Eliminating X Terms To Solve For Y In Linear Equations

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In mathematics, solving systems of linear equations is a fundamental skill. One common method involves eliminating one variable to solve for the other. This article will delve into the process of eliminating the $x$ terms in a system of equations to solve for $y$ in the most efficient way possible. We will use a specific example to illustrate the steps and reasoning involved. Let's consider the following system of equations:

First equation: $6x - 5y = 17$

Second equation: $7x + 3y = 11$

Our goal is to determine the constants by which each equation should be multiplied so that when the equations are added together, the $x$ terms cancel out, leaving us with an equation solely in terms of $y$. This method streamlines the process of solving for $y$, making it more straightforward and less prone to errors. The key is to find the least common multiple (LCM) of the coefficients of $x$ in both equations. The LCM will serve as the target value for the coefficients of $x$ after multiplication, ensuring they cancel out upon addition.

Finding the Right Multipliers

To effectively eliminate the $x$ terms and solve for $y$ in the fewest steps, we need to determine the appropriate constants to multiply each equation by before adding them together. The core concept here is to manipulate the equations so that the coefficients of the $x$ terms become additive inverses (i.e., they have the same magnitude but opposite signs). This ensures that when we add the equations, the $x$ terms will cancel each other out, leaving us with an equation in terms of $y$ only.

Let's revisit the equations:

First equation: $6x - 5y = 17$

Second equation: $7x + 3y = 11$

To eliminate $x$, we need to find the least common multiple (LCM) of the coefficients of $x$, which are 6 and 7. The LCM of 6 and 7 is 42. Our goal is to multiply each equation by a constant such that the coefficient of $x$ in both equations becomes 42 or -42. To achieve this, we need to consider the signs as well. We can choose to make one coefficient 42 and the other -42, so they cancel out upon addition.

Let's multiply the first equation by -7. This will give us a coefficient of -42 for the $x$ term:

-7 * (6x - 5y) = -7 * 17

-42x + 35y = -119

Now, let's multiply the second equation by 6. This will give us a coefficient of 42 for the $x$ term:

6 * (7x + 3y) = 6 * 11

42x + 18y = 66

Now we have two new equations:

Modified First Equation: -42x + 35y = -119

Modified Second Equation: 42x + 18y = 66

Notice that the coefficients of the $x$ terms are now -42 and 42, which are additive inverses. When we add these equations together, the $x$ terms will cancel out, leaving us with an equation solely in terms of $y$.

The Elimination Process: Adding the Equations

With the multipliers determined, the next step involves adding the modified equations together. This process is the heart of the elimination method, as it allows us to get rid of one variable and simplify the system to a single equation that can be easily solved. The key here is to align the like terms (i.e., the $x$ terms, the $y$ terms, and the constants) and then add them column by column.

Let's recap the modified equations we obtained in the previous step:

Modified First Equation: -42x + 35y = -119

Modified Second Equation: 42x + 18y = 66

Now, we add the left-hand sides and the right-hand sides of the equations separately:

(-42x + 35y) + (42x + 18y) = -119 + 66

Combine the like terms:

(-42x + 42x) + (35y + 18y) = -53

The $x$ terms cancel out, as intended:

0x + 53y = -53

This simplifies to:

53y = -53

Now we have a simple equation in terms of $y$, which we can easily solve in the next step.

Solving for $y$

After eliminating the $x$ terms, we are left with an equation in a single variable, $y$. Solving for $y$ is now a straightforward algebraic process. This step typically involves isolating $y$ on one side of the equation by performing the same operation on both sides, maintaining the equation's balance. Let's revisit the equation we obtained after adding the modified equations:

53y = -53

To isolate $y$, we need to divide both sides of the equation by the coefficient of $y$, which is 53:

(53y) / 53 = -53 / 53

This simplifies to:

y = -1

So, we have found the value of $y$ to be -1. This is one part of the solution to the system of equations. Now that we have the value of $y$, we can substitute it back into one of the original equations to solve for $x$. This substitution step is crucial for finding the complete solution to the system.

Finding $x$ by Substitution

Now that we have found the value of $y$, we can substitute it back into either of the original equations to solve for $x$. This substitution step is essential to complete the solution of the system of equations. It allows us to use the value of one variable to find the value of the other. Let's choose the first original equation for simplicity:

First equation: 6x - 5y = 17

We know that $y = -1$, so we substitute -1 for $y$ in the equation:

6x - 5(-1) = 17

Now, simplify the equation:

6x + 5 = 17

To isolate the $x$ term, subtract 5 from both sides of the equation:

6x + 5 - 5 = 17 - 5

6x = 12

Finally, divide both sides by 6 to solve for $x$:

6x / 6 = 12 / 6

x = 2

So, we have found the value of $x$ to be 2. With both $x$ and $y$ values determined, we can now express the solution to the system of equations as an ordered pair.

The Complete Solution

We have successfully eliminated the $x$ terms and solved for $y$, and then used the value of $y$ to solve for $x$. Now, we can present the complete solution to the system of equations. The solution is an ordered pair $(x, y)$ that satisfies both equations simultaneously. In our case, we found:

x = 2

y = -1

Therefore, the solution to the system of equations is the ordered pair (2, -1). This means that the point (2, -1) is the intersection of the two lines represented by the equations. To ensure the accuracy of our solution, we can substitute these values back into both original equations to verify that they hold true. Let's check our solution:

First equation: 6x - 5y = 17

Substitute x = 2 and y = -1:

6(2) - 5(-1) = 12 + 5 = 17

The first equation is satisfied.

Second equation: 7x + 3y = 11

Substitute x = 2 and y = -1:

7(2) + 3(-1) = 14 - 3 = 11

The second equation is also satisfied. Since the solution (2, -1) satisfies both equations, we can confidently conclude that it is the correct solution to the system.

Conclusion: Mastering Elimination for Efficiency

In conclusion, to eliminate the $x$ terms and solve for $y$ in the fewest steps in the given system of equations, the first equation should be multiplied by -7 and the second equation should be multiplied by 6. This approach leverages the elimination method, a powerful technique for solving systems of linear equations. By strategically multiplying the equations, we created additive inverses for the $x$ coefficients, allowing us to eliminate $x$ by adding the equations. This simplified the problem to a single equation in $y$, which we easily solved. Furthermore, by substituting the value of $y$ back into one of the original equations, we efficiently found the value of $x$, completing the solution. Mastering this method not only enhances your problem-solving skills but also provides a valuable tool for various mathematical applications. The key takeaway is to identify the LCM of the coefficients you want to eliminate and then use appropriate multipliers to create additive inverses. This systematic approach ensures efficiency and accuracy in solving systems of equations.

This step-by-step guide has demonstrated the process of eliminating the $x$ terms and solving for $y$ in a system of equations. By understanding the underlying principles and following the outlined steps, you can confidently tackle similar problems and master the art of solving systems of linear equations efficiently.