Solving Linear Equations Determining Solutions For -5x = 35

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In the realm of mathematics, solving equations is a fundamental skill. Linear equations, in particular, form the bedrock of algebraic problem-solving. This article delves into the process of determining the nature of solutions for a given linear equation. We will explore the equation βˆ’5x=35-5x = 35, meticulously analyzing it to ascertain whether it possesses a unique solution, no solution, or an infinite number of solutions. Furthermore, we will not only identify the type of solution but also provide concrete examples by finding two values of xx that substantiate our conclusion. This approach ensures a comprehensive understanding of linear equations and their solutions, equipping you with the tools to tackle similar problems with confidence. Understanding the concept of solutions in linear equations is crucial for various applications, ranging from simple real-world problems to complex mathematical models. By mastering this fundamental skill, you can unlock a wide array of mathematical concepts and problem-solving techniques.

The given equation is βˆ’5x=35-5x = 35. To determine the nature of its solutions, we must first analyze its structure. This equation is a linear equation in one variable, xx. Linear equations are characterized by the variable being raised to the power of 1, and they can be represented in the general form ax+b=cax + b = c, where aa, bb, and cc are constants. In our specific equation, βˆ’5x=35-5x = 35, we can identify aa as βˆ’5-5, bb as 00, and cc as 3535. The goal is to isolate the variable xx on one side of the equation to find its value. This can be achieved by performing algebraic operations on both sides of the equation while maintaining the equality. The key is to understand that whatever operation you perform on one side, you must also perform on the other side to preserve the balance of the equation. This principle ensures that the solution obtained remains valid. The properties of equality allow us to manipulate the equation in a systematic way, leading us to the solution or solutions. Linear equations are fundamental in algebra, and their solutions provide valuable insights into various mathematical and real-world problems.

To solve the equation βˆ’5x=35-5x = 35, we need to isolate xx. This means getting xx by itself on one side of the equation. The coefficient of xx is βˆ’5-5, which indicates that xx is being multiplied by βˆ’5-5. To undo this multiplication, we can perform the inverse operation, which is division. We will divide both sides of the equation by βˆ’5-5. This step is crucial because it ensures that the equation remains balanced. Dividing both sides by the same non-zero number maintains the equality. So, we have:

rac{-5x}{-5} = rac{35}{-5}

On the left side, βˆ’5-5 in the numerator and denominator cancel each other out, leaving us with just xx. On the right side, 3535 divided by βˆ’5-5 is βˆ’7-7. Therefore, the equation simplifies to:

x=βˆ’7 x = -7

This result tells us that there is only one value of xx that satisfies the equation, which is x=βˆ’7x = -7. This indicates that the equation has a unique solution. The process of isolating the variable is a cornerstone of solving algebraic equations, and it's essential to understand this technique to solve more complex equations.

From the previous step, we found that x=βˆ’7x = -7 is the solution to the equation βˆ’5x=35-5x = 35. This indicates that the equation has exactly one solution. A linear equation can have one solution, no solutions, or an infinite number of solutions. In this case, we have a unique solution because there is only one value of xx that makes the equation true. To verify this, we can substitute x=βˆ’7x = -7 back into the original equation:

βˆ’5(βˆ’7)=35 -5(-7) = 35

35=35 35 = 35

Since the equation holds true, our solution is correct. An equation with one solution is called a conditional equation because it is true only under specific conditions (in this case, when x=βˆ’7x = -7). If we had arrived at a contradiction (e.g., 0=10 = 1), the equation would have no solutions. If we had arrived at an identity (e.g., 0=00 = 0), the equation would have infinitely many solutions. Understanding the different types of solutions is crucial for solving and interpreting linear equations in various contexts.

To ensure the accuracy of our solution, we need to verify it. Verification involves substituting the value we found for xx back into the original equation. This process confirms that the solution satisfies the equation, meaning that when we replace xx with βˆ’7-7, the left side of the equation equals the right side. Let's substitute x=βˆ’7x = -7 into the original equation βˆ’5x=35-5x = 35:

βˆ’5(βˆ’7)=35 -5(-7) = 35

Multiplying βˆ’5-5 by βˆ’7-7 gives us 3535:

35=35 35 = 35

Since both sides of the equation are equal, the solution x=βˆ’7x = -7 is verified. This step is essential because it eliminates the possibility of errors in our calculations. It confirms that the value we found for xx is indeed the correct solution to the equation. Verification is a fundamental part of the problem-solving process in mathematics, providing a crucial check on the accuracy of our work. This step gives us confidence in our solution and ensures that we have correctly solved the equation.

To further support our conclusion that the equation βˆ’5x=35-5x = 35 has one solution, we can demonstrate that only x=βˆ’7x = -7 satisfies the equation. We can do this by testing other values of xx and showing that they do not result in a true statement. Let's try two different values for xx, say x=0x = 0 and x=1x = 1.

  1. Testing x = 0:

    Substitute x=0x = 0 into the equation βˆ’5x=35-5x = 35:

    βˆ’5(0)=35 -5(0) = 35

    0=35 0 = 35

    This statement is false, so x=0x = 0 is not a solution.

  2. Testing x = 1:

    Substitute x=1x = 1 into the equation βˆ’5x=35-5x = 35:

    βˆ’5(1)=35 -5(1) = 35

    βˆ’5=35 -5 = 35

    This statement is also false, so x=1x = 1 is not a solution.

These examples clearly show that only x=βˆ’7x = -7 makes the equation true. By testing other values, we reinforce the conclusion that the equation has only one solution. This method of testing different values is a valuable tool for understanding the nature of solutions in equations. It helps to visualize why a specific value is the solution and why other values are not.

In conclusion, the equation βˆ’5x=35-5x = 35 has one solution, which is x=βˆ’7x = -7. We arrived at this conclusion by isolating the variable xx and solving for its value. Furthermore, we verified our solution by substituting x=βˆ’7x = -7 back into the original equation, confirming that it satisfies the equation. To further support our conclusion, we tested two other values, x=0x = 0 and x=1x = 1, and demonstrated that they do not satisfy the equation. This comprehensive analysis provides a clear understanding of the nature of solutions for the given linear equation. The ability to determine the number of solutions and find their values is a fundamental skill in algebra, essential for solving various mathematical problems. By mastering these concepts, you can confidently approach and solve linear equations, paving the way for more advanced mathematical studies. Understanding the properties of linear equations and their solutions is a crucial step in building a strong foundation in mathematics.