Solving Algebraic Equations A Step By Step Guide
1. 2z = 6
In the realm of algebra, solving equations is a fundamental skill. This section will help you in understanding how to find the value of unknown variables. Let's start with the equation 2z = 6. Our goal is to isolate the variable z on one side of the equation. To do this, we need to undo the operation that's being applied to z, which in this case is multiplication by 2. The inverse operation of multiplication is division, so we divide both sides of the equation by 2.
When we divide both sides of the equation 2z = 6 by 2, we get:
(2z) / 2 = 6 / 2
This simplifies to:
z = 3
Therefore, the solution to the equation 2z = 6 is z = 3. This means that if we substitute 3 for z in the original equation, we get a true statement: 2 * 3 = 6. This simple example illustrates the core principle of solving algebraic equations: performing the same operation on both sides of the equation to maintain equality while isolating the variable.
To solidify your understanding, consider the implications of this solution. The equation 2z = 6 represents a linear relationship where the value of z is directly proportional to the result when multiplied by 2. Finding the value of z that satisfies this equation allows us to understand the specific point on this linear relationship where the result is exactly 6. This basic concept forms the foundation for solving more complex algebraic problems, including those involving multiple variables and operations. By mastering these fundamentals, you'll be well-equipped to tackle a wide range of mathematical challenges.
2. 7a = 63
Continuing our journey into the world of algebraic equations, let's tackle the equation 7a = 63. Similar to the previous example, our primary objective is to isolate the variable a to determine its value. In this equation, a is being multiplied by 7. To isolate a, we need to perform the inverse operation, which is division. We will divide both sides of the equation by 7 to maintain the balance and equality.
Dividing both sides of the equation 7a = 63 by 7, we have:
(7a) / 7 = 63 / 7
This simplifies to:
a = 9
Thus, the solution to the equation 7a = 63 is a = 9. This signifies that when we substitute 9 for a in the original equation, the equation holds true: 7 * 9 = 63. This simple yet crucial step of verification ensures the accuracy of our solution and reinforces the fundamental principles of algebraic manipulation.
Let's delve deeper into the practical implications of this solution. The equation 7a = 63 can be interpreted as a linear relationship where the value of a determines the result when multiplied by 7. Finding the solution a = 9 pinpoints the exact value of a that yields 63 in this linear relationship. This understanding is vital for various real-world applications, from calculating proportions to solving complex mathematical models. By consistently applying the principles of inverse operations and maintaining equation balance, you'll develop a robust skill set for solving algebraic problems efficiently and accurately. This foundation is crucial for advancing to more challenging topics in mathematics and related fields.
3. 6y = 30
Now, let's move on to the equation 6y = 30. As with the previous examples, the core strategy remains the same: isolate the variable. In this case, we need to isolate y. Currently, y is being multiplied by 6. To undo this multiplication and isolate y, we apply the inverse operation, which is division. We must divide both sides of the equation by 6 to maintain the equality.
Dividing both sides of the equation 6y = 30 by 6 gives us:
(6y) / 6 = 30 / 6
Simplifying this, we get:
y = 5
Therefore, the solution to the equation 6y = 30 is y = 5. To confirm the correctness of this solution, we can substitute 5 for y in the original equation: 6 * 5 = 30, which verifies our solution. This practice of verifying solutions is an essential step in problem-solving, ensuring accuracy and building confidence in your algebraic skills.
Consider the broader implications of this solution. The equation 6y = 30 represents a proportional relationship between y and the result when y is multiplied by 6. The solution y = 5 identifies the specific value of y that results in 30. This type of proportional relationship is fundamental in many areas of mathematics and science, from calculating rates and ratios to understanding linear functions. By mastering the techniques to solve these simple equations, you are building a foundation for more advanced mathematical concepts. The consistent application of inverse operations and the habit of verifying solutions will prove invaluable as you progress in your mathematical studies.
