Numerical and Word Problems, FAQs, and a Summary on the Algebraic Equations
Table of Contents:
- Numerical Problems Based on the Algebraic Equations
- Word Problems based on Algebraic Expressions
- FAQs Based on The Topics of Algebraic Equations
- Summary Of Algebraic Equations
Numerical Problems Based on the Algebraic Equations
Q1: Simplify the given expression, which is of teh form: 7a + 4a – 3a.
A1: In such scenarios wherein we encounter like terms (all have the variable a), we combine them in the following manner: (7 + 4 – 3)a = 8a.
Explanation: The best example would be: suppose that we examine the three terms in question: 7a, 4a, and –3a.
In such an instance, they all contain the variable a, which means they are like terms. The explanation being: they have identical variable parts. The usage of combining coefficients is such that it helps to understand actual-world situations. We add and subtract the numerical coefficients: 7 + 4 = 11; then 11 – 3 = 8.
Q2. Simplify the expression that is given as: 7a + 4a – 3a.
A2: Please note the use of like terms (that is, all having the variable a). In such types of instances, we combine them in the following manner: (7 + 4 – 3)a = 8a.
Explanation: Start by examining the three terms, which are: 7a, 4a, and –3a. You will also notice that with the same variable a, like terms--including identical variable parts. Then, we move on to bring together the coefficients that help to solve the problem quickly. Further, we move on to add and subtract the numerical coefficients in the following manner: 7 + 4 = 11; then 11 – 3 = 8.
Q3. Simplify the given equation given as: 3n + 5 = 20.
A3: We need to start by separating the variable in hand. Then, we have to proceed as follows: 3n = 20 – 5, which presents us 15. This also means that n = 15 ÷ 3 equalling to 5.
Explanation: The equation 3n + 5 = 20 in words means: when three times a number gets added to five will equal twenty. Next, we move on to isolate n and we further subtract 5 from both sides. This helps to balance the equation: 3n = 15 (L.H.S. and R.H.S in equation). Then we proceed to divide both sides by the number 3, which provides us n = 5.
Word Problems based on Algebraic Expressions
- A shop sells t notebooks, each at ₹ 12. If the total revenue is ₹ (12t), and the shop owner pays ₹ 90 as rent, what is the profit expression if the cost per notebook is ₹ 8? Then find profit when t =10.
A1: We need to start by determining the revenue, which presents us with 12t. Next, we have to proceed to calculate the cost, that is, 8t + 90.
In such scenarios wherein we need to find profit, we ought to subtract the cost from revenue in the following manner: profit = 12t – (8t + 90), which ultimately provides us with 4t – 90.
When we move on to substitute t = 10, we proceed as follows: profit = 4×10 – 90 = 40 – 90, which fetches us –₹ 50 (that is, a loss of ₹ 50).
Explanation: In such instances wherein we encounter algebraic expressions, we need to represent everything using the variable t.
Then, we have to proceed to form the profit expression by following the fundamental principle. The best example would be: we examine the revenue calculation, which is 12×10=120; then we move on to calculate the cost that helps us understand the actual situation: 8×10+90=80+90=170.
Further, we proceed to determine profit=120–170=–50, which means a ₹ 50 loss. Note that when costs exceed revenue, we encounter a loss scenario, that is, the business operates at a deficit.
Q2. Sara is 4 years younger than her brother. If her brother's age is x years, express Sara's age. Then, if brother is 15, how old is Sara?
A2: We need to proceed by understanding that Sara's age will be x – 4. However, in such instances where x is equal to 15 is considered. Then, we move on to find that Sara's age will automatically become = 15 – 4 = 11 years.
Explanation: We have to start by identifying the variable x, and this is equivalent to the brother's age (that is, the given variable in the problem).
Sara is 4 years younger, which means that we need to proceed with the operation of subtraction by 4: x – 4. In such scenarios, we move on to plug in 15 for x, which presents us with 11.
Q3. Suppose a factory is able to produce n gadgets per day. However, if each gadget requires 0.5 litres of paint and costs ₹ 20 to manufacture (not counting paint). The paint costs ₹ 120 per litre. Write the expression for the total daily cost (paint + manufacturing). Then compute the cost of painting when n becomes 100.
A3: We need to proceed as follows: Paint cost per gadget is given by 0.5 litre × ₹ 120 = ₹ 60. In such instances, we move on to find that the total cost per gadget = ₹ 20 + ₹ 60 = ₹ 80. For n gadgets, we have to proceed with the expression: cost = 80n. Putting n=100, which presents us with → cost = ₹ 80×100 = ₹ 8,000.
