Variable Substitution
At this point in your study of mathematics, you are likely familiar with the idea of substitution. To start this lesson, we will review some common places we use substitution in math by first looking at expressions, and then looking at substitution in an equation.
Substitution in an Expression
If we consider the expression \(a+b+c\), and we are given \(a=-2\), \(b=1\), and \(c=7\), we can substitute these numerical values in for \(a\), \(b\), and \(c\), and evaluate as follows:
\(-2+1+7=6\)
If \(a=x-3\), \(b=x+1\), and \(c=2-x\), we can substitute these expressions in for \(a\), \(b\), and \(c\), and simplify to:
\[\begin{align*} &\, a+b+c\\ =&\, (x-3)+(x+1)+(2-x)\\ =&\, x+x-x-3+1+2\\ =&\, x \end{align*}\]
Even though it's uncommon, there's no reason we're limited to substituting with just numbers or letters. If we are given \(a=\square\), \(b= \diamondsuit\), and \(c=\triangle \), we can substitute in for \(a\), \(b\), and \(c\), and get the result:
\[\square +\diamondsuit + \triangle\]
In fact, we can substitute in any assigned value for \(a\), \(b\),and \(c\).
Substitution in an Equation
Next, let's consider that the cost of three apples and two avocados is \($3\), as shown here:
\(+\)\(+\)\(+\)
\(+\)\(=3\)
where
represents the cost of one apple (in dollars), and
represents the cost of one avocado (in dollars).
If we are given that \(= 0.50\), then we can calculate the cost of an avocado by substituting the cost of and isolating .
\(0.50+0.50+0.50\) \(+\)\(+\)
\(1.50\) \(+\) \(2\times\)
\(2\times\)
\(2\times\)
The cost of one avocado is \($0.75\).
In this case, we were given a value for one of the variables, and we solved for the second.
What if we are instead given that \(=\)?
We can then determine the cost of an avocado by substitution because we will only be left with one unknown:
\(5\times\)
Source: Apple and Avocado - Sudowoodo/iStock/Getty Images
This gives us that the cost of an avocado is \($0.60\), and because the cost of an apple is equal to the cost of an avocado in this instance, the cost of one apple is also \($0.60\).
As we move forward in this lesson, we look at how we can use the idea of substitution to help us solve a given linear system of equations with two unknowns. Before we look at some examples, let's first recall the following from the previous lesson.
Recall
When we are asked to solve a system of equations, we are being asked to find values of the variables that satisfy all of the equations simultaneously.
To solve a system of equations having \(n\) variables, we typically need \(n\) equations written in those variables.
This means, if we have two variables, we can solve for those variables if we are given two equations involving those variables.