Gold pieces the first leprechaun has: x
Gold pieces the second leprechaun has: y
<span>
The first leprechaun says to the other, ‘Give me seven of your gold pieces and I will have twice as many as you!’:
The first leprechaun would have x+7
The second leprechaun would have y-7
</span>I will have twice as many as you:
(1) x+7=2(y-7)
<span>(1) x+7=2y-14
(1) x+7-7-2y=2y-14-7-2y
(1) x-2y=-21
</span>The other one replies, ‘No way! Give me seven of yours and we’ll have the same number’
The first leprechaun would have x-7
The second leprechaun would have y+7
We’ll have the same number:
(2) x-7=y+7
(2) x-7+7-y=y+7+7-y
(2) x-y=14
<span>How many gold pieces does the first leprechaun have?
x=?
We have a system with 2 equations and two unknows (x and y). We need to solve for x:
(1) x-2y=-21
(2) x-y=14
Using the method of substitution, we can isolating y in the second equation:
(2) x-y=14
(2) x-y+y-14=14+y-14
(2) x-14=y
(2) y=x-14
And we can replace y in the first equation by x-14, and solve for x:
(1) x-2y=-21
(1) x-2(x-14)=-21
(1) x-2x+28=-21
(1) -x+28=-21
(1) -x+28-28=-21-28
(1) -x=-49
(1) (-1)*(-x=-49)
(1) x=49
Answer: T</span>he first leprechaun has 49 gold pieces
The area of your figure equals to about 67
The answer is A I think sorry if I’m wrong
There are nine possible combinations as follows: bb, bc, bt, cb, cc, ct, tb, tc, tt, each having a probability of 1/9. The four combinations with a cab exactly one time are: bc, cb, ct, tc. Therefore the probability that she'll use a cab exactly one time is: 4/9.
