Lydia baked a total of 144 chocolate chip cookies and peanut butter treats for valentine's day. initially, the ratio of chocolate chip cookies to peanut butter treats was 5: 3. after lydia's friends ate 2/5 of her chocolate chip cookies and some of her peanut butter treats, the cookies outnumbered the treats 6 to 1. how many peanut butter treats did she have left? show your work with your answer!

Let  be the chocolate chip cookies, so the peanut butter treats will be .
We know that the cookies and the treats are in a ratio of 5:3, so:
Now we can solve for :







We now know Lydia has 90 chocolate chip cookies, and  peanut butter treats.

Then, Lydia's friend ate  of her cookies, so her friend ate  cookies. Therefore, Lydia has cookies left.
Now we can calculate the total amount of baked goods after her friend ate the cookies:

Therefore, our remainder treats will be:


We also now that after her friend ate 54 cookies and some treats, the new ratio is 6:1, and that's all we need to set up our new equation and solve it to find how many treats she ate:







Finally, if she ate 81 out 90 treats, we can conclude the Lydia has left with 9 peanut butter treats.

Let:
 x: chip cookies
 y: peanut butter
 We have the following system of equations:
 x + y = 144
 x / y = 5/3
 We solve the system of equations:
 Step 1:
 x + y = 144
 y = (3/5) x
 Step 2:
 x + (3/5) x = 144
 (8/5) x = 144
 x = (5/8) * (144) = 90
 Step 3:
 y = (3/5) x
 y = (3/5) * (90) = 54
 Lydia's friends ate 2/5 of her chocolate chip cookies:
 x '= (2/5) * (90) = 36
 the cookies outnumbered the treats 6 to 1
 x '/ y' = 6/1
 Clearing y '
 y '= (1/6) * x'
 y '= (1/6) * 36
 y '= 6
 
 she had left 6 butter treats

Intersecting secant-tangent theorem states that if a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.so to break it down: 1. pq acts as the tangent and ps as the secant with an intersection at p.sr = 21rp = 3x + 3pq = 4x + 4following to equation: tangent squared = the two segments multiplied by each otherso (4x + 4) ^ 2 = (3x + 3) • (21) simplifies to 16x^2 - 31x - 47 = 0 or (16x - 47)(x + 1) x = -1 or x = 47/162. pq acts as the tangent and ps as the secant with an intersection at p (again! )sr = 16rp = xpq = 15(15) ^ 2 = 16 • x225 = 16xx = 225/163. the interesting chords are proportional to each other so a ratio is possible to set up: ap dp 10 3 + x = = —— = pc pb 8 xcross multiple to get 10x = (3 + x)(8)x = 12