The discussion revolves around solving linear algebra equations, specifically the equations 2x – 1 = 9 – 3x and 4(3b-1)+6=5(2b+4). Participants seek assistance in understanding the steps involved in solving these equations.
There is no consensus on the solution to the first equation, as one participant initially concludes x=4, while Dan later corrects this to x=2. The second equation appears to be agreed upon with b=9.
Participants' solutions depend on the accuracy of their algebraic manipulations, and there are unresolved steps in the reasoning process that could affect the final answers.
gazparkin said:Hi,
Could someone help me with this question:
2x – 1 = 9 – 3x
Thank you in advance!
Step 1: Get all the x's on one side: Subtract 2x on both sides:gazparkin said:Hi,
Could someone help me with this question:
2x – 1 = 9 – 3x
Thank you in advance!
Expand the parentheses:gazparkin said:Sorry, there's another one that I'm stuck on too. Really interested to understand the working out on this.
4(3b-1)+6=5(2b+4)
Thanks Dan - makes this really clear. The first one I get x=4 and the 2nd one I get b=9.topsquark said:Step 1: Get all the x's on one side: Subtract 2x on both sides:
2x - 1 - 2x = 9 - 3x - 2x
-1 = 9 - 5x
Now do the same with the numbers. Can you finish?
-Dan
- - - Updated - - -Expand the parentheses:
4(3b - 1) + 6 = 5(2b + 4)
[math]4 \cdot 3b + 4 \cdot (-1) + 6 = 5 \cdot 2b + 5 \cdot 4[/math]
Now combine terms:
12b - 4 + 6 = 10b + 20
12b + 2 = 10b + 20
Now it's like your first problem. Can you finish?
-Dan
The second one is okay. Let's take another look at the first one. I left off withgazparkin said:Thanks Dan - makes this really clear. The first one I get x=4 and the 2nd one I get b=9.