What is a System of Equations?
A system of equations is a set of two or more equations that share the same variables. The goal of solving a system of equations is to find the values of the variables that satisfy all equations in the system simultaneously. Systems can be classified as:
- Consistent: There is at least one solution.
- Inconsistent: There are no solutions.
- Dependent: There are infinitely many solutions.
The Substitution Method Explained
The substitution method is one of the most straightforward techniques for solving systems of equations. This method involves solving one of the equations for one variable and substituting that expression into the other equation. Here’s a step-by-step approach to using the substitution method:
Step 1: Solve One Equation for One Variable
Choose one of the equations and solve for one variable in terms of the other. For example, if you have the following system:
\[
\begin{align}
1. & \quad y = 2x + 3 \\
2. & \quad 3x + 2y = 12
\end{align}
\]
In this case, the first equation is already solved for \(y\).
Step 2: Substitute into the Other Equation
Take the expression you found in Step 1 and substitute it into the other equation. Using the above example, substitute \(y\) in the second equation:
\[
3x + 2(2x + 3) = 12
\]
Step 3: Solve for the Remaining Variable
Now, simplify and solve for the remaining variable:
\[
3x + 4x + 6 = 12 \\
7x + 6 = 12 \\
7x = 6 \\
x = \frac{6}{7}
\]
Step 4: Substitute Back to Find the Other Variable
Substitute the value of \(x\) back into the expression you derived for \(y\):
\[
y = 2\left(\frac{6}{7}\right) + 3 \\
y = \frac{12}{7} + 3 \\
y = \frac{12}{7} + \frac{21}{7} = \frac{33}{7}
\]
Thus, the solution to the system of equations is:
\[
\left(x, y\right) = \left(\frac{6}{7}, \frac{33}{7}\right)
\]
Practice Problems
To help reinforce the substitution method, here are some practice problems. Try to solve these systems of equations using the substitution method.
\[
\begin{align}
1. & \quad y = x + 4 \\
2. & \quad 2x - y = 1
\end{align}
\]
\[
\begin{align}
1. & \quad y = 3x - 2 \\
2. & \quad 4x + 2y = 10
\end{align}
\]
\[
\begin{align}
1. & \quad x + y = 7 \\
2. & \quad 2x - y = 3
\end{align}
\]
\[
\begin{align}
1. & \quad y = -2x + 6 \\
2. & \quad 3x + y = 9
\end{align}
\]
Answer Key for Practice Problems
Here are the solutions for the practice problems listed above:
Problem 1
1. From the first equation: \(y = x + 4\)
2. Substitute into the second equation:
\[
2x - (x + 4) = 1 \implies 2x - x - 4 = 1 \implies x = 5
\]
3. Substitute \(x = 5\) back into \(y = x + 4\):
\[
y = 5 + 4 = 9
\]
Solution: \((5, 9)\)
Problem 2
1. From the first equation: \(y = 3x - 2\)
2. Substitute into the second equation:
\[
4x + 2(3x - 2) = 10 \implies 4x + 6x - 4 = 10 \implies 10x = 14 \implies x = \frac{14}{10} = \frac{7}{5}
\]
3. Substitute \(x = \frac{7}{5}\) back into \(y = 3x - 2\):
\[
y = 3\left(\frac{7}{5}\right) - 2 = \frac{21}{5} - \frac{10}{5} = \frac{11}{5}
\]
Solution: \(\left(\frac{7}{5}, \frac{11}{5}\right)\)
Problem 3
1. From the first equation: \(y = 7 - x\)
2. Substitute into the second equation:
\[
2x - (7 - x) = 3 \implies 2x - 7 + x = 3 \implies 3x = 10 \implies x = \frac{10}{3}
\]
3. Substitute \(x = \frac{10}{3}\) back into \(y = 7 - x\):
\[
y = 7 - \frac{10}{3} = \frac{21}{3} - \frac{10}{3} = \frac{11}{3}
\]
Solution: \(\left(\frac{10}{3}, \frac{11}{3}\right)\)
Problem 4
1. From the first equation: \(y = -2x + 6\)
2. Substitute into the second equation:
\[
3x + (-2x + 6) = 9 \implies 3x - 2x + 6 = 9 \implies x + 6 = 9 \implies x = 3
\]
3. Substitute \(x = 3\) back into \(y = -2x + 6\):
\[
y = -2(3) + 6 = -6 + 6 = 0
\]
Solution: \((3, 0)\)
Conclusion
Solving systems of equations by substitution answer key provides not only the answers but also a comprehensive understanding of the substitution method. By practicing the steps outlined in this article, learners can enhance their mathematical skills and gain confidence in solving systems of equations. Whether you're preparing for exams or simply looking to improve your understanding, mastering the substitution method will serve you well in your math journey.
Frequently Asked Questions
What is the substitution method for solving systems of equations?
The substitution method involves solving one of the equations for one variable and then substituting that expression into the other equation to find the value of the second variable.
Can you provide an example of a system of equations solved by substitution?
Sure! For the system: y = 2x + 3 and x + y = 10, you substitute y in the second equation: x + (2x + 3) = 10, which simplifies to 3x + 3 = 10. Solving this gives x = 7/3, and substituting back gives y = 2(7/3) + 3 = 19/3.
What should you do if one equation is already solved for a variable?
If one equation is already solved for a variable, you can directly substitute that expression into the other equation to find the value of the remaining variable.
What are some common mistakes when using substitution to solve systems of equations?
Common mistakes include incorrectly substituting the expression, making calculation errors, or forgetting to solve for the second variable after substitution.
How do you verify your solution after solving a system of equations by substitution?
To verify your solution, substitute the values of both variables back into the original equations to ensure that both equations are satisfied.
What types of systems of equations can be solved using substitution?
Substitution can be used for all types of systems, including consistent, inconsistent, and dependent systems, as long as at least one equation can be solved for one variable in terms of the other.
