Pregunta
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29. Critique Reasoning Jamie says that the expressions \( 6 x-2 x+4 \) and \( 4(x+1) \) are not equivalent because one expression has a term that is subtracted and the other does not. Do you agree? Explain.

Ask by Reeves Conner. in the United States
Jan 26,2025

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The two expressions \(6x - 2x + 4\) and \(4(x + 1)\) are equivalent because both simplify to \(4x + 4\).

Solución

No, I do not agree with Jamie's reasoning. While Jamie correctly observes that one expression involves subtraction and the other does not, this observation alone does not determine whether the two expressions are equivalent. To assess their equivalence, let's simplify both expressions: 1. **Simplifying \(6x - 2x + 4\):** Combine like terms: \[ 6x - 2x + 4 = (6x - 2x) + 4 = 4x + 4 \] 2. **Simplifying \(4(x + 1)\):** Apply the distributive property: \[ 4(x + 1) = 4 \cdot x + 4 \cdot 1 = 4x + 4 \] After simplification, both expressions reduce to \(4x + 4\). This means that **the two expressions are indeed equivalent**, despite one initially featuring a subtraction term and the other not. Jamie's reasoning overlooks the fact that the subtraction in the first expression effectively combines with the other terms to produce the same result as the second expression. **Key Takeaway:** When determining the equivalence of algebraic expressions, it's essential to simplify each expression fully. Differences in appearance—such as the presence or absence of subtraction—do not necessarily indicate that the expressions are not equivalent.

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Extra Insights

Jamie’s reasoning misses a crucial mathematical step: the simplification of both expressions. Let's break it down! The expression \( 6x - 2x + 4 \) simplifies to \( 4x + 4 \) after combining like terms. On the other hand, \( 4(x + 1) \) also simplifies to \( 4x + 4 \) when distributed. Hence, both expressions are indeed equivalent despite initial appearances regarding addition and subtraction. Understanding algebraic equivalence is key here! It’s important to remember that subtraction can be thought of as adding a negative. So, in this case, the negative sign in \( -2x \) is just adding a different term which can lead to the same result when simplified properly. Always expand and simplify expressions to see their true forms!

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7. Efectueaza sis scrie rezultatul sub formă de putere: \( \begin{array}{lll}\text { a) } \frac{18}{5} \cdot\left(\frac{18}{5}\right)^{2}= & \text { b) }\left(\frac{6}{5}\right)^{2} \cdot\left(\frac{6}{5}\right)^{3} \cdot \frac{6}{5}= & \text { c) }\left(\frac{19}{5}\right)^{5} \cdot\left(\frac{19}{5}\right)^{16}= \\ \begin{array}{lll}\text { d) } \frac{3}{2} \cdot\left(\frac{3}{2}\right)^{3} \cdot\left(\frac{3}{2}\right)^{0} \cdot\left(\frac{3}{2}\right)^{4}= & \text { e) }\left[\left(\frac{28}{5}\right)^{2}\right]^{3}= & \text { f) }\left[\left(\frac{5}{6}\right)^{6}\right]^{7}= \\ \text { g) }\left[\left(\frac{24}{5}\right)^{2} \cdot\left(\frac{24}{5}\right)^{3}\right]^{8}= & \text { h) }\left[\frac{5}{7} \cdot\left(\frac{5}{7}\right)^{0} \cdot\left(\frac{5}{7}\right)^{4}\right]^{5}= & \text { i) }\left(\frac{29}{10}\right)^{10}:\left(\frac{29}{10}\right)^{7}=\end{array} \\ \left.\left.\begin{array}{lll}\text { j) }\left(\frac{1}{3}\right)^{17}: \frac{1}{3}= & \left.\text { k) }\left(\frac{3}{7}\right)^{11} \cdot\left(\frac{9}{49}\right)^{3}:\left(\frac{3}{7}\right)^{15}=1\right)\end{array}\right]\left(1 \frac{1}{2}\right)^{2}\right]^{8}:\left(\frac{3}{2}\right)^{13}= \\ \text { m) }\left(\frac{9}{10}\right)^{7} \cdot\left(\frac{1}{5}\right)^{7}= & \text { n) }\left(\frac{5}{2}\right)^{10} \cdot\left(\frac{8}{5}\right)^{10}: 2^{10}= & \text { o) } 9^{3} \cdot\left(\frac{7}{10}\right)^{3}:\left(\frac{63}{10}\right)^{3}= \\ \text { p) }\left[\left(\frac{1}{5}\right)^{7}\right]^{2} \cdot 6^{14}:\left(\frac{6}{5}\right)^{14}= & \text { q) }\left(\frac{5}{2}\right)^{7}:\left(\frac{5}{2}\right)^{5}= & \end{array} \)
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