To find the first error that Liz made in her work, we need to examine each step of the given solution carefully.
Liz's Work:
[tex]\[
\begin{array}{ll}
& 27(36) \\
\text{Step 1} & 27(3+12) \\
\text{Step 2} & 27(3) + 27(12) \\
\text{Step 3} & 81 + 324 \\
\text{Step 4} & 405 \\
\end{array}
\][/tex]
We start by analyzing each step:
1. Step 1: Liz writes [tex]\(27(36)\)[/tex] as [tex]\(27(3+12)\)[/tex].
- Here, 36 is decomposed incorrectly. If we want to use friendlier numbers, a more appropriate decomposition of 36 would be into numbers like [tex]\(27(6+30)\)[/tex] or [tex]\(27(30+6)\)[/tex]. Let's keep this point in mind.
2. Step 2: Liz applies the distributive property to [tex]\(27(3+12)\)[/tex]:
[tex]\[
27(3) + 27(12)
\][/tex]
- This step correctly follows the distributive property, [tex]\(a(b+c) = ab + ac\)[/tex], assuming that [tex]\(3 + 12\)[/tex] was correctly decomposed. Since Step 1 has an incorrect decomposition, this step is based on an incorrect assumption but doesn't contain any new errors.
3. Step 3: Liz calculates:
[tex]\[
27(3) = 81 \quad \text{and} \quad 27(12) = 324 \implies 81+324
\][/tex]
- The calculations here [tex]\(27 \times 3 = 81\)[/tex] and [tex]\(27 \times 12 = 324\)[/tex] are correct.
4. Step 4: Adding the results from Step 3:
[tex]\[
81 + 324 = 405
\][/tex]
- The final addition here is correct.
The root of the problem lies in Step 1. Liz decomposed 36 incorrectly. For a correct decomposition to leverage the distributive property better, a possible way would have been recognizing numbers closer to the operations involved with more accessible arithmetic.
Hence, the first error Liz made was in Step 1, and the correction would have been:
[tex]\[
27(36) \rightarrow 27(30 + 6)
\][/tex]
Therefore, the corrected steps should be:
1. [tex]\(27(30 + 6)\)[/tex]
2. [tex]\(27(30) + 27(6)\)[/tex]
3. [tex]\(810 + 162\)[/tex]
4. [tex]\(972\)[/tex]
Conclusion:
The first error that Liz made was in the decomposition in Step 1. The corrected Step 1 should have been [tex]\(27(30 + 6)\)[/tex].
