Respuesta :
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
Answer:
she had left 6 butter treats
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
Answer:
she had left 6 butter treats
Let [tex]x[/tex] be the chocolate chip cookies, so the peanut butter treats will be [tex]144-x[/tex].
We know that the cookies and the treats are in a ratio of 5:3, so:[tex] \frac{cookies}{treats} = \frac{5}{3} = \frac{x}{144-x} [/tex]
Now we can solve for [tex]x[/tex]:
[tex] \frac{5}{3} = \frac{x}{144-x} [/tex]
[tex]5(144-x)=3x[/tex]
[tex]720-5x=3x[/tex]
[tex]8x=720[/tex]
[tex]x= \frac{720}{8} [/tex]
[tex]x=90[/tex]
We now know Lydia has 90 chocolate chip cookies, and [tex]144-x=144-90=54[/tex] peanut butter treats.
Then, Lydia's friend ate [tex] \frac{3}{5} [/tex] of her cookies, so her friend ate [tex] \frac{2}{5} (90)=36[/tex] cookies. Therefore, Lydia has [tex]90-36=54[/tex] cookies left.
Now we can calculate the total amount of baked goods after her friend ate the cookies:
[tex]144-54=90[/tex]
Therefore, our remainder treats will be:
[tex]90-x[/tex]
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:
[tex] \frac{cookies}{treats} = \frac{6}{1} = \frac{54}{90-x} [/tex]
[tex]6(90-x)=54[/tex]
[tex]540-6x=54[/tex]
[tex]6x=486[/tex]
[tex]x= \frac{486}{6} [/tex]
[tex]x=81[/tex]
Finally, if she ate 81 out 90 treats, we can conclude the Lydia has left with 9 peanut butter treats.
We know that the cookies and the treats are in a ratio of 5:3, so:[tex] \frac{cookies}{treats} = \frac{5}{3} = \frac{x}{144-x} [/tex]
Now we can solve for [tex]x[/tex]:
[tex] \frac{5}{3} = \frac{x}{144-x} [/tex]
[tex]5(144-x)=3x[/tex]
[tex]720-5x=3x[/tex]
[tex]8x=720[/tex]
[tex]x= \frac{720}{8} [/tex]
[tex]x=90[/tex]
We now know Lydia has 90 chocolate chip cookies, and [tex]144-x=144-90=54[/tex] peanut butter treats.
Then, Lydia's friend ate [tex] \frac{3}{5} [/tex] of her cookies, so her friend ate [tex] \frac{2}{5} (90)=36[/tex] cookies. Therefore, Lydia has [tex]90-36=54[/tex] cookies left.
Now we can calculate the total amount of baked goods after her friend ate the cookies:
[tex]144-54=90[/tex]
Therefore, our remainder treats will be:
[tex]90-x[/tex]
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:
[tex] \frac{cookies}{treats} = \frac{6}{1} = \frac{54}{90-x} [/tex]
[tex]6(90-x)=54[/tex]
[tex]540-6x=54[/tex]
[tex]6x=486[/tex]
[tex]x= \frac{486}{6} [/tex]
[tex]x=81[/tex]
Finally, if she ate 81 out 90 treats, we can conclude the Lydia has left with 9 peanut butter treats.