Solving For X In 2y=5-3x When Y Is 1 A Step By Step Guide
In the realm of mathematics, particularly in algebra, solving for variables in an equation is a fundamental skill. This article delves into a step-by-step solution for finding the value of x when y is given as 1 in the linear equation 2y = 5 - 3x. This is a common type of problem encountered in introductory algebra courses, emphasizing the manipulation of equations to isolate a specific variable. Understanding the process not only helps in solving similar problems but also builds a strong foundation for more advanced algebraic concepts.
Understanding the Equation: 2y = 5 - 3x
Before we dive into solving for x, let's first understand the equation itself. The equation 2y = 5 - 3x is a linear equation in two variables, x and y. A linear equation represents a straight line when plotted on a graph. The equation states that twice the value of y is equal to 5 minus three times the value of x. Our goal is to find the specific value of x that satisfies this equation when y is equal to 1.
Substituting the Value of y
The first step in solving for x is to substitute the given value of y, which is 1, into the equation. This substitution simplifies the equation by eliminating one variable, leaving us with an equation in terms of x only. Replacing y with 1, the equation becomes:
2(1) = 5 - 3x
This simplifies to:
2 = 5 - 3x
Now, we have a simpler equation that we can manipulate to isolate x.
Isolating the Term with x
To isolate the term with x (-3x), we need to get it by itself on one side of the equation. We can do this by subtracting 5 from both sides of the equation. This maintains the equality of the equation while moving the constant term to the left side:
2 - 5 = 5 - 3x - 5
This simplifies to:
-3 = -3x
Now we have -3 on the left side and -3x on the right side. The next step is to isolate x itself.
Solving for x
To solve for x, we need to get rid of the coefficient -3 that is multiplying x. We can do this by dividing both sides of the equation by -3. This operation will isolate x on the right side and give us its value:
-3 / -3 = -3x / -3
This simplifies to:
1 = x
Therefore, the value of x when y is 1 in the equation 2y = 5 - 3x is 1. This means that the point (1, 1) lies on the line represented by the equation.
Verification of the Solution
To ensure our solution is correct, we can substitute x = 1 back into the original equation along with y = 1 and check if the equation holds true. Substituting x = 1 and y = 1 into 2y = 5 - 3x, we get:
2(1) = 5 - 3(1)
This simplifies to:
2 = 5 - 3
Which further simplifies to:
2 = 2
Since the equation holds true, our solution x = 1 is correct. This verification step is crucial in problem-solving to avoid errors and build confidence in the solution.
Alternative Methods and Insights
While we have solved for x using a direct algebraic method, it's worth noting that there are other approaches to solving this type of problem. For instance, one could rearrange the original equation to express x in terms of y before substituting the value of y. This approach can be particularly useful if you need to solve for x for multiple values of y. Furthermore, understanding the graphical representation of the equation can provide valuable insights into the relationship between x and y.
Rearranging the Equation
Before substituting y = 1, we could rearrange the equation 2y = 5 - 3x to solve for x in terms of y. To do this, we first isolate the term with x by adding 3x to both sides:
2y + 3x = 5 - 3x + 3x
This simplifies to:
2y + 3x = 5
Next, we subtract 2y from both sides to isolate the 3x term:
2y + 3x - 2y = 5 - 2y
This simplifies to:
3x = 5 - 2y
Finally, we divide both sides by 3 to solve for x:
3x / 3 = (5 - 2y) / 3
This gives us:
x = (5 - 2y) / 3
Now, we can substitute y = 1 into this equation to find x:
x = (5 - 2(1)) / 3
x = (5 - 2) / 3
x = 3 / 3
x = 1
As we found before, the value of x is 1 when y is 1. This alternative method demonstrates the flexibility in algebraic manipulation and provides a different perspective on solving the problem.
Graphical Interpretation
Linear equations can be visually represented as straight lines on a coordinate plane. The equation 2y = 5 - 3x represents a specific line. The solution we found, x = 1 when y = 1, corresponds to the point (1, 1) on this line. By plotting the line, we can visually confirm that this point lies on it. This graphical interpretation adds another layer of understanding to the algebraic solution.
To plot the line, we can find two points that satisfy the equation. We already know one point: (1, 1). To find another point, we can choose a different value for y and solve for x. For example, let's choose y = 0:
2(0) = 5 - 3x
0 = 5 - 3x
3x = 5
x = 5/3
So, another point on the line is (5/3, 0). Plotting these two points (1, 1) and (5/3, 0) and drawing a line through them gives us the graphical representation of the equation 2y = 5 - 3x. The point (1, 1) is clearly on this line, visually confirming our solution.
Real-World Applications
Understanding how to solve linear equations is not just an academic exercise; it has numerous real-world applications. Linear equations are used to model relationships between two variables in various fields, such as physics, economics, and engineering. For example, they can represent the relationship between distance and time for an object moving at a constant speed, the relationship between supply and demand in economics, or the relationship between voltage and current in electrical circuits. Being able to manipulate and solve these equations is crucial for making predictions and solving problems in these fields.
Example: Distance and Time
Consider an object moving at a constant speed. The distance d traveled by the object is related to the time t by the equation d = vt, where v is the constant speed. If we know the speed and the distance, we can solve for the time it takes to travel that distance. Similarly, if we know the speed and the time, we can solve for the distance. This is a simple example of how linear equations are used to model real-world phenomena.
Example: Supply and Demand
In economics, the relationship between the supply and demand for a product can often be modeled using linear equations. The quantity demanded by consumers typically decreases as the price increases, while the quantity supplied by producers typically increases as the price increases. The equilibrium price is the price at which the quantity demanded equals the quantity supplied. By setting the supply and demand equations equal to each other and solving for the price, we can find the equilibrium price. This is a fundamental concept in economics, and linear equations play a crucial role in its analysis.
Example: Electrical Circuits
In electrical circuits, Ohm's Law states that the voltage V across a resistor is related to the current I flowing through it by the equation V = IR, where R is the resistance. If we know any two of these quantities, we can solve for the third. For example, if we know the voltage and the resistance, we can solve for the current. This is a fundamental law in electrical engineering, and linear equations are essential for its application.
Conclusion
In conclusion, solving for x when y = 1 in the equation 2y = 5 - 3x is a fundamental algebraic problem that demonstrates the importance of equation manipulation and variable isolation. The step-by-step solution involves substituting the value of y, isolating the term with x, and then solving for x. Verification of the solution is crucial to ensure accuracy. Furthermore, understanding alternative methods, such as rearranging the equation or using graphical interpretation, provides a more comprehensive understanding of the problem. The ability to solve linear equations is a valuable skill with numerous real-world applications in various fields. By mastering these concepts, students can build a strong foundation for more advanced mathematical studies and problem-solving in general. The problem-solving approach discussed in this article can be applied to a wide range of algebraic problems, making it a valuable tool for students and professionals alike. Remember to always double-check your work and consider different approaches to ensure a thorough understanding of the concepts involved. Understanding the application of linear equations extends beyond the classroom, impacting various real-world scenarios and professional fields. The ability to solve algebraic equations accurately is crucial for success in many disciplines.
