas in context

Out in the real world all sorts of amazing relationships play out every day: Air temperatures change with ocean temperatures;  Populations of species rise and fall depending on seasons, food availability and the number of predators; The surface area of a human body can even be measured fairly accurately according to your height and weight. 

One of the most powerful things about mathematics is its ability to describe and measure these patterns and relationships exactly. Given a mathematical formula for the relationship between, say, the weight of a patient and how much medication they should be given, we can find one quantity by substituting a value for the other. 

We have come across so many different formulas in mathematics that allow us to measure quantities such as Area, Volume, Speed, etc. Let's have a look at the process of substituting values into these formulas to find a particular unknown.

 

Worked examples

example 1

The perimeter of a triangle is defined by the formula $P=x+y+z$P=x+y+z. Find $P$P if the length of each of its three sides are $x=5$x=5 cm, $y=6$y=6 cm and $z=3$z=3 cm.

Think: By inserting the number values of $x$x, $y$y and $z$z we have a new equation that we can use to find the value of $P$P.

Do:

$P=5+6+3$P=5+6+3

$P=14$P=14 cm

So we can see the triangle has a perimeter of $14$14 cm.

Example 2

To find the velocity of an object we can use the formula $v=u+at$v=u+at, where $u$u is the initial velocity, $a$a is its acceleration and $t$t is the time taken. 

Find the acceleration of an object if it started travelling at a speed of $5$5 m/s, and accelerated for $4$4 seconds until it had a velocity of $25$25 m/s.

Think: We know the initial velocity, final velocity and the time taken, but we want to find the acceleration. We can substitute in the known values and then rearrange to solve for $a$a.

Do:

$v$v	$=$=	$u+at$u+at	The original formula
$25$25	$=$=	$5+a\times4$5+a×4	Substituting in the known values
$20$20	$=$=	$a\times4$a×4	Subtracting $5$5 from both sides of the equation
$\frac{20}{4}$204​	$=$=	$a$a	Dividing both sides of the equation by $4$4
$a$a	$=$=	$5$5	Evaluating the quotient and rearranging so that $a$a is on the left hand side

 

We can see the object had an acceleration of $5$5 ms^-2
 

Reflect: In this case we substituted in the known values and then rearranged to solve for the unknown value. We could just have easily rearranged the formula first to make a the subject, and then substituted in the known values. In most cases both methods require the same amount of steps, but some times one method may lead to an easier experience.

Let's investigate this method now:

$v$v	$=$=	$u+at$u+at	The original formula
$v-u$v−u	$=$=	$at$at	Subtracting $u$u from both sides of the equation
$\frac{v-u}{t}$v−ut​	$=$=	$a$a	Dividing both sides of the equation by $t$t
$a$a	$=$=	$\frac{25-5}{4}$25−54​	$a$a is now the subject, so we can substitute in the known values
$a$a	$=$=	$5$5	Evaluating the quotient

 

As expected we arrived at the same answer of $a=5$a=5. 

Practice questions

Question 1

Question 2

Question 3