Solving Systems Of Linear Equations Step By Step Guide
When faced with a system of linear equations, finding the solution involves identifying the ordered pair (x, y) that satisfies all equations simultaneously. In simpler terms, we're looking for the point where the lines represented by these equations intersect on a graph. In this article, we'll explore a practical approach to solving systems of linear equations, using the given example as a guide.
Understanding the Problem
We are given the following system of linear equations:
3x + y = 1
5x + y = 3
Our goal is to determine which of the provided ordered pairs, (A) (-2, 1), (B) (-0.5, 0.5), (C) (0.5, -0.5), or (D) (1, -2), is the solution to this system. To do this, we will substitute the x and y values from each ordered pair into the equations and see if they hold true for both.
Method 1: Substitution
Step 1: Isolating a Variable
The substitution method begins with isolating one variable in one of the equations. Looking at our system, it seems easiest to isolate y in both equations:
From the first equation, 3x + y = 1, we can isolate y:
y = 1 - 3x
Similarly, from the second equation, 5x + y = 3, we isolate y:
y = 3 - 5x
Step 2: Setting the Expressions Equal
Since both equations now express y in terms of x, we can set these expressions equal to each other:
1 - 3x = 3 - 5x
This creates a new equation with only one variable, x, which we can solve.
Step 3: Solving for x
To solve for x, we'll add 5x to both sides of the equation:
1 - 3x + 5x = 3 - 5x + 5x
Simplifying, we get:
1 + 2x = 3
Next, subtract 1 from both sides:
2x = 2
Finally, divide by 2:
x = 1
Step 4: Solving for y
Now that we have the value of x, we can substitute it back into either of the original equations (or the rearranged ones) to find y. Let's use the first rearranged equation, y = 1 - 3x:
y = 1 - 3(1)
y = 1 - 3
y = -2
So, we find that y = -2.
Step 5: Verifying the Solution
Our solution is the ordered pair (1, -2). To ensure this is correct, we substitute these values into both original equations:
For the first equation, 3x + y = 1:
3(1) + (-2) = 3 - 2 = 1
This holds true.
For the second equation, 5x + y = 3:
5(1) + (-2) = 5 - 2 = 3
This also holds true. Therefore, the ordered pair (1, -2) is indeed the solution to the system of equations.
Method 2: Elimination
The elimination method offers another approach to solving systems of linear equations. This method involves manipulating the equations to eliminate one variable, making it easier to solve for the other.
Step 1: Aligning the Equations
Ensure that the equations are aligned, with the x and y terms in the same columns. Our system is already aligned:
3x + y = 1
5x + y = 3
Step 2: Eliminating a Variable
To eliminate a variable, we want the coefficients of either x or y to be opposites (e.g., 3 and -3). In this case, the y terms have the same coefficient (1). We can eliminate y by subtracting the first equation from the second:
(5x + y) - (3x + y) = 3 - 1
Step 3: Simplifying and Solving for x
Simplifying the equation, we get:
5x + y - 3x - y = 2
2x = 2
Dividing both sides by 2:
x = 1
Step 4: Solving for y
Now that we have x = 1, we substitute this value into either of the original equations to find y. Let's use the first equation, 3x + y = 1:
3(1) + y = 1
3 + y = 1
Subtracting 3 from both sides:
y = -2
Thus, we find that y = -2.
Step 5: Verifying the Solution
As before, we verify the solution (1, -2) by substituting it into both original equations. We've already done this in Method 1, and we know it holds true.
Method 3: Testing the Answer Choices
In some cases, particularly when dealing with multiple-choice questions, the most efficient approach is to test the provided answer choices. This involves substituting the x and y values from each ordered pair into the system of equations to see which pair satisfies both.
Step 1: Testing Option A (-2, 1)
Substitute x = -2 and y = 1 into the first equation, 3x + y = 1:
3(-2) + 1 = -6 + 1 = -5
Since -5 ≠1, option A is not the solution.
Step 2: Testing Option B (-0.5, 0.5)
Substitute x = -0.5 and y = 0.5 into the first equation, 3x + y = 1:
3(-0.5) + 0.5 = -1.5 + 0.5 = -1
Since -1 ≠1, option B is not the solution.
Step 3: Testing Option C (0.5, -0.5)
Substitute x = 0.5 and y = -0.5 into the first equation, 3x + y = 1:
3(0.5) + (-0.5) = 1.5 - 0.5 = 1
This satisfies the first equation. Now, test it in the second equation, 5x + y = 3:
5(0.5) + (-0.5) = 2.5 - 0.5 = 2
Since 2 ≠3, option C is not the solution.
Step 4: Testing Option D (1, -2)
Substitute x = 1 and y = -2 into the first equation, 3x + y = 1:
3(1) + (-2) = 3 - 2 = 1
This satisfies the first equation. Now, test it in the second equation, 5x + y = 3:
5(1) + (-2) = 5 - 2 = 3
This also holds true. Therefore, option D (1, -2) is the solution.
Conclusion
In summary, the ordered pair (1, -2) is the solution to the given system of linear equations. We arrived at this conclusion using three different methods: substitution, elimination, and testing answer choices. Each method provides a valid approach to solving systems of equations, and the most suitable one may depend on the specific problem and the solver's preference. For this particular problem, testing the answer choices proved to be the most direct and efficient method. However, understanding both substitution and elimination enhances your problem-solving toolkit for various mathematical scenarios. Whether you prefer the algebraic precision of substitution and elimination or the directness of testing choices, the key is to apply the method that you find most effective and reliable for each given problem. By mastering these techniques, you'll be well-equipped to tackle a wide range of linear systems and related mathematical challenges. Remember, practice is crucial for solidifying your understanding and improving your speed and accuracy in solving such problems. So, keep practicing, and you'll become more proficient in solving systems of linear equations.
