+2446 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=294∗1+252 | 294 now becomes your a and 252 becomes your new b. Now, solve again. |
| 294=252q+r | Do the same process. How many times does 252 go into q without going over? Once. The remainder goes to r. |
| 294=252∗1+42 | |
| 252=42q+r | |
| 252=42∗6+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. |
+2446 To find the equation that passes through (−4,0) and is parallel to the line y=34x−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=34x−2 is 34. As I stated above, parallel lines have the same slope, so the equation of this unknown line is 34.
We have deduced already that this unknown line is in the form of y=34x+b. The only thing to find now is the b:
| y=34x+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=34∗−41+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=34x+3
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