Question

Solve each system of equations. If the system has no solution, state that it is inconsistent. $$ \left\\{\begin{aligned} 2 x-y &=-1 \\ x+\frac{1}{2} y &=\frac{3}{2} \end{aligned}\right. $$

Short answer

x = \frac{1}{2}, y = 2

Step-by-step solution

Write the system of equations

Consider the given system of equations:1) \[ 2x - y = -1 \]2) \[ x + \frac{1}{2}y = \frac{3}{2} \]

Multiply the second equation to eliminate fractions

Multiply the second equation by 2 to eliminate the fraction:\[ 2\left(x + \frac{1}{2}y \right) = 2 \left(\frac{3}{2}\right) \]Hence, the second equation becomes:\[ 2x + y = 3\]

Add the equations

Add the modified second equation to the first equation:\[ (2x - y) + (2x + y) = -1 + 3 \]This simplifies to:\[ 4x = 2 \]Therefore,\[ x = \frac{2}{4} = \frac{1}{2} \]

Substitute the value of x back into one of the original equations

Substitute \(x = \frac{1}{2} \) into the first original equation:\[ 2\left(\frac{1}{2}\right) - y = -1 \]This simplifies to:\[ 1 - y = -1 \]Thus, solving for y:\[ -y = -2 \]\[ y = 2 \]

Verify the solution

Substitute \(x = \frac{1}{2}\) and \(y = 2\) into the second original equation to check the solution:\[ \left(\frac{1}{2}\right) + \frac{1}{2}(2) = \frac{3}{2} \]This holds true, so the solution is verified.

Key concepts

linear equations

Linear equations are equations that make straight lines when graphed on a coordinate plane. They can be written in various forms, such as the **standard form**: \( Ax + By = C \), where \(A\), \(B\) and \(C\) are constants, and \(x\) and \(y\) are variables.
In this exercise, we are given a system of two linear equations:
1) \( 2x - y = -1 \)
2) \( x + \frac{1}{2}y = \frac{3}{2} \)
To solve a system of linear equations, we need to find values for the variables \(x\) and \(y\) that satisfy both equations simultaneously.
Linear equations are fundamental in algebra, and solving systems of linear equations helps in understanding how different lines interact on a graph.

substitution method

The substitution method involves solving one of the equations for one variable and then substituting that expression into the other equation. This method is quite useful when one of the equations is already solved for a single variable.
Although we didn't use the substitution method in the solution provided, understanding it is essential. Here’s a quick example of how you would use the substitution method for this system:
Start with the second equation:
1) \( x + \frac{1}{2}y = \frac{3}{2} \)
Solve for \(x\):
\( x = \frac{3}{2} - \frac{1}{2}y \)
Now, substitute \(x\) in the first equation:
\( 2(\frac{3}{2} - \frac{1}{2}y) - y = -1 \)
Simplify and solve for \(y\).

elimination method

The elimination method focuses on eliminating one of the variables by combining equations, making it easier to solve for the remaining variable.
In the given exercise, we use the elimination method by first multiplying the second equation by 2 to eliminate the fraction:
\( 2(x + \frac{1}{2}y) = 2(\frac{3}{2}) \)
This results in:
\( 2x + y = 3 \)
Now we have:
1) \( 2x - y = -1 \)
2) \( 2x + y = 3 \)
Add these equations to eliminate \(y\):
\( (2x - y) + (2x + y) = -1 + 3 \), Simplifying yields:
\( 4x = 2 \)
So, \( x = \frac{1}{2} \)
Then, substitute \(x\) back into one of the original equations to find \(y\):
\( 2(\frac{1}{2}) - y = -1 \), which leads to \( y = 2 \).

inconsistent systems

An inconsistent system of equations is one where there are no solutions; the lines represented by the equations never intersect.
If we simplify two equations and find a contradiction like \( 0 = 1 \), then the system is inconsistent.
Inconsistent systems usually show up when the lines are parallel (have the same slope but different intercepts).
For example, if instead of finding a valid solution like in our given system, we found something like:
\( 2x - y = -1 \)
\( 2x - y = 3 \)
No \(x\) and \(y\) can satisfy both equations simultaneously, indicating they never intersect.
Recognizing an inconsistent system is crucial as it tells us when equations have no common solution.