Solving Systems Of Equations 7x - 4y = -14 And -4x + 3y = 3
This article will guide you through the process of solving the system of linear equations:
7x - 4y = -14
-4x + 3y = 3
We will explore two common methods: substitution and elimination. Understanding these methods is crucial for various mathematical and real-world applications. Let's dive in!
Method 1: Elimination Method
The elimination method involves manipulating the equations to eliminate one variable, allowing us to solve for the other. Our main keyword here is elimination method. The elimination method works by manipulating the given equations so that either the x coefficients or the y coefficients are additive inverses (i.e., they add up to zero). This is usually achieved by multiplying one or both equations by a constant. When we add the modified equations together, one of the variables will be eliminated, leaving us with a single equation in a single variable, which we can easily solve. Once we find the value of one variable, we can substitute it back into either of the original equations to find the value of the other variable. This systematic approach ensures a clear and concise solution, making the elimination method a powerful tool in solving systems of equations. In this case, we can eliminate either x or y. Let's choose to eliminate x. To do this, we need to find the least common multiple (LCM) of the coefficients of x, which are 7 and 4. The LCM of 7 and 4 is 28. Therefore, we will multiply the first equation by 4 and the second equation by 7 to make the coefficients of x equal to 28 and -28 respectively. When using the elimination method, the goal is to manipulate the equations so that when they are added together, one of the variables cancels out. This often involves multiplying each equation by a suitable constant. The choice of constant depends on the coefficients of the variables you want to eliminate. For instance, if you have equations with x coefficients of 2 and 3, you might multiply the first equation by 3 and the second equation by -2 so that the x coefficients become 6 and -6, which will cancel when added. The key is to find the smallest constants that will make the coefficients of one variable additive inverses. Once you've eliminated a variable, solving for the remaining variable is straightforward. The value you obtain can then be substituted back into one of the original equations to find the value of the eliminated variable, providing a complete solution to the system of equations. Remember, the power of the elimination method lies in its ability to simplify complex systems of equations into more manageable forms, making it an essential technique in algebra and beyond.
Step 1: Multiply the equations
Multiply the first equation by 4:
4 * (7x - 4y) = 4 * (-14)
28x - 16y = -56
Multiply the second equation by 7:
7 * (-4x + 3y) = 7 * 3
-28x + 21y = 21
Step 2: Add the equations
Now, add the two modified equations together:
(28x - 16y) + (-28x + 21y) = -56 + 21
28x - 28x - 16y + 21y = -35
5y = -35
Step 3: Solve for y
Divide both sides by 5:
y = -35 / 5
y = -7
Step 4: Substitute y back into one of the original equations
Let's use the first original equation:
7x - 4y = -14
7x - 4(-7) = -14
7x + 28 = -14
Step 5: Solve for x
Subtract 28 from both sides:
7x = -14 - 28
7x = -42
Divide both sides by 7:
x = -42 / 7
x = -6
Therefore, the solution to the system of equations using the elimination method is x = -6 and y = -7.
Method 2: Substitution Method
The substitution method involves solving one equation for one variable and then substituting that expression into the other equation. Our focus keyword is the substitution method. The substitution method is a versatile technique for solving systems of equations, especially when one of the equations can be easily solved for one variable in terms of the other. The basic idea is to isolate one variable in one of the equations and then substitute that expression into the other equation. This results in a single equation with one variable, which can then be solved using standard algebraic techniques. Once the value of this variable is found, it can be substituted back into either of the original equations (or the expression obtained during the initial isolation) to find the value of the other variable. This step-by-step process ensures accuracy and clarity in finding the solution to the system. The substitution method is particularly useful when one of the equations has a variable with a coefficient of 1 or -1, as this simplifies the isolation step. However, it can be applied to any system of equations, making it a valuable tool in your algebraic arsenal. When employing the substitution method, careful attention to detail is crucial. Ensure you accurately isolate the variable and correctly substitute the expression into the other equation. Double-check your calculations to avoid errors that can lead to an incorrect solution. The beauty of the substitution method lies in its adaptability and its ability to break down complex systems into simpler, solvable parts. By mastering this technique, you'll be well-equipped to tackle a wide range of algebraic problems. This method is particularly useful when one of the equations can be easily solved for one variable. In our case, neither equation immediately lends itself to easy isolation, but we can still apply the method. Let's solve the second equation for y.
Step 1: Solve one equation for one variable
Let's solve the second equation, -4x + 3y = 3, for y:
-4x + 3y = 3
3y = 4x + 3
y = (4x + 3) / 3
Step 2: Substitute the expression into the other equation
Substitute this expression for y into the first equation, 7x - 4y = -14:
7x - 4((4x + 3) / 3) = -14
Step 3: Solve for x
Multiply both sides by 3 to eliminate the fraction:
3 * (7x - 4((4x + 3) / 3)) = 3 * (-14)
21x - 4(4x + 3) = -42
21x - 16x - 12 = -42
5x - 12 = -42
Add 12 to both sides:
5x = -30
Divide both sides by 5:
x = -6
Step 4: Substitute x back into the expression for y
Substitute x = -6 into the expression we found for y, y = (4x + 3) / 3:
y = (4(-6) + 3) / 3
y = (-24 + 3) / 3
y = -21 / 3
y = -7
Therefore, the solution to the system of equations using the substitution method is x = -6 and y = -7.
Conclusion
Both the elimination method and the substitution method lead to the same solution: x = -6 and y = -7. You can choose the method that you find easier or more efficient for a particular system of equations. In summary, mastering these techniques is essential for solving systems of equations, a fundamental skill in algebra and various applications. By understanding both the elimination method and the substitution method, you are equipped to tackle a wide range of problems, from academic exercises to real-world scenarios. Remember, practice is key to proficiency. Work through various examples, and you'll develop a strong intuition for which method is best suited for a given problem. Whether you prefer the systematic approach of elimination or the flexibility of substitution, the ability to solve systems of equations is a valuable asset in your mathematical toolkit.
