This method is best utilized when at least one linear equation is already solved for one of the variables. For example, the system below $$\begin{array}{c} y=3x-8 \\ 4x+7y=19 \end{array}$$ has the first equation solved for the \(y\) variable. As the name Substitution suggest, a replacement of \(y\) in the second equation with \(3x-8\) from the first equation is exactly the point of the Substitution Method. This creates one equation with only the \(x\) variable:
| \(\begin{array}{c} y=\color{blue}{3x-8} \\ 4x+7\color{blue}{y}=19 \end{array}\) | \(\implies4x+7(\color{blue}{3x-8})=19\) |
The result of the substitution is an equation in the \(x\) variable only, and can be solved using normal algebra methods. Once the \(x\)-value has been solved for, then pick either original equation and plug this value back in to find the \(y\)-value. Recall that a solution to a system is a point in the plane: \((x,y)\).
Solve the system by substitution: \(\begin{array}{c} y-2x=14 \\ y=4-3x \end{array}\).
Not all systems are going to have one equation solved for one of the variables. Some algebra might be needed to solve one equation for one of the variables. The following examples will show different systems of equations and how to implement the Substitution Method.
Solve the system by substitution: \(\begin{array}{c} 3s=-5t-3 \\ 2s-6=t+5 \end{array}\).
The following system is built with equations in standard form. While substitution may not be the best method to solve this system, just focus on the rewriting of one equation to start the substitution:
Solve the system by substitution: \(\begin{array}{c} 5x-4y=23 \\ 7x+8y=5 \end{array}\).