4. u/4 = 8
Moving on to a slightly different type of equation, let's consider u/4 = 8. Here, the variable u is being divided by 4. To isolate u, we need to perform the inverse operation of division, which is multiplication. We will multiply both sides of the equation by 4 to maintain balance and solve for u.
Multiplying both sides of the equation u/4 = 8 by 4, we get:
(u/4) * 4 = 8 * 4
This simplifies to:
u = 32
So, the solution to the equation u/4 = 8 is u = 32. To ensure our solution is correct, we substitute 32 for u in the original equation: 32 / 4 = 8, which confirms our answer. This verification step is crucial for reinforcing the understanding of equation solving and ensuring accuracy.
The equation u/4 = 8 represents a scenario where u is divided into four equal parts, and each part equals 8. Finding the value of u helps us understand the total quantity before division. This type of problem is common in various applications, such as calculating total quantities from fractional parts or determining initial values before a reduction. By understanding how to solve equations involving division, you are expanding your ability to tackle a wider range of mathematical problems. The key takeaway is to always apply the inverse operation to isolate the variable and maintain the balance of the equation. This method, coupled with verification, is a powerful tool in your algebraic toolkit.
5. u/3 = 6
Let's examine another equation involving division: u/3 = 6. Similar to the previous example, our goal is to isolate the variable u. In this equation, u is being divided by 3. To isolate u, we must perform the inverse operation, which is multiplication. We will multiply both sides of the equation by 3 to maintain equality and solve for u.
Multiplying both sides of the equation u/3 = 6 by 3, we have:
(u/3) * 3 = 6 * 3
This simplifies to:
u = 18
Thus, the solution to the equation u/3 = 6 is u = 18. To verify this solution, we substitute 18 for u in the original equation: 18 / 3 = 6, which confirms our result. This step of verification is essential for ensuring the accuracy of our solution and reinforcing our understanding of algebraic principles.
The equation u/3 = 6 illustrates a situation where u is divided into three equal parts, and each part is equal to 6. Finding the value of u allows us to determine the original quantity before it was divided. This concept is widely applicable in various real-world scenarios, such as calculating the total amount when given a fraction of it or determining the original price after a discount. By mastering the technique of solving equations involving division, you are enhancing your ability to solve a broader range of mathematical problems. The fundamental principle remains consistent: apply the inverse operation to isolate the variable while maintaining the equation's balance. This approach, combined with verification, is a powerful tool in your algebraic toolkit, enabling you to tackle more complex challenges with confidence.
6. 5c = 50
Now, let's consider the equation 5c = 50. Our objective, as always, is to isolate the variable c. In this equation, c is being multiplied by 5. To isolate c, we need to perform the inverse operation, which is division. We will divide both sides of the equation by 5 to maintain the equality and solve for c.
Dividing both sides of the equation 5c = 50 by 5, we obtain:
(5c) / 5 = 50 / 5
This simplifies to:
c = 10
Therefore, the solution to the equation 5c = 50 is c = 10. To verify the accuracy of our solution, we substitute 10 for c in the original equation: 5 * 10 = 50, which confirms our answer. This verification step is crucial for reinforcing our understanding of algebraic principles and ensuring the correctness of our solutions.
The equation 5c = 50 represents a proportional relationship where the value of c determines the result when multiplied by 5. The solution c = 10 pinpoints the specific value of c that yields 50 in this proportional relationship. This type of relationship is commonly encountered in various applications, such as calculating the cost of multiple items or determining the total amount based on a unit rate. By mastering the technique of solving equations involving multiplication, you are expanding your ability to tackle a wider range of mathematical problems. The key principle is to apply the inverse operation to isolate the variable while maintaining the balance of the equation. This approach, combined with the practice of verification, is a valuable tool in your algebraic toolkit, empowering you to solve more complex problems with confidence.
7. 2c = 8
Let's examine the equation 2c = 8. Our primary goal remains consistent: to isolate the variable c. In this equation, c is being multiplied by 2. To isolate c, we need to perform the inverse operation, which is division. We will divide both sides of the equation by 2 to maintain the equality and solve for c.
