few_shots: 
Example:
Concept/skill: 
Apply and extend previous understandings of multiplication and division and of fractions to multiply and divide rational numbers.

Problem/activity:
For each pair of numbers, determine if the values are equivalent. Explain your reasoning for each one.
a.
$$-0.25$$
and
$$\frac{1}{-4}$$
b.
$$-\left ( \frac{1}{4} \right )$$
and
$$\frac{-1}{-4}$$
c.
$$\frac{-1}{4}$$
and
$$-\frac{1}{4}$$


Answer: 
Yes. This problem involves understanding fractions and division to divide rational numbers, such as 1, -4, 4, and -1. This problem also involves an understanding of multiplication: if p and q are integers, then -(p/q) = (-p)/q = p/(-q).

Now, determine whether the following problem or activity can enable students to understand the concepts or skills in the provided description.

shot_info: fl_problem_001097_pos_0

7
standard_description: Prove that, given a system of two equations in two variables, replacing one equation by the sum of that equation and a multiple of the other produces a system with the same solutions.
problem_activity: Task
Lisa is working with the system of equations $x + 2y = 7$ and $2x - 5y = 5$. She multiplies the first equation by 2 and then subtracts the second equation to find $9y = 9$, telling her that $y = 1$. Lisa then finds that $x = 5$. Thinking about this procedure, Lisa wonders
There are lots of ways I could go about solving this problem. I could add 5 times the first equation and twice the second or I could multiply the first equation by -2 and add the second. I seem to find that there is only one solution to the two equations but I wonder if I will get the same solution if I use a different method?
What is the answer to Lisa's question? Explain.
Does the answer to (a) change if we have a system of two equations in two unknowns with no solutions? What if there are infinitely many solutions?


aligns: yes