+2447 We can use the Euclidean algorithm for this:
| \(a=bq+r\) | This is where (a,b) is the number we are trying to find the GCF of. |
| \(546=294q+r\) | How many times can 294 go into 546 without going over? Once! The remainder goes to r. Continue this process until r becomes 0. |
| \(546=\textcolor{blue}{294}*1+\textcolor{red}{252}\) | 294 now becomes your a and 252 becomes your new b. Now, solve again. |
| \(\textcolor{blue}{294}=\textcolor{red}{252}q+r\) | Do the same process. How many times does 252 go into q without going over? Once. The remainder goes to r. |
| \(294=\textcolor{blue}{252}*1+\textcolor{red}{42}\) | |
| \(\textcolor{blue}{252}=\textcolor{red}{42}q+r\) | |
| \(252=\textcolor{blue}{42}*6+\textcolor{red}{0}\) | Now that r=0, look at the r that was immediately previous to, which happens to be 42. This means that the GCF of 294 and 546 is 42. |
+2447 To find the equation that passes through \((-4,0)\) and is parallel to the line \(y=\frac{3}{4}x-2\), we must understand a few properties.
1) Parallel lines have the same slope.
This fact, alone, can help us do half the problem. The slope of the line \(y=\frac{3}{4}x-2\) is \(\frac{3}{4}\). As I stated above, parallel lines have the same slope, so the equation of this unknown line is \(\frac{3}{4}\).
We have deduced already that this unknown line is in the form of \(y=\frac{3}{4}x+b \). The only thing to find now is the b:
| \(y=\frac{3}{4}x+b \) | Now, plug in a coordinate that we know is on the line. In this case, we only know that \((-4,0)\) lies on the line. Plug it in for x and y. |
| \(0=\frac{3}{4}*\frac{-4}{1}+b\) | Simplif the right hand side. |
| \(0=-3+b\) | Add 3 to both sides of the equation. |
| \(b=3\) | |
We have found both of the mystery values to construct the proper equation of a line. It is \(y=\frac{3}{4}x+3\)
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