Explanation: We have to start by expressing the paint cost and the manufacturing cost in one expression (that is, combining both cost components). Please note that because the paint cost is variable with n, we ought to include it in our calculations. In such scenarios, we move on to substitute n=100, which helps to get the actual number that we require for the solution.
FAQs Based on The Topics of Algebraic Equations
Q1: What exactly is a variable in algebra?
A1: A variable is a letter or symbol that is used to represent a quantity. The reason being: this quantity can change or remain unknown in a given situation.
The best example would be: suppose that you encounter the letter 'x' in an algebraic expression. In such scenarios, this letter serves as a placeholder for a number that we don't know yet.
Thus, it helps to understand the problem application in real-world situations. The usage of these letters and symbols is such that they can help to represent actual-world situations and apply concepts easily.
Q2: Why do we call some terms "like terms"?
A2: Terms are called like wherein they have the same variable(s) raised to the same powers. The explanation for this is that similarity allows them to be combined by adding or subtracting their coefficients.
The best example would be: suppose that you have terms like 3x and 5x. In such scenarios, both terms have the variable 'x' raised to the same power (which is 1), so they can be combined to fetch us 8x. Additionally, the variable parts are identical; only the numerical coefficients change when we combine them.
Q3: Can we add unlike terms directly in an expression?
A3: No--unlike terms have different variables or powers, so they cannot be merged. The reason being: they stay separate in the result. The best example that could be thought is: suppose that you have terms like 2x and 3y, or 4x² and 5x.
In such scenarios, you cannot combine them into a single term, as doing so will not fetch a mathematically correct outcome as expected. Also, unlike terms have different variable parts--a combination of different types of terms. Instead, you keep unlike ones separate; each maintains its identity in the final expression.
Q4: What is the difference between an expression and an equation?
A4: An expression is a combination of terms wherein it contains variables and numbers. For example, 2x + 5) without an equals sign.
In other words, it represents a mathematical phrase. An equation, on the other hand, contains the equals sign "=" and states that two expressions are equal.
The best example would be: 2x + 5 = 11. In such scenarios, an equation has the left-hand side (L.H.S.) and the right-hand side (R.H.S.)--these need to balanced compulsorily. This is mainly due to the fact that equations on both sides should represent the same value.
Q5: How do we solve a simple equation like 4x + 3 = 19?
A5: We isolate the variable with inverse operations--we undo what has been done to the variable. Reason being: this helps to find the unknown value.
One good example would be: suppose that we start with an equation, 4x + 3 = 19. In such scenarios, we first subtract 3 from both sides, which yields 4x = 16.
Then, the next step involves dividing both sides by the number 4. This is done in order to simplify the problem, and this will fetch us the answer: x = 4. All in all, we performed inverse operations to isolate the variable, and the equation remains balanced throughout the process.
Summary Of Algebraic Equations
1. Algebra is not as difficult as one thinks, and this is what differentiates it from others. Importantly, it only introduces symbols-for-numbers (variables) and fixed numbers (constants). This is done in such a way that unknown variables (not known factors) can be found out more easily. Moreover, this kind of mathematical system ensures that learners can grasp complex problems with enhanced capability.
- Expressions are built from terms in a highly sophisticated manner, where each term has a coefficient and a variable part or just a constant.
- Like-terms have identical variable parts and can be combined easily; unlike-terms cannot be processed in this manner.
Accordingly, this kind of classification system incorporated within algebraic expressions ensures that any kind of mathematical operation is worked out. In addition, learners can work on their computational skills based on this systematic approach.
- You form expressions from real-life statements in a timely manner, which enhances the retention of mathematical concepts.
Moreover, you then add/subtract expressions by combining like-terms and keeping unlike-terms separate. This kind of system greatly enhances the capability of students to solve practical problems, and this mainly happens as students start getting a grasp of real-world applications.
- Importantly, algebra uncovers patterns (number-patterns, shape-patterns) and general rules for many cases. This kind of pattern recognition system ensures that learners can identify mathematical relationships that are deeply rooted in their minds.
- An equation states equality between two expressions in such a way that mathematical balance is maintained. That is, solving it means finding the variable value that balances both sides greatly.
There is much evidence that has shown that this algebraic approach works, in the sense that systematic solving helps with mathematical reinforcement and enhances learning capabilities significantly.
