Verify Solutions For Systems Of Equations

Welcome to the world of algebra, where we often encounter systems of equations. These are sets of two or more equations that share the same variables. Today, we're going to dive into a specific task: verifying if given values are indeed solutions to a given system of equations. It's like being a detective, checking if the clues (the proposed solutions) fit the crime scene (the equations). We'll be working with a system involving two linear equations with two variables, x and y.

Our system of equations for today is:

$\begin{array}{l} \left\{\begin{array}{r} 5 x-8 y=11 \\ \frac{1}{2} x-2 y=\frac{1}{2} \end{array}\right. \end{array}

And the proposed solution we need to verify is x=3, y=rac{1}{2}. We can also represent this proposed solution as an ordered pair: (3, rac{1}{2}). To verify if this pair of values is a solution, we need to substitute these values into both equations in the system. If the values satisfy both equations simultaneously, then it is a valid solution. If it fails to satisfy even one of the equations, it's not a solution.

The First Equation: A Closer Look

Let's start with the first equation: $5x - 8y = 11$. Our proposed solution is $x=3$ and y=rac{1}{2}. We will substitute these values into the equation and see if the left side equals the right side.

Substituting $x=3$ and y=rac{1}{2} into the first equation, we get:

5(3) - 8(rac{1}{2})

First, let's perform the multiplication: $5 imes 3 = 15$. So, the equation becomes 15 - 8(rac{1}{2}).

Next, we handle the subtraction involving the fraction: 8 imes rac{1}{2} = rac{8}{1} imes rac{1}{2} = rac{8 imes 1}{1 imes 2} = rac{8}{2} = 4. Now our expression simplifies to $15 - 4$.

Finally, we perform the subtraction: $15 - 4 = 11$.

So, the left side of the first equation evaluates to $11$. The right side of the first equation is also $11$. Since $11 = 11$, the first equation is satisfied by our proposed solution (3, rac{1}{2}). This is a good sign! However, it's crucial to remember that a solution must satisfy all equations in the system. So, we must proceed to check the second equation.

The Second Equation: The Final Test

Now, let's move on to the second equation in our system: rac{1}{2}x - 2y = rac{1}{2}. Again, we will substitute our proposed values $x=3$ and y=rac{1}{2} into this equation.

Substituting $x=3$ and y=rac{1}{2} into the second equation, we get:

rac{1}{2}(3) - 2(rac{1}{2})

Let's simplify the terms. The first term is rac{1}{2} imes 3 = rac{3}{2}.

The second term is 2 imes rac{1}{2} = rac{2}{1} imes rac{1}{2} = rac{2 imes 1}{1 imes 2} = rac{2}{2} = 1.

Now, our expression becomes rac{3}{2} - 1.

To subtract these, we need a common denominator. We can rewrite $1$ as rac{2}{2}. So, the expression is rac{3}{2} - rac{2}{2}.

Performing the subtraction: rac{3-2}{2} = rac{1}{2}.

The left side of the second equation evaluates to rac{1}{2}. The right side of the second equation is also rac{1}{2}. Since rac{1}{2} = rac{1}{2}, the second equation is also satisfied by our proposed solution (3, rac{1}{2}).

Conclusion: A Verified Solution!

Since the values $x=3$ and y=rac{1}{2} satisfy both equations in the system:

$5x - 8y = 11$

and

rac{1}{2}x - 2y = rac{1}{2}

we can confidently conclude that (3, rac{1}{2}) is indeed a solution to the given system of equations. We have successfully verified the solution by performing the necessary substitutions and calculations. This process is fundamental when working with systems of equations, whether you are solving them or checking proposed answers.

Understanding how to verify solutions is a critical skill in algebra. It reinforces the concept that a solution to a system of equations is a point that lies on the graph of all equations in the system. If you're looking to deepen your understanding of systems of equations and how to solve them, exploring resources like Khan Academy's section on systems of linear equations can be incredibly helpful.