I do not understand
- Thread starter Cofend
- Start date
Zermelo
Junior Member
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- Jan 7, 2021
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Not sure I understand your questions completely, but I'll try to answer:
What you have here is an equation. 4x means [MATH]4 \cdot x[/MATH]Aka this equation expressed with words would be: "We have 4 times a number x, and then we added 4 to it and got a result of 12. What is that number x?"
If you tried to figure it out by guessing, you might have stumbled across x = 2, which works:
4 * 2 + 4 = 8 + 4 = 12, so x = 2 is the solution.
As for the other questions, your goal is to "isolate" x, to take away everything from x and put it on the other side of the "=" sign. We achieve that by "doing the same thing to the left and right side of the "=" sign ".
For example, we could have added +2 to the both sides, and would have gotten 4x + 4 + 2 = 12 + 2, aka 4x + 6 = 14, which is equivalent to the starting equation but we didn't achieve anything. We want to get rid of the +4 on the left side, so we are going to subtract 4 from both sides:
4x + 4 - 4 = 12 - 4, aka 4x=8. Now, how can we get rid of the 4? Well, the 4 is being multiplied by x, so we need to do the "inverse" operation, we are going to divide by 4, and that's where fractions appear:
If we divide both sides by 4, we get [MATH] \frac{4x}{4}= \frac{8}{4} \Longrightarrow \frac{4}{4} \cdot x= 2[/MATH]. Hope this makes sense!
What you have here is an equation. 4x means [MATH]4 \cdot x[/MATH]Aka this equation expressed with words would be: "We have 4 times a number x, and then we added 4 to it and got a result of 12. What is that number x?"
If you tried to figure it out by guessing, you might have stumbled across x = 2, which works:
4 * 2 + 4 = 8 + 4 = 12, so x = 2 is the solution.
As for the other questions, your goal is to "isolate" x, to take away everything from x and put it on the other side of the "=" sign. We achieve that by "doing the same thing to the left and right side of the "=" sign ".
For example, we could have added +2 to the both sides, and would have gotten 4x + 4 + 2 = 12 + 2, aka 4x + 6 = 14, which is equivalent to the starting equation but we didn't achieve anything. We want to get rid of the +4 on the left side, so we are going to subtract 4 from both sides:
4x + 4 - 4 = 12 - 4, aka 4x=8. Now, how can we get rid of the 4? Well, the 4 is being multiplied by x, so we need to do the "inverse" operation, we are going to divide by 4, and that's where fractions appear:
If we divide both sides by 4, we get [MATH] \frac{4x}{4}= \frac{8}{4} \Longrightarrow \frac{4}{4} \cdot x= 2[/MATH]. Hope this makes sense!
If [MATH]\hspace1ex [4×x+4] \hspace3ex \text{ is }\hspace3ex [12][/MATH]
then
[MATH][4×x+4]+(-4) \hspace3ex \text{ is }\hspace3ex [12]+(-4)[/MATH]
[MATH][4×x+(4+(-4))] \hspace3ex \text{ is }\hspace3ex [12+(-4)][/MATH]
[MATH][4×x+0] \hspace3ex \text{ is }\hspace3ex [12+(-4)][/MATH]
[MATH][4×x] \hspace3ex \text{ is }\hspace3ex [12+(-4)][/MATH]
[MATH]\tfrac{1}{4}×[4×x] \hspace3ex \text{ is } \hspace3ex \tfrac{1}{4}×[12+(-4)][/MATH]
[MATH]\left(\tfrac{1}{4}×4\right)×x \hspace3ex \text{ is } \hspace3ex \tfrac{1}{4}×\left(12+(-4)\right)[/MATH]
[MATH]1×x \hspace3ex \text{ is } \hspace3ex \tfrac{1}{4}×\left(12+(-4)\right)[/MATH]
[MATH]x \hspace3ex \text{ is } \hspace3ex \tfrac{1}{4}×\left(12+(-4)\right)[/MATH]
then
[MATH][4×x+4]+(-4) \hspace3ex \text{ is }\hspace3ex [12]+(-4)[/MATH]
[MATH][4×x+(4+(-4))] \hspace3ex \text{ is }\hspace3ex [12+(-4)][/MATH]
[MATH][4×x+0] \hspace3ex \text{ is }\hspace3ex [12+(-4)][/MATH]
[MATH][4×x] \hspace3ex \text{ is }\hspace3ex [12+(-4)][/MATH]
[MATH]\tfrac{1}{4}×[4×x] \hspace3ex \text{ is } \hspace3ex \tfrac{1}{4}×[12+(-4)][/MATH]
[MATH]\left(\tfrac{1}{4}×4\right)×x \hspace3ex \text{ is } \hspace3ex \tfrac{1}{4}×\left(12+(-4)\right)[/MATH]
[MATH]1×x \hspace3ex \text{ is } \hspace3ex \tfrac{1}{4}×\left(12+(-4)\right)[/MATH]
[MATH]x \hspace3ex \text{ is } \hspace3ex \tfrac{1}{4}×\left(12+(-4)\right)[/MATH]