Solving System Of Equations: Y = -0.5x + 13 And Y = 8 + X
Let's dive into solving a system of equations! We've got two equations here:
- y = -0.5x + 13
- y = 8 + x
Our goal is to find the values of x and y that satisfy both equations simultaneously. There are a couple of ways to tackle this, but the most straightforward method in this case is using substitution. Ready? Let's get started!
Method 1: Substitution
The substitution method works by solving one equation for one variable and then substituting that expression into the other equation. Since both equations are already solved for y, this makes our job super easy! We can set the two expressions for y equal to each other:
-0. 5x + 13 = 8 + x
Now, we have a single equation with just one variable, x. Let's solve for x.
Step 1: Combine Like Terms
First, let's get all the x terms on one side of the equation and the constants on the other side. Add 0.5x to both sides:
13 = 8 + x + 0.5x
13 = 8 + 1.5x
Next, subtract 8 from both sides:
13 - 8 = 1.5x
5 = 1.5x
Step 2: Solve for x
Now, to isolate x, divide both sides by 1.5:
x = 5 / 1.5
x = 10 / 3
So, x = 10/3, which is approximately 3.33.
Step 3: Solve for y
Now that we have the value of x, we can plug it back into either of the original equations to find y. Let's use the second equation, y = 8 + x, because it looks a bit simpler:
y = 8 + (10 / 3)
y = (24 / 3) + (10 / 3)
y = 34 / 3
So, y = 34/3, which is approximately 11.33.
Step 4: Verify the Solution
To make sure our solution is correct, let's plug both x and y values into both original equations:
Equation 1: y = -0.5x + 13
(34 / 3) = -0.5 * (10 / 3) + 13
(34 / 3) = - (5 / 3) + (39 / 3)
(34 / 3) = (34 / 3) (This checks out!)
Equation 2: y = 8 + x
(34 / 3) = 8 + (10 / 3)
(34 / 3) = (24 / 3) + (10 / 3)
(34 / 3) = (34 / 3) (This also checks out!)
Since our values for x and y satisfy both equations, our solution is correct.
Solution
The solution to the system of equations is:
x = 10 / 3 y = 34 / 3
Or, approximately:
x ≈ 3.33 y ≈ 11.33
In summary, we used the substitution method to solve this system of equations. We found the values of x and y that make both equations true. Always remember to verify your solution by plugging the values back into the original equations. Great job, guys! This is how you nail these types of problems.
Visualizing the Solution
Another cool way to think about this problem is graphically. Each equation represents a straight line. The solution to the system is the point where these two lines intersect. If you were to plot these two lines on a graph, they would intersect at the point (10/3, 34/3), or approximately (3.33, 11.33).
Why is Visualizing Helpful?
Visualizing the problem can give you a better understanding of what you're actually doing when you solve the equations. It's also a great way to check your work. If you graph the lines and they don't intersect at the point you calculated, you know you've made a mistake somewhere.
Method 2: Elimination (Just for Fun!)
While substitution was the easiest method for this particular problem, let's briefly touch on the elimination method. It's a valuable tool to have in your algebraic toolkit!
What is Elimination?
The elimination method involves manipulating the equations so that when you add or subtract them, one of the variables cancels out (is eliminated). Then you can solve for the remaining variable.
Applying Elimination to Our Equations
- y = -0.5x + 13
- y = 8 + x
To eliminate y, we can subtract equation (2) from equation (1):
(y) - (y) = (-0.5x + 13) - (8 + x)
0 = -0.5x + 13 - 8 - x
0 = -1.5x + 5
Now, solve for x:
- 5x = 5
x = 5 / 1.5
x = 10 / 3
As you can see, we arrive at the same value for x as we did with the substitution method. Then, we can plug x back into either original equation to solve for y, just as before.
When to Use Elimination
Elimination is particularly useful when the equations are in standard form (Ax + By = C) or when it's easy to manipulate the equations to cancel out a variable. For example, if we had the equations:
3x + 2y = 7
2x - 2y = 3
We could simply add these equations together to eliminate y:
5x = 10
x = 2
Then, plug x = 2 back into either equation to solve for y.
Key Takeaways
- Systems of equations involve finding values that satisfy multiple equations simultaneously.
- The substitution method is great when one equation is already solved for a variable.
- The elimination method is useful when equations are in standard form or easily manipulated to cancel out a variable.
- Visualizing the solution as the intersection of lines can provide a deeper understanding.
- Always verify your solution by plugging the values back into the original equations.
Keep practicing, and you'll become a pro at solving systems of equations! Algebra can be fun, especially when you break it down step by step. Whether you prefer substitution, elimination, or even graphing, having these tools in your toolbox will help you conquer any algebraic challenge.