Thread: Linear Equations - I Need Someone to Check my Work

1. Linear Equations - I Need Someone to Check my Work

Hello. I was wondering if someone would be so kind as to check my work on some linear equations. I think I am getting a better idea on how to do them, but I would like to have someone double check me

I have attached a Word document that has the problems. Thank you in advance for your time.

2. First it would be far better to type the problems in yourself rather than link to a word document- they are notorious for having viruses. Many people will not open "Word" attachements. I did only because I have very strong virus protection on this computer.

Second, the problem asks you to determine how many solutions a system of equations has without solving the system- yet you have solved the first system. That said, your answers to the first two questions are correct and your method is good. Why did you not do the same with the third problem?

3. Hi, I am sorry, I wasn't aware that Word attachments are frowned upon, I will remember that in the future. I am not sure what you are referring to about the third problem? The third problems states

3. Solve the following linear equations (5 points)

x/3 + y/4 = 5
x + 3y/8 = 12

Could you tell me what I did wrong? This is so confusing...thank you for your help.

4. Originally Posted by dclary
I am not sure what you are referring to about the third problem? The third problems states

3. Solve the following linear equations (5 points)

x/3 + y/4 = 5
x + 3y/8 = 12

Could you tell me what I did wrong? This is so confusing...thank you for your help.
I'm not speaking for HallsofIvy here, but too me you took a lot of steps. I personally wouldn't have gotten rid of the fractions for the first step because the fractions are simple enough. I would have multiplied the first equation by -3, and then add the two equations to get rid of the x:
\displaystyle \begin{aligned} \frac{x}{3} + \frac{y}{4} &= 5 \\ x + \frac{3y}{8} &= 12 \end{aligned}

\displaystyle \begin{aligned} -x - \frac{3y}{4} &= -15 \\ x + \frac{3y}{8} &= 12 \\ \end{aligned}
--------------------
\displaystyle \begin{aligned} -\frac{3y}{8} &= -3 \\ -\frac{3y}{8}\left( -\frac{8}{3} \right) &= -3\left( -\frac{8}{3} \right) \\ y &= 8 \end{aligned}

Or, if you really want to get rid of all of the fractions first, multiply the first equation by 12 and the second equation by -8. After combining you'll get rid of the y.

5. Originally Posted by eumyang
I'm not speaking for HallsofIvy here, but too me you took a lot of steps. I personally wouldn't have gotten rid of the fractions for the first step because the fractions are simple enough. I would have multiplied the first equation by -3, and then add the two equations to get rid of the x:
\displaystyle \begin{aligned} \frac{x}{3} + \frac{y}{4} &= 5 \\ x + \frac{3y}{8} &= 12 \end{aligned}

\displaystyle \begin{aligned} -x - \frac{3y}{4} &= -15 \\ x + \frac{3y}{8} &= 12 \\ \end{aligned}
--------------------
\displaystyle \begin{aligned} -\frac{3y}{8} &= -3 \\ -\frac{3y}{8}\left( -\frac{8}{3} \right) &= -3\left( -\frac{8}{3} \right) \\ y &= 8 \end{aligned}

Or, if you really want to get rid of all of the fractions first, multiply the first equation by 12 and the second equation by -8. After combining you'll get rid of the y.
Thank you so much, you have been extremely helpful. Those fractions always throw me off, sometimes I wonder if I am ever going to catch on. Thanks again

6. I didn't scroll down far enough to realize that the solution to the third problem was on a different page!

7. Originally Posted by HallsofIvy
I didn't scroll down far enough to realize that the solution to the third problem was on a different page!
No problem :-)