This paper demonstrates how the properties of real numbers are applied to simplify three algebraic expressions using the distributive property and combining like terms. Through worked examples, the paper shows step-by-step how parentheses are cleared by multiplying factors through terms, how like terms are identified and combined, and how logical reasoning guides the order of operations. A concluding discussion explains why understanding real number properties is foundational to algebra and how algebraic thinking appears in everyday situations such as calculating paint coverage or managing time.
In mathematics, a real number is assigned to each statement written in a language, within a range from 0 to 1, where 1 means that the statement is completely true and 0 means that the statement is completely false, while values between 0 and 1 represent that the statement is partly true to a given, quantifiable extent. This framework makes it possible to analyze a distribution of statements for their truth content, identify data patterns, make inferences and predictions, and model how processes operate. Because variables in algebra simply represent real numbers, understanding the properties of real numbers is essential before evaluating or simplifying any algebraic expression.
The first expression to simplify is 2a(a β 5) + 4(a β 5).
Beginning with the distributive property, multiply 2a by each term inside the first set of parentheses:
2aΒ² β 10a + 4(a β 5)
Next, multiply 4 by each term inside the second set of parentheses. The distributive property removes the parentheses:
2aΒ² β 10a + 4a β 20
Since β10a and 4a are like terms, combine them by adding their coefficients:
β10a + 4a = β6a
The expression is now fully simplified:
2aΒ² β 6a β 20
The second expression to simplify is 3(w β 4) β 5(w β 6).
Multiply 3 by each term inside the first set of parentheses:
3w β 12 β 5(w β 6)
Then multiply β5 by each term inside the second set of parentheses. The distributive property removes the parentheses:
3w β 12 β 5w + 30
Identify and combine like terms. The terms 3w and β5w are like terms:
3w β 5w = β2w
The constant terms β12 and 30 are also like terms:
β12 + 30 = 18
The expression is now fully simplified:
β2w + 18
Note on the worked example in the source: The original problem includes the step 2w β 3 + 3(w β 4) β 5(w β 6), suggesting an additional term of 2w β 3 was present. Following that version through: combining 2w, 3w, and β5w gives 0w (the w terms cancel); combining the constants β3, β12, and +30 gives 15. That version simplifies to 15.
The third expression to simplify is 0.05(0.3m + 35n) β 0.8(β0.09n β 22m).
Begin by multiplying 0.05 by each term inside the first set of parentheses:
0.015m + 1.75n β 0.8(β0.09n β 22m)
"Decimal coefficients simplified using distribution"
"Logic, reasoning, and real-world algebraic applications"
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