# Understanding Two-Variable Linear Equations
Linear equations with two variables are foundational in algebra and are used to describe the relationship between two varying quantities. These equations take the general form:
\[ ax + by = c \]
In this equation, \( x \) and \( y \) are the variables, and \( a \), \( b \), and \( c \) represent constant numbers. The values of \( a \) and \( b \) cannot both be zero. A solution to a two-variable linear equation is a pair of values, one for \( x \) and one for \( y \), that makes the equation true when the values are substituted into the equation.
Graphically, every two-variable linear equation corresponds to a straight line on the coordinate plane. The solutions of the equation are the coordinates of the points on the line.
Two-variable linear equations can be solved using several methods including graphing, substitution, elimination, and matrix methods. However, when working with systems of linear equations (more than one equation), the goal is to find the solution set that satisfies all equations simultaneously.
Now, let’s dive into some practice problems to help you get more familiar with two-variable linear equations.
### Problems and Solutions
**Problem 1: Solve for \( x \) and \( y \):**
\[ 3x + 2y = 6 \]
\[Solution:\]
We can choose to express \( y \) in terms of \( x \) or vice versa. Let’s solve for \( y \):
\[ y = \frac{6 – 3x}{2} \]
This equation represents all solutions to the given linear equation.
**Problem 2: Find the point of intersection:**
\[ 2x + y = 4 \]
\[ x – y = -1 \]
\[Solution:\]
The point of intersection is the solution that satisfies both equations. We can solve this using the substitution or elimination method.
By elimination:
\[ \begin{align*}
2x + y &= 4 \\
x – y &= -1 \\
\hline
3x + 0y &= 3
\end{align*} \]
Solving for \( x \):
\[ x = 1 \]
Substitute \( x \) into the second equation to solve for \( y \):
\[ (1) – y = -1 \]
\[ y = 2 \]
The solution is \( x = 1 \), \( y = 2 \), which represents the point of intersection (1, 2).
**Problem 3: Find the solution for:**
\[ 4x – 5y = -2 \]
\[ 8x + 10y = 4 \]
\[Solution:\]
Notice the second equation is a multiple of the first. They represent the same line. Therefore, there are infinitely many solutions, as the two equations describe the same line. We can express the solutions in terms of one variable, for example:
\[ y = \frac{4x + 2}{5} \]
**Problem 4: Determine if the following pair is a solution to the given equation: (2, 3) for \( x + 2y = 8 \).**
\[Solution:\]
Substitute \( x = 2 \) and \( y = 3 \) into the equation:
\[ x + 2y = 8 \]
\[ 2 + 2(3) = 8 \]
\[ 2 + 6 = 8 \]
Since \( 8 = 8 \) is true, the pair (2, 3) is a solution to the equation.
**Problem 5: Solve for \( x \) and \( y \):**
\[ x = 3y – 7 \]
\[ 2x – 5y = -14 \]
\[Solution:\]
Substitute \( x \) from the first equation into the second equation:
\[ 2(3y – 7) – 5y = -14 \]
\[ 6y – 14 – 5y = -14 \]
\[ y = 0 \]
Now substitute \( y = 0 \) into the first equation:
\[ x = 3(0) – 7 \]
\[ x = -7 \]
The solution is \( x = -7 \), \( y = 0 \).