Dividing both sides of the equation 2c = 8 by 2, we get:
(2c) / 2 = 8 / 2
This simplifies to:
c = 4
Thus, the solution to the equation 2c = 8 is c = 4. To ensure the accuracy of our solution, we substitute 4 for c in the original equation: 2 * 4 = 8, which confirms our result. This step of verification is crucial for reinforcing our understanding of algebraic principles and ensuring the correctness of our solutions.
The equation 2c = 8 represents a proportional relationship where the value of c determines the result when multiplied by 2. The solution c = 4 identifies the specific value of c that yields 8 in this proportional relationship. This type of relationship is commonly encountered in various applications, such as calculating the total cost when given the price per unit or determining the total quantity based on a rate. By mastering the technique of solving equations involving multiplication, you are expanding your ability to tackle a wider range of mathematical problems. The fundamental principle is to apply the inverse operation to isolate the variable while maintaining the balance of the equation. This approach, combined with the practice of verification, is a valuable tool in your algebraic toolkit, empowering you to solve more complex problems with confidence.
8. 7y = 70
Moving on to the equation 7y = 70, our objective remains to isolate the variable y. In this case, y is being multiplied by 7. To isolate y, we need to perform the inverse operation, which is division. We will divide both sides of the equation by 7 to maintain the equality and solve for y.
Dividing both sides of the equation 7y = 70 by 7, we obtain:
(7y) / 7 = 70 / 7
This simplifies to:
y = 10
Therefore, the solution to the equation 7y = 70 is y = 10. To verify the accuracy of our solution, we substitute 10 for y in the original equation: 7 * 10 = 70, which confirms our answer. This verification step is crucial for reinforcing our understanding of algebraic principles and ensuring the correctness of our solutions.
The equation 7y = 70 represents a proportional relationship where the value of y determines the result when multiplied by 7. The solution y = 10 pinpoints the specific value of y that yields 70 in this proportional relationship. This type of relationship is commonly encountered in various applications, such as calculating the total cost when given the price per unit or determining the total quantity based on a rate. By mastering the technique of solving equations involving multiplication, you are expanding your ability to tackle a wider range of mathematical problems. The key principle is to apply the inverse operation to isolate the variable while maintaining the balance of the equation. This approach, combined with the practice of verification, is a valuable tool in your algebraic toolkit, empowering you to solve more complex problems with confidence.
9. u/4 = 7
Now, let's consider the equation u/4 = 7. Here, the variable u is being divided by 4. To isolate u, we need to perform the inverse operation of division, which is multiplication. We will multiply both sides of the equation by 4 to maintain balance and solve for u.
Multiplying both sides of the equation u/4 = 7 by 4, we get:
(u/4) * 4 = 7 * 4
This simplifies to:
u = 28
So, the solution to the equation u/4 = 7 is u = 28. To ensure our solution is correct, we substitute 28 for u in the original equation: 28 / 4 = 7, which confirms our answer. This verification step is crucial for reinforcing the understanding of equation solving and ensuring accuracy.
The equation u/4 = 7 represents a scenario where u is divided into four equal parts, and each part equals 7. Finding the value of u helps us understand the total quantity before division. This type of problem is common in various applications, such as calculating total quantities from fractional parts or determining initial values before a reduction. By understanding how to solve equations involving division, you are expanding your ability to tackle a wider range of mathematical problems. The key takeaway is to always apply the inverse operation to isolate the variable and maintain the balance of the equation. This method, coupled with verification, is a powerful tool in your algebraic toolkit.
10. c/7 = 6
Let's address the equation c/7 = 6. In this equation, the variable c is being divided by 7. To isolate c, we need to perform the inverse operation of division, which is multiplication. We will multiply both sides of the equation by 7 to maintain the balance and solve for c.
