Question

9. Consider this system of equations, which has one solution:2x+2y = 1800.1x + 7y= 78Here are some equivalent systems. Each one is a step in solving the original system.Step 1:Step 2:Step 3:x = 807x + 7y = 6300.1x + 7y = 78{oix6.9x = 5520.1x + y = 780.1x + 7y = 78a. Look at the original system and the system in Step 1.i. What was done to the original system to get the system in Step 1?ii. Explain why the system in Step 1 shares a solution with the originalsystem.b. Look at the system in Step 1 and the system in Step 2.i. What was done to the system in Step 1 to get the system in Step 2?ii. Explain why the system in Step 2 shares a solution with that in Step 1.c. What is the solution to the original system?

Answer 1

Explanation:

ok

Answer 2

Given the system of equations;

`\begin{gathered} 2x+2y=180\ldots\ldots\ldots\ldots\ldots\text{equation}1 \\ 0.1x+7y=78\ldots\ldots\ldots\ldots\text{.}\mathrm{}\text{equation}2 \end{gathered}`

Also, some equivalent systems;

Step 1:

`\begin{gathered} 7x+7y=630\ldots\ldots\ldots\ldots\ldots\text{.equation}3 \\ 0.1x+7y=78\ldots\ldots\ldots\ldots\ldots\text{.equation}4 \end{gathered}`

Step 2:

`\begin{gathered} 6.9x=552\ldots\ldots\ldots\ldots\ldots\ldots\ldots..equation5 \\ 0.1x+7y=78\ldots\ldots\ldots\ldots\ldots equation6 \end{gathered}`

Step 3:

`\begin{gathered} x=80\ldots\ldots\ldots\ldots\ldots\ldots equation7 \\ 0.1x+7y=78\ldots\ldots\ldots..equation8 \end{gathered}`

Statement Problem(a):

(i) What was done to the original system to get the system in step 1?

ANSWER:

Multiply equation 1 by 3.5 to produce equation3, and equation 2 is unchanged. We have;

`\begin{gathered} 3.5(2x+2y)=3.5(180)_{} \\ 7x+7y=630\ldots\ldots\ldots\ldots\ldots\text{.equation}3 \end{gathered}`

(ii) Explain why the system in step 1 shares the a solution with the original system.

ANSWER:

Systems of equations that have the same solution are called equivalent systems. They are equivalent systems because one of the equations is only transformed by multiplying 3.5 by it.

Statement Problem (b):

(i) What was done to the system in step 1 to get the system in step 2?

ANSWER:

Subtract equation 4 from equation 3 to produce equation 5 and equation 4 is unchanged. We have;

`\begin{gathered} (7-0.1)x+(7-7)y=630-78 \\ 6.9x=552\ldots\ldots\ldots\ldots\text{.equation}5 \end{gathered}`

(ii) Explain why the system in step 2 shares the a solution with that in step 1.

ANSWER:

Systems of equations that have the same solution are called equivalent systems. They are equivalent systems because equation 5 is formed by subtracting equation 3 from equation 4.

Statement Problem (c):

What is the solution to the original system?

Solution:

`\begin{gathered} (2x+2y=180)*0.1 \\ (0.1x+7y=78)*2 \\ 0.2x+0.2y=18\ldots\ldots\ldots equation9 \\ 0.2x+14y=156\ldots..equation10 \\ \text{equation}10-\text{equation}9 \\ 13.8y=138 \\ \text{Divide both sides by 13.8,} \\ (13.8y)/(13.8)=(138)/(13.8) \\ y=10 \end{gathered}`

Then, we solve for x by substituting into equation9 9, we have;

`\begin{gathered} 0.2x+0.2y=18 \\ 0.2x+0.2(10)=18 \\ 0.2x+2=18 \\ 0.2x=18-2 \\ 0.2x=16 \\ \text{Divide both sides by 0.2,} \\ (0.2x)/(0.2)=(16)/(0.2) \\ x=80 \end{gathered}`

Hence, the solution of the original system is;

`x=80,y=10`