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Algebraic substitution

Algebraic substitution is a powerful technique for simplifying algebraic expressions and solving equations by replacing the pronumerals with known values. This skill is essential for tackling complex problems in mathematics, science, and engineering, enabling you to evaluate and manipulate expressions with ease.

Algebraic expressions

An algebraic term consists of a number, pronumeral or both. When algebraic terms are combined using addition and/or subtraction, they become an algebraic expression.

For example, \(3x^{3}\), \(pq\) and \(8ab^{2}c\) are algebraic terms. \(\dfrac{a+b}{2}\), \(5x-2\) and \(3p+v\) are algebraic expressions.

Some examples of algebraic equations, like V equals I times R, and y equals 3x plus 2.

Algebraic equations

When algebraic expressions are written as a mathematical statement, this is an algebraic equation. One algebraic equation you might have encountered before is the equation for the area of a circle:

\[A=\pi r^{2}\]

where \(A\) is the area of a circle, \(\pi\) (pi) is a constant equal to approximately \(3.14\) and \(r\) is the radius of the circle.

Circle with a line showing the radius, labelled with an r

Video tutorial – algebraic substitution

Watch this video to learn about algebraic substitution.

Hi, this is Martin Lindsay from the Study and Learning Centre at RMIT University. This is a short movie on substitution in algebra. Replacing a pro-numeral or variable or letter in an expression or formula with a number is called substitution. Instance, here’s an expression, three P, and we want to work this out, in other words evaluate, if P equals minus five. So substituting P equals minus five into the expression, three P is the same as three times P, which is three times minus five which is minus 15.

Here are two more examples. Evaluate A plus five over B, if A equals minus nine and B equals two, here you notice we have two pro-numerals A and B, so substituting A and B into the formula or, sorry, into the expression, we have minus nine plus five over two which gives us minus four over two which is minus two.

Here’s another one. We want to evaluate W squared minus two Z, if W equals minus one and Z equals five, so substituting into the expression we have minus one all squared minus two times five which gives us one minus one times minus one is one minus 10 is minus 9.

Let’s look at a more complicated expression. Here we have the formula C equals five brackets F minus 32 close brackets over nine and that’s the formula that converts a temperature in Fahrenheit into one in Centigrade, so here we want to convert a temperature of 100 degrees in Fahrenheit, so substituting F equals 100 into the formula we get 37.78 degrees Celsius, so again, as with the previous examples, we’re just substituting the unknown, or the pro-numeral, which in this case is F, into the formula.

Now try some questions for yourself. The answers to these questions are on the next slide. Thanks for watching this short movie.

Substitution

Substitution is when you put a number into an algebraic expression or formula in place of the pronumerals.

Example 1 – substituting values into algebraic expressions

Calculate the area of a circle with a radius of \(5\textrm{ cm}\).

We can substitute the number for the radius of the circle into the algebraic equation \(A=\pi r^{2}\) to calculate the area of the circle. In this question, \(r=5\textrm{ cm}\).

\[\begin{align*} A & =\pi r^{2}\\
& =\pi\times(5)^{2}\\
& =25\pi\\
& \approx78.54\textrm{ cm}^{2}
\end{align*}\]

Remember that area is represented in square units, thus \(\textrm{cm}^{2}\).

Evaluate \(\dfrac{a+5}{b}\) if \(a=-9\) and \(b=2\).
\[\begin{align*} \frac{a+5}{b} & =\frac{(-9)+5}{(2)}\\
& =\frac{-4}{2}\\
& =-2
\end{align*}\]

In this example, there are two pronumerals being substituted with numbers. Both \(a\) and \(b\) have been replaced, and the calculation is completed.

