Question

Which statement is true about their work? Neither student solved for k correctly because K = 2 and StartFraction 1 over 8 EndFraction. Only Adler solved for k correctly because the inverse of addition is subtraction. Only Erika solved for k correctly because the opposite of One-half is Negative one-half. Both Adler and Erika solved for k correctly because either the addition property of equality or the subtraction property of equality can be used to solve for k.

Answer 1

Question:

Adler and Erika solved the same equation using the calculations below.

Adler’s Work

$\frac{13}{8}= k + \frac{1}{2}$

$\frac{13}{8} - \frac{1}{2} = k + \frac{1}{2} -\frac{1}{2}$

$\frac{9}{8}= k$

Erika’s Work

$\frac{13}{8}= k + \frac{1}{2}$

$\frac{13}{8}+ (-\frac{1}{2}) = k + \frac{1}{2} + (-\frac{1}{2}).$

$\frac{9}{8}= k$

Which statement is true about their work?

Answer:

Both Adler and Erika solved for k correctly because either the addition property of equality or the subtraction property of equality can be used to solve for k.

Step-by-step explanation:

Given

$\frac{13}{8}= k + \frac{1}{2}$

Required

What is true about Adler and Erika's workings

Analyzing Adler's work;

$\frac{13}{8}= k + \frac{1}{2}$

Adler subtracted 1/2 from both sides

$\frac{13}{8} - \frac{1}{2} = k + \frac{1}{2} -\frac{1}{2}$

Solving the expression on the left-hand side

$\frac{13}{8} - \frac{1}{2} = \frac{13 - 4}{8} = \frac{9}{8}$

Solving the expression on the right-hand side

$k + \frac{1}{2} - \frac{1}{2} = k$

Hence:

$\frac{9}{8}= k$

So, Adler's workings is correct

Analyzing Erika's work;

$\frac{13}{8}= k + \frac{1}{2}$

Erika added -1/2 to both sides

$\frac{13}{8}+ (-\frac{1}{2}) = k + \frac{1}{2} + (-\frac{1}{2}).$

Solving the expression on the left-hand side

$\frac{13}{8} + (-\frac{1}{2}) = \frac{13}{8} - \frac{1}{2} = \frac{13 - 4}{8} = \frac{9}{8}$

Solving the expression on the right-hand side

$k + \frac{1}{2} + (-\frac{1}{2}) =k + \frac{1}{2} - \frac{1}{2} = k$

Hence:

$\frac{9}{8}= k$

So, Erika's workings is correct

Both workings are correct