The borders are shown in the picture attached.
As you can see, starting with border 1, we have 6 daises (white squares) surrounded by 10 tulips (colored squares). Through Jerry's expression we expected:
<span>8(b − 1) + 10 =
</span>8(1 − 1) + 10 =
0 + 10 =
10 tulips.
When considering border 2, we expect:
<span>8(b − 1) + 10 =
</span>8(2 − 1) + 10 =
8 + 10 =
<span>18 tulips.
Indeed, we have the 10 tulips from border 1 and 8 additional tulips, for a total of 18 tulips.
Then, consider border 3, we expect:
</span><span>8(b − 1) + 10 =
</span>8(3 − 1) + 10 =
16 + 10 =
26<span> tulips.
Again, this is correct: we have the 10 tulips used in border 1 plus other 16 tulips, for a total of 26.
Therefore, Jerry's expression is
correct.</span>
The best way to solve is by using elimination method.
20x = -58 - 2y
17x = -49 - 2y
Multiply second equation by -1
20x = -58 - 2y
-17x = 49 + 2y
Add equations.
3x = -9
Divide.
x = -3
Plug in -3 into one of the equations.
17(-3) = -49 - 2y
-51 = -49 - 2y
Add 49 to both sides.
-2 = -2y
Divide.
1 = y
So your solution is (-3, 1).
I hope this helps love! :)
Supposing the sides with 6 and 8 is a right angle, you can create a new line from C and P, and find the length using the equation of a²+b²=c² or 6²+8²=c², with c equaling the radius of the circle.
After finding c, you will have to find the length from C to the midpoint of AC, using the same equation a²+b²=c². If both the lengths of C to the midpoint of AC, and A to the midpoint of AC are equal, you can do b+b to find the length of AC.
Using the same approach, you can find AB. Hope this makes sense, if not, I can clarify more.
Find the LCM (least common multiple) of the denominators, and use that as the denominator for the two fractions.
Ex.

LCM (3, 4) is 12

=