Which best explains why the equation 4x+5=4x-1 has no solutions? A. the equation has a variable on both sides. B. The equation is true for any value of x. C. The equation is not true for any value of x.
+3454 Let's see here. We have three options.
A. The equation has a variable on both sides.
B. The equation is true for any value of x.
C. The equation is not true for any value of x.
A would not be true, because you can still solve for x even if there is x (a variable) on both sides.
Think about it, if you had something like 5x+3=2x-9, you could still solve for x. This equation has a variable on both sides.
B would not be true, and we could prove that by putting a couple values of x into the equation. Let's try 4 and 15.
\begin{array}{rll} 4x+5&=&4x-1\\ 4(4)+5&=&4(4)-1\\ 16+5&=&16-1\\ 21&\neq&15 \end{array} \boxed{21&\neq&15}\\ \begin{array}{rll} 4x+5&=&4x-1\\ 4(15)+5&=&4(15)-1\\ 60+5&=&60-1\\ 65&\neq&59 \end{array} \boxed{65\neq59}
Finally, we can just prove our answer is C by putting in values for x.
Nomatter which value you put in for x, there will never be a solution for x.
+3454 Let's see here. We have three options.
A. The equation has a variable on both sides.
B. The equation is true for any value of x.
C. The equation is not true for any value of x.
A would not be true, because you can still solve for x even if there is x (a variable) on both sides.
Think about it, if you had something like 5x+3=2x-9, you could still solve for x. This equation has a variable on both sides.
B would not be true, and we could prove that by putting a couple values of x into the equation. Let's try 4 and 15.
\begin{array}{rll} 4x+5&=&4x-1\\ 4(4)+5&=&4(4)-1\\ 16+5&=&16-1\\ 21&\neq&15 \end{array} \boxed{21&\neq&15}\\ \begin{array}{rll} 4x+5&=&4x-1\\ 4(15)+5&=&4(15)-1\\ 60+5&=&60-1\\ 65&\neq&59 \end{array} \boxed{65\neq59}
Finally, we can just prove our answer is C by putting in values for x.
Nomatter which value you put in for x, there will never be a solution for x.
+118703 hi Ninja,
You are starting to use LaTex in ways that I have not used it. 
I think I know how you put the boxed results on the right but I will need to play before I am sure. Thanks.
I will be learning all sorts of things off you soon. I am very impressed !!
+3454 Hey Melody,
I actually didn't mean for the boxed numbers to be on the side like they did, it actually came by accident. Lol!
Maybe you can explain why this happened. I basically just ended the array, then put in the box command and it appeared on the side like so. I was hoping to box the not equal sign that's already centered, but when I tried to box it, it just kept spooling and didn't give me a result in the output box.
The highlighted is what I was trying to box:
Here's my code:
\begin{array}{rll}
4x+5&=&4x-1\\
4(4)+5&=&4(4)-1\\
16+5&=&16-1\\
21&\neq&15
\end{array}
\boxed{21&\neq&15}\\
\begin{array}{rll}
4x+5&=&4x-1\\
4(15)+5&=&4(15)-1\\
60+5&=&60-1\\
65&\neq&59
\end{array}
\boxed{65\neq59}
Any help is appreciated. :)
