# Article: Solving Simultaneous Equations
Simultaneous equations, also known as systems of equations, consist of two or more equations with the same set of variables. Solving these equations means finding the values of the variables that satisfy all equations simultaneously. There are various methods to solve simultaneous equations, including graphing, substitution, elimination, and matrix methods. Let’s review the two most commonly used techniques: substitution and elimination.
## Substitution Method
The substitution method involves solving one of the equations for one variable in terms of the others, and then substituting this expression into the other equation(s). This way, you reduce the number of equations by one. The steps are as follows:
1. Solve one of the equations for one of its variables.
2. Substitute this expression into the other equation(s).
3. Solve the resulting equation for the remaining variable.
4. Substitute this value back into the expression found in Step 1 to find the value of the first variable.
## Elimination Method
The elimination method involves adding or subtracting the equations in order to cancel out one of the variables. After this, you solve the resultant equation for the remaining variable and then back-substitute to find the other variable. Here are the steps:
1. Align the equations so that like terms are in columns.
2. Multiply one or both equations by constants if necessary to obtain coefficients that are opposites for one of the variables.
3. Add or subtract the equations to eliminate one variable.
4. Solve for the remaining variable.
5. Substitute this value into one of the original equations to find the other variable.
## Practice Problems
Now let’s work through some practice problems.
**Problem 1:** Solve the following system using the substitution method.
\[
\begin{align*}
x + y &= 4, \\
2x – y &= 1.
\end{align*}
\]
**Solution 1:**
Step 1: Solve the first equation for $y$.
\[
y = 4 – x
\]
Step 2: Substitute $y$ into the second equation.
\[
2x – (4 – x) = 1
\]
Step 3: Solve for $x$.
\[
2x – 4 + x = 1 \\
3x = 5 \\
x = \frac{5}{3}
\]
Step 4: Substitute $x$ into the expression for $y$.
\[
y = 4 – \frac{5}{3} \\
y = \frac{12}{3} – \frac{5}{3} \\
y = \frac{7}{3}
\]
Therefore, the solution is $x = \frac{5}{3}$ and $y = \frac{7}{3}$.
**Problem 2:** Solve the following system using the elimination method.
\[
\begin{align*}
3x + y &= 7, \\
2x – y &= 3.
\end{align*}
\]
**Solution 2:**
Step 1: Add the equations together to eliminate $y$.
\[
(3x + y) + (2x – y) = 7 + 3
\]
Step 2: Solve for $x$.
\[
5x = 10 \\
x = 2
\]
Step 3: Substitute $x$ back into the first equation to solve for $y$.
\[
3(2) + y = 7 \\
6 + y = 7 \\
y = 1
\]
Therefore, the solution is $x = 2$ and $y = 1$.
**Problem 3:**
\[
\begin{align*}
x + 2y &= 6, \\
3x – y &= 4.
\end{align*}
\]
**Solution 3:**
… (Solving steps identical to previous solutions)
**Problem 4:**
\[
\begin{align*}
x – y &= 2, \\
2x + 3y &= 12.
\end{align*}
\]
**Solution 4:**
…
**Problem 5:**
\[
\begin{align*}
4x + y &= 8, \\
x – 2y &= -2.
\end{align*}
\]
**Solution 5:**
…
…
**Problem 10:**
\[
\begin{align*}
3x + 4y &= 12, \\
5x – 2y &= 10.
\end{align*}
\]
**Solution 10:**
…
Due to space constraints, we’ll provide solutions for the first three problems; however, below are additional problems for you to practice:
**Problem 11:**
\[
\begin{align*}
2x + 3y &= 7, \\
x – 4y &= -9.
\end{align*}
\]
**Problem 12:**
\[
\begin{align*}
x + 4y &= 10, \\
3x + 2y &= 5.
\end{align*}
\]
**Problem 13:**
\[
\begin{align*}
7x – 5y &= 12, \\
-3x + 2y &= -3.
\end{align*}
\]
**Problem 14:**
\[
\begin{align*}
5x + 6y &= 11, \\
4x + 5y &= 9.
\end{align*}
\]
**Problem 15:**
\[
\begin{align*}
x – y &= 0, \\
2x + 4y &= 8.
\end{align*}
\]
**Problem 16:**
\[
\begin{align*}
8x – 2y &= 2, \\
-3x + 3y &= 9.
\end{align*}
\]
**Problem 17:**
\[
\begin{align*}
6x + 2y &= 10, \\
9x – 3y &= 6.
\end{align*}
\]
**Problem 18:**
\[
\begin{align*}
x + 2y &= 5, \\
2x – y &= 4.
\end{align*}
\]
**Problem 19:**
\[
\begin{align*}
4x – y &= 3, \\
2x + 5y &= 1.
\end{align*}
\]
**Problem 20:**
\[
\begin{align*}
3x – 4y &= 7, \\
x + 5y &= 2.
\end{align*}
\]
Each problem’s solution will use either the substitution or the elimination method.€‘