Multiplying both sides of the equation c/7 = 6 by 7, we obtain:
(c/7) * 7 = 6 * 7
This simplifies to:
c = 42
Therefore, the solution to the equation c/7 = 6 is c = 42. To verify this solution, we substitute 42 for c in the original equation: 42 / 7 = 6, which confirms our result. This verification step is essential for reinforcing our understanding of algebraic principles and ensuring the accuracy of our solutions.
The equation c/7 = 6 represents a situation where c is divided into seven equal parts, and each part is equal to 6. Finding the value of c allows us to determine the original quantity before it was divided. This concept is widely applicable in various real-world scenarios, such as calculating the total amount when given a fraction of it or determining the original price after a discount. By mastering the technique of solving equations involving division, you are enhancing your ability to solve a broader range of mathematical problems. The fundamental principle remains consistent: apply the inverse operation to isolate the variable while maintaining the equation's balance. This approach, combined with verification, is a powerful tool in your algebraic toolkit, enabling you to tackle more complex challenges with confidence.
11. 7x = 42
As we progress in our exploration of algebraic equations, let's consider the equation 7x = 42. Our goal remains consistent: to isolate the variable, in this case, x. Here, x is being multiplied by 7. To isolate x, we need to perform the inverse operation, which is division. We will divide both sides of the equation by 7 to maintain equality and solve for x.
Dividing both sides of the equation 7x = 42 by 7, we get:
(7x) / 7 = 42 / 7
This simplifies to:
x = 6
Therefore, the solution to the equation 7x = 42 is x = 6. To ensure the accuracy of our solution, we substitute 6 for x in the original equation: 7 * 6 = 42, which confirms our result. This verification step is crucial for reinforcing our understanding of algebraic principles and ensuring the correctness of our solutions.
The equation 7x = 42 represents a proportional relationship where the value of x determines the result when multiplied by 7. The solution x = 6 pinpoints the specific value of x that yields 42 in this proportional relationship. This type of relationship is commonly encountered in various applications, such as calculating the total cost when given the price per unit or determining the total quantity based on a rate. By mastering the technique of solving equations involving multiplication, you are expanding your ability to tackle a wider range of mathematical problems. The key principle is to apply the inverse operation to isolate the variable while maintaining the balance of the equation. This approach, combined with the practice of verification, is a valuable tool in your algebraic toolkit, empowering you to solve more complex problems with confidence.
12. 7x = 21
Now, let's tackle the equation 7x = 21. Similar to the previous examples, our objective is to isolate the variable x. In this equation, x is being multiplied by 7. To isolate x, we need to perform the inverse operation, which is division. We will divide both sides of the equation by 7 to maintain the equality and solve for x.
Dividing both sides of the equation 7x = 21 by 7, we obtain:
(7x) / 7 = 21 / 7
This simplifies to:
x = 3
Therefore, the solution to the equation 7x = 21 is x = 3. To verify the accuracy of our solution, we substitute 3 for x in the original equation: 7 * 3 = 21, which confirms our answer. This verification step is crucial for reinforcing our understanding of algebraic principles and ensuring the correctness of our solutions.
The equation 7x = 21 represents a proportional relationship where the value of x determines the result when multiplied by 7. The solution x = 3 pinpoints the specific value of x that yields 21 in this proportional relationship. This type of relationship is commonly encountered in various applications, such as calculating the total cost when given the price per unit or determining the total quantity based on a rate. By mastering the technique of solving equations involving multiplication, you are expanding your ability to tackle a wider range of mathematical problems. The key principle is to apply the inverse operation to isolate the variable while maintaining the balance of the equation. This approach, combined with the practice of verification, is a valuable tool in your algebraic toolkit, empowering you to solve more complex problems with confidence.
13. 3z = 6
Let's consider the equation 3z = 6. As with our previous examples, the goal is to isolate the variable z. In this equation, z is being multiplied by 3. To isolate z, we need to perform the inverse operation, which is division. We will divide both sides of the equation by 3 to maintain the equality and solve for z.
