+0  
 
0
524
12
avatar

33-2x=5-9x 

how do you work this out

Guest May 16, 2015

Best Answer 

 #2
avatar+4664 
+11

Your goal is to get x all alone.

 

Even i had trouble and still do with these equations, but every time you get rid of something you have to think of the consequences that's why you should take away the small things so that the equations is closer to x and its less complicated.

 

In the coding, i have shown what i added or took away.

Look at that and speak to me after if you're still having trouble answering it.

 

When i added 9x if instead +2x i would have gotten a negative but it's still right and i still would have gotten an answer.

 

$$\begin{array}{rrl}
33-2x&=&5-9x\qquad(- 5)\\
28-2x&=&-9x\qquad(+2x)\\
28&=&-7x\qquad(28\div(-7)=-4)\\
x&=&-4\\
\end{array}$$

 

Also correct!

MathsGod1  May 16, 2015
 #1
avatar+4664 
+11

$$\begin{array}{rrlll}33-2x&=&5-9x\qquad&(- 5)\\
28-2x&=&-9x\qquad&(+9x)\\
28+7x&=&0\qquad&(which\;means)\\
28&=&-7x\qquad&(28\div-7=4)\\
x&=&-4\\
\end{array}$$

MathsGod1  May 16, 2015
 #2
avatar+4664 
+11
Best Answer

Your goal is to get x all alone.

 

Even i had trouble and still do with these equations, but every time you get rid of something you have to think of the consequences that's why you should take away the small things so that the equations is closer to x and its less complicated.

 

In the coding, i have shown what i added or took away.

Look at that and speak to me after if you're still having trouble answering it.

 

When i added 9x if instead +2x i would have gotten a negative but it's still right and i still would have gotten an answer.

 

$$\begin{array}{rrl}
33-2x&=&5-9x\qquad(- 5)\\
28-2x&=&-9x\qquad(+2x)\\
28&=&-7x\qquad(28\div(-7)=-4)\\
x&=&-4\\
\end{array}$$

 

Also correct!

MathsGod1  May 16, 2015
 #3
avatar+93333 
+5

Hi MG

Your understanding of equations and your LaTex presentations are coming along brilliantly.

All the adults on the forum a are very pleased that you are learning here with us!

 

Your first post here was incorrect (you did not add 9x correctly) and you have redone it so you should delete your first post so it will not be confusing.

 

Better comments for the correct post would have been  

(-5 from both sides)

(+2x to both sides)

(divide both sides by -7)

Melody  May 16, 2015
 #4
avatar+4664 
+5

Thanks Melody, but why was it wrong to + 9x

 

How did you know that it was wrong and you had to add 2x?

MathsGod1  May 16, 2015
 #5
avatar+26971 
+5

When you add 9x to both sides MG you should get 28 + 7x = 0, not 28 = 7x

.

Alan  May 16, 2015
 #6
avatar+4664 
+5

I know that Alan, but i just skipped it or is that part vital for the equation?

MathsGod1  May 16, 2015
 #7
avatar+26971 
+5

In that case, in your first answer you got a sign wrong when you skipped it!

 

28+ 7x = 0 isn't the same as 28 = 7x, it's the same as 28 = -7x

 

.

Alan  May 16, 2015
 #8
avatar+4664 
+5

I've edited my post. :) 

 

Thanks Melody and Alan!

MathsGod1  May 16, 2015
 #9
avatar+26971 
+5

It needs a further edit MG!  28 + 7x means 28 = minus 7x 

.

Alan  May 16, 2015
 #10
avatar+4664 
+5

Huh I'm not sure whether you mean negative 7x as its placed there or the operation sign minus?

MathsGod1  May 16, 2015
 #11
avatar+26971 
+5

I mean the fourth line of your first solution should look the same as the third line of your second solution.

.

Alan  May 16, 2015
 #12
avatar+4664 
+5

Ok, I'll do that but how does it turn from a positive to a negative number?

(Answer then is also -4)

MathsGod1  May 16, 2015

15 Online Users

avatar
avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.