Question

unit 2 lesson 16: solving systems by elimination (part 3) practice
here is a system of equations with a solution: $\begin{cases}p + 8q=-8\\frac{1}{2}p + 5q=-5end{cases}$
a. which system of equations is equivalent to this system?
system a: $\begin{cases}p + 8q=-8\\p + 10q=-10end{cases}$
system b: $\begin{cases}p + q=-1\\p + 10q=-5end{cases}$
b. true or false: the following statements correctly explain how you know the equivalent system has the same solution as the original system.
multiplying both sides of an equation by the same number keeps the two sides equal, so the $(p,q)$ pair that makes the second equation true in the original system also makes the new equation true. this means that the same $(p,q)$ pair also makes the new system true.

Explanation:

Step1: Manipulate the second - original equation

Multiply the equation $\frac{1}{2}p + 5q=-5$ by 2. We get $p + 10q=-10$. The first original equation is $p + 8q=-8$.
The original system

$$\begin{cases}p + 8q=-8\\\frac{1}{2}p+5q = - 5\end{cases}$$

is equivalent to the system

$$\begin{cases}p + 8q=-8\\p + 10q=-10\end{cases}$$

(System A) because multiplying an equation in a system of linear equations by a non - zero number does not change the solution of the system.

Step2: Analyze the statement about equivalent systems

The statement "Multiplying both sides of an equation by the same number keeps the two sides equal, so the $(p,q)$ pair that makes the second equation true in the original system also makes the new equation true. This means that the same $(p,q)$ pair also makes the new system true" is a correct principle for equivalent systems of linear equations.

Answer:

a. System A:

$$\begin{cases}p + 8q=-8\\p + 10q=-10\end{cases}$$

b. True