Lesson 2: Solving Systems of Equations by Substitution

New Concept

Graphing systems can be imprecise. The substitution method offers an exact algebraic solution. By isolating a variable in one equation and substituting its expression into the other, we can find the precise point of intersection for any system.

What’s next

Next up, you'll see this method in action through interactive examples. Then, you'll tackle a series of practice problems to master the technique.

Property

Step 1. Solve one of the equations for either variable.
Step 2. Substitute the expression from Step 1 into the other equation.
Step 3. Solve the resulting equation.
Step 4. Substitute the solution in Step 3 into one of the original equations to find the other variable.
Step 5. Write the solution as an ordered pair.
Step 6. Check that the ordered pair is a solution to both original equations.

Examples

• Solve the system $\begin{cases} y = x + 3 \\ 3x + 2y = 19 \end{cases}$. Substitute $x+3$ for $y$ in the second equation: $3x + 2(x+3) = 19$. This simplifies to $5x+6=19$, so $5x=13$ and $x=\frac{13}{5}$. Then $y = \frac{13}{5} + 3 = \frac{28}{5}$. The solution is $(\frac{13}{5}, \frac{28}{5})$.

• Solve the system $\begin{cases} 2x - y = 8 \\ x + 3y = 11 \end{cases}$. From the first equation, solve for $y$: $y = 2x - 8$. Substitute this into the second equation: $x + 3(2x-8) = 11$. This gives $7x-24=11$, so $7x=35$ and $x=5$. Then $y=2(5)-8=2$. The solution is $(5, 2)$.

Property

Step 1. Read the problem. Make sure all the words and ideas are understood.
Step 2. Identify what we are looking for.
Step 3. Name what we are looking for. Choose variables to represent those quantities.
Step 4. Translate into a system of equations.
Step 5. Solve the system of equations using good algebra techniques.
Step 6. Check the answer in the problem and make sure it makes sense.
Step 7. Answer the question with a complete sentence.

Examples

• The sum of two numbers is 25. One number is 5 more than the other. Find them. Let the numbers be $x$ and $y$. The system is $\begin{cases} x+y=25 \\ x=y+5 \end{cases}$. Substituting gives $(y+5)+y=25$, so $2y=20$ and $y=10$. Then $x=10+5=15$. The numbers are 10 and 15.

• The perimeter of a rectangle is 40 inches. The length is 4 inches less than twice the width. Find the dimensions. Let $L$ be length and $W$ be width. The system is $\begin{cases} 2L+2W=40 \\ L=2W-4 \end{cases}$. Substituting gives $2(2W-4)+2W=40$, so $6W-8=40$, $6W=48$, and $W=8$. Then $L=2(8)-4=12$. The length is 12 inches and the width is 8 inches.

Property

If solving a system of equations by substitution results in a true statement without variables, such as $0=0$, the equations are dependent. The graphs of the two equations are the same line, and the system has infinitely many solutions.

Examples

• Solve the system $\begin{cases} y = 2x - 3 \\ 4x - 2y = 6 \end{cases}$. Substitute $y$ in the second equation: $4x - 2(2x-3) = 6$. This simplifies to $4x - 4x + 6 = 6$, which gives $6=6$. This is a true statement, so there are infinitely many solutions.

• Solve the system $\begin{cases} x = 3y + 1 \\ 2x - 6y = 2 \end{cases}$. Substitute $x$ in the second equation: $2(3y+1) - 6y = 2$. This simplifies to $6y + 2 - 6y = 2$, which results in $2=2$. This is a true statement, indicating infinitely many solutions.

Property

If solving a system of equations by substitution results in a false statement, such as $0=-10$, the equations are inconsistent. The graphs of the two equations would be parallel lines, and the system has no solution.

Examples

• Solve the system $\begin{cases} y = 5x + 2 \\ y = 5x - 1 \end{cases}$. Set the expressions for $y$ equal: $5x + 2 = 5x - 1$. Subtracting $5x$ from both sides gives $2 = -1$. This is a false statement, so there is no solution.

• Solve the system $\begin{cases} x = -4y + 1 \\ 2x + 8y = 5 \end{cases}$. Substitute $x$ in the second equation: $2(-4y+1) + 8y = 5$. This simplifies to $-8y + 2 + 8y = 5$, which gives $2=5$. This is false, so there is no solution.