Dividing both sides of the equation 3z = 6 by 3, we get:
(3z) / 3 = 6 / 3
This simplifies to:
z = 2
Therefore, the solution to the equation 3z = 6 is z = 2. To verify the accuracy of our solution, we substitute 2 for z in the original equation: 3 * 2 = 6, which confirms our answer. This verification step is crucial for reinforcing our understanding of algebraic principles and ensuring the correctness of our solutions.
The equation 3z = 6 represents a proportional relationship where the value of z determines the result when multiplied by 3. The solution z = 2 pinpoints the specific value of z that yields 6 in this proportional relationship. This type of relationship is commonly encountered in various applications, such as calculating the total cost when given the price per unit or determining the total quantity based on a rate. By mastering the technique of solving equations involving multiplication, you are expanding your ability to tackle a wider range of mathematical problems. The key principle is to apply the inverse operation to isolate the variable while maintaining the balance of the equation. This approach, combined with the practice of verification, is a valuable tool in your algebraic toolkit, empowering you to solve more complex problems with confidence.
14. v/6 = 4
Let's examine the equation v/6 = 4. Here, the variable v is being divided by 6. To isolate v, we need to perform the inverse operation of division, which is multiplication. We will multiply both sides of the equation by 6 to maintain balance and solve for v.
Multiplying both sides of the equation v/6 = 4 by 6, we get:
(v/6) * 6 = 4 * 6
This simplifies to:
v = 24
So, the solution to the equation v/6 = 4 is v = 24. To ensure our solution is correct, we substitute 24 for v in the original equation: 24 / 6 = 4, which confirms our answer. This verification step is crucial for reinforcing the understanding of equation solving and ensuring accuracy.
The equation v/6 = 4 represents a scenario where v is divided into six equal parts, and each part equals 4. Finding the value of v helps us understand the total quantity before division. This type of problem is common in various applications, such as calculating total quantities from fractional parts or determining initial values before a reduction. By understanding how to solve equations involving division, you are expanding your ability to tackle a wider range of mathematical problems. The key takeaway is to always apply the inverse operation to isolate the variable and maintain the balance of the equation. This method, coupled with verification, is a powerful tool in your algebraic toolkit.
15. u/7 = 6
Finally, let's consider the equation u/7 = 6. In this equation, the variable u is being divided by 7. To isolate u, we need to perform the inverse operation of division, which is multiplication. We will multiply both sides of the equation by 7 to maintain the balance and solve for u.
Multiplying both sides of the equation u/7 = 6 by 7, we obtain:
(u/7) * 7 = 6 * 7
This simplifies to:
u = 42
Therefore, the solution to the equation u/7 = 6 is u = 42. To verify this solution, we substitute 42 for u in the original equation: 42 / 7 = 6, which confirms our result. This verification step is essential for reinforcing our understanding of algebraic principles and ensuring the accuracy of our solutions.
The equation u/7 = 6 represents a situation where u is divided into seven equal parts, and each part is equal to 6. Finding the value of u allows us to determine the original quantity before it was divided. This concept is widely applicable in various real-world scenarios, such as calculating the total amount when given a fraction of it or determining the original price after a discount. By mastering the technique of solving equations involving division, you are enhancing your ability to solve a broader range of mathematical problems. The fundamental principle remains consistent: apply the inverse operation to isolate the variable while maintaining the equation's balance. This approach, combined with verification, is a powerful tool in your algebraic toolkit, enabling you to tackle more complex challenges with confidence.
In conclusion, mastering the art of solving algebraic equations is a fundamental skill in mathematics. Throughout this comprehensive guide, we've explored various equations and applied the core principle of isolating the variable by using inverse operations. Whether it's dealing with multiplication or division, the key is to maintain the balance of the equation by performing the same operation on both sides. The importance of verifying solutions cannot be overstated, as it ensures accuracy and reinforces understanding. By consistently applying these techniques, you'll build a strong foundation in algebra, empowering you to tackle more complex problems with confidence. This skill set is not only crucial for academic success but also for real-world applications where mathematical reasoning plays a pivotal role. Keep practicing, and you'll find that solving algebraic equations becomes second nature.
