Lesson 21: Solving Systems of Equations Using the Substitution Method

New Concept

The substitution method is a method used to solve systems of equations by solving an equation for one variable and substituting the resulting expression into the other equation.

Why it matters

The substitution method is your first major tool for finding the exact intersection of two relationships, proving more powerful than simple graphing. Mastering this technique of replacement is crucial for solving more complex systems you'll encounter in science, engineering, and economics.

What’s next

Next, you'll apply this method to solve for specific coordinate solutions and learn to recognize systems with no solution or infinite solutions.

Property

The substitution method is a method used to solve systems of equations by solving an equation for one variable and substituting the resulting expression into the other equation.

Solve $\begin{cases} y = 2x + 1 \\ 2y - x = -6 \end{cases}$. Substitute $(2x+1)$ for y in the second equation: $2(2x+1) - x = -6$, which simplifies to $x = -\frac{8}{3}$.
Solve $\begin{cases} x = 3y - 1 \\ 2x + 4y = 16 \end{cases}$. Substitute $(3y-1)$ for x in the second equation: $2(3y-1) + 4y = 16$, which simplifies to $y = \frac{9}{5}$.

Think of substitution as a clever disguise for a variable. First, you solve one equation to find out what a variable (like 'y') is equal to. Then, you sneak that expression into the other equation, replacing the original 'y'. This move eliminates one variable, letting you easily solve for the other and crack the system's code.

Property

When solving dependent systems algebraically, the result is a true numerical statement (one without variables) such as $0 = 0$ or $5 = 5$.

Solve $\begin{cases} y = 3x - 1 \\ 3y = 9x - 3 \end{cases}$. Substitute y: $3(3x-1) = 9x - 3 \rightarrow 9x - 3 = 9x - 3 \rightarrow 0 = 0$. This is always true, so there are infinite solutions.
Solve $\begin{cases} 4x + 2y = 8 \\ y = -2x + 4 \end{cases}$. Substitute y: $4x + 2(-2x+4) = 8 \rightarrow 4x - 4x + 8 = 8 \rightarrow 8 = 8$. This is always true, so there are infinite solutions.

This is the 'secretly the same line' scenario! After substituting, if all your variables magically vanish and you are left with a statement that is obviously true, like $8=8$, it means the lines are identical. They overlap everywhere, so there are infinite solutions, and every point on the line is an answer. It is a mathematical mic drop.

Property

When attempting to solve systems with no solution, the result is a false numerical statement such as $0 = 6$ or $-4 = 4$.

Solve $\begin{cases} y = 4x + 1 \\ -4x + y = 3 \end{cases}$. Substitute y: $-4x + (4x+1) = 3 \rightarrow 1 = 3$. This is false, so there is no solution.
Solve $\begin{cases} 2y = 6x - 10 \\ y = 3x + 2 \end{cases}$. Substitute y: $2(3x+2) = 6x - 10 \rightarrow 6x + 4 = 6x - 10 \rightarrow 4 = -10$. This is false, so there is no solution.

What happens when two lines are parallel and never meet? In algebra, it looks like a contradiction. After you substitute, the variables cancel out, but you are left with a statement that makes no sense, like $2 = -2$. This mathematical impossibility is the system’s way of screaming, 'We never cross!' which means there is no solution.

Property

To eliminate a denominator when solving, multiply every term on both sides of the equation by the denominator.

Solve $\begin{cases} 3x + 2y = -3 \\ 4x - 3y = 13 \end{cases}$. Isolate y: $y = \frac{-3x-3}{2}$. Substitute into the second equation: $4x - 3(\frac{-3x-3}{2}) = 13$. Multiply all terms by 2 to get: $8x - 3(-3x-3) = 26$.
Solve $\begin{cases} 2x + 5y = 7 \\ 4y - 3x = 1 \end{cases}$. Isolate x: $x = \frac{4y-1}{3}$. Substitute into the first equation: $2(\frac{4y-1}{3}) + 5y = 7$. Multiply all terms by 3 to get: $2(4y-1) + 15y = 21$.

Don't let fractions freak you out. When substitution leaves you with an equation containing a fraction, you have a secret weapon: the denominator. Multiply every single term on both sides of the equation by that denominator. This simple trick eliminates the fraction completely, leaving you with a much cleaner and friendlier equation that is far easier to solve.