Evaluate \(w^{2}-2z\) if \(w=-1\) and \(z=5\).
\[\begin{align*} w^{2}-2z & = (-1) \times (-1) -2 \times 5\\
& =1-10\\
& =-9
\end{align*}\]

The formula to convert degrees Fahrenheit to degrees Celsius is:
\[C=\frac{5(F-32)}{9}\]
What is \(212^{\circ}\textrm{F}\) in degrees Celsius (\(^{\circ}\textrm{C}\))?
\[\begin{align*}C=\frac{5(F-32)}{9} & = \frac{5((212)-32)}{9}\\
& = \frac{5\times 180}{9}\\
& = \frac{900}{9}\\
& = 100^{\circ}\textrm{C}
\end{align*}\]

The current in an electrical circuit is given by \(V=IR\) where \(V\) is the voltage (in volts), \(I\) is the current (in amps) and \(R\) is the resistance (in ohms).
A circuit diagram showing a resistor connected in series with a battery. The battery is giving V volts and a current of I is moving towards the resistor.
If the resistance is \(5\textrm{ ohms}\) and the current is \(2\textrm{ amps}\), what is the voltage, \(V\)?
\[\begin{align*} V & = IR\\
& = (2)\times (5)\\
& = 10\textrm{ V}
\end{align*}\]

The volume \(V\) of a right circular cone with base radius \(r\) and height \(h\) is given by:
\[V=\frac{1}{3}\pi r^{2}h\]
A cone of radius r and height h
If the radius of a cone is \(5\textrm{ cm}\) and its height is \(15\textrm{ cm}\), what is the volume of the cone?
\[\begin{align*} V & =\frac{1}{3}\pi r^{2}h\\
& =\frac{1}{3}\times\pi\times(5)^{2}\times(15)\\
& \approx392.7\textrm{ cm}^{3}
\end{align*}\]

The formula relating distance traveled \(s\) to initial speed \(u\), acceleration \(a\) and time \(t\) is:
\[s=ut+\frac{1}{2}at^{2}\] A car traveling at a speed of \(4\textrm{ m/s}\) accelerates at a rate of \(2\textrm{ m/s}^{2}\) for \(5\textrm{ s}\). How far does it travel during this time?

In this case, \(u=4\), \(a=2\) and \(t=5\). The distance traveled is:

\[\begin{align*} s & =ut+\frac{1}{2}at^{2}\\
& =(4)\times(5)+\frac{1}{2}\times(2)\times(5)^{2}\\
& =20+\frac{1}{2}\times2\times25\\
& =45\textrm{ m}
\end{align*}\]

The area of a circle \(A\) is given by:
\[ A=\pi r^{2} \] where \(r\) is the radius. If a circle has an area of \(30\textrm{ cm}^{2}\), what is its radius?

Substituting \(A=30\), we have:

\[ (30) =\pi r^{2}\]

We want \(r\). First, we divide both sides by \(\pi\) to get:

\[\begin{align*} \frac{30}{\pi} & =\frac{\pi r^{2}}{\pi}\\
& =r^{2}\\
r & =\sqrt{\frac{30}{\pi}}\\
& \approx3.1\textrm{ cm}
\end{align*}\]

Your turn – substituting values into algebraic expressions

  1. Evaluate the following:
    1. \(-4k\) if \(k=7\)
    2. \(2mn\) if \(m=4\) and \(n=-2\)
    3. \(e^{2}-5\) if \(e=2\)
    4. \(5-b-b^{2}\) if \(b=3\)
    5. \(2k^{2}+4\) if \(k=-6\)
    6. \(-3ab^{2}\) if \(a=4\) and \(b=2\)
    7. \(\dfrac{n}{4}+2\) if \(n=10\)
    8. \(\dfrac{u}{5v}\) if \(u=-20\) and \(v=2\).
  2. Evaluate the following if \(a=-1\), \(b=6\), \(c=3\), \(m=-2\) and \(n=2\).
    1. \(3a^{2}-7\)
    2. \(3a-b^{2}\)
    3. \((2m+1)^{2}\)
    4. \(3(a-b^{2})\)
    5. \(\dfrac{2m}{n}\)
    6. \((m-n)^{2}\)

    1. \(-28\)
    2. \(-16\)
    3. \(-1\)
    4. \(7\)
    5. \(76\)
    6. \(-48\)
    7. \(4.5\)
    8. \(-2\)
    1. \(-4\)
    2. \(-39\)
    3. \(9\)
    4. \(147\)
    5. \(-2\)
    6. \(16\)

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