# Thread: Two Problems to Solve

1. ## Two Problems to Solve

Hello,
I am sorry that I do not have a better title for these two problems. I am an adult taking an Algebra I class equivalent to ninth grade algebra.
If I could learn to solve these two, they would be helpful as examples to solve the rest.
Thank you so much for your willingness to help me!

1. A box has length (l) 3 inches less than the height (h) and width (w) 9 inches less than h. The volume is 324 cubic inches.
a. Write an equation you can use to solve for the dimensions.
b. What are the dimensions of the box?

2. Write a = c in 5 different ways.
b d

2. Hello,

Originally Posted by VAP
Hello,
I am sorry that I do not have a better title for these two problems. I am an adult taking an Algebra I class equivalent to ninth grade algebra.
If I could learn to solve these two, they would be helpful as examples to solve the rest.
Thank you so much for your willingness to help me!

1. A box has length (l) 3 inches less than the height (h) and width (w) 9 inches less than h. The volume is 324 cubic inches.
a. Write an equation you can use to solve for the dimensions.
b. What are the dimensions of the box?
Translate the text :
l is 3 inches less than h.
--> $\displaystyle \boxed{l=h-3}$

w is 9 inches less than h.
--> $\displaystyle \boxed{w=h-9}$

The volume V is : $\displaystyle V=l \cdot w \cdot h$

Now replace such that you get V with respect to h.
Then, solve for h

3. Thank you so much! Can you help with the other ?

4. Originally Posted by VAP
...

1. A box has length (l) 3 inches less than the height (h) and width (w) 9 inches less than h. The volume is 324 cubic inches.
a. Write an equation you can use to solve for the dimensions.
b. What are the dimensions of the box?

2. Write a = c in 5 different ways.
b d
to #1:

$\displaystyle l = h-3$
$\displaystyle w = h-9$

Since $\displaystyle V = l\cdot w \cdot h$ the volume can be calculated by:

$\displaystyle V= 324 = (h-3)(h-9)\cdot h = h^3-12h^2+27h$ Solve for h:

$\displaystyle h^3-12h^2+27h -324=0$

$\displaystyle h^2(h-12) + 27(h-12) = 0$

$\displaystyle (h^2+27)(h-12)=0$

A product of 2 factors is zero if one factor equals zero:

$\displaystyle h^2+27>0$

Therefore

$\displaystyle h-12 = 0~\implies~ h = 12$

The demensions of the box are: $\displaystyle l=9; w=3; h=12$

to #2:

I assume that you mean:

\displaystyle \begin{aligned}\frac ab=\frac cd & \implies & ad=bc \\ & \implies & \frac ac=\frac db \\ & \implies & \frac ca=\frac bd \\ & \implies & \frac{ad}b=c \\ & \implies & \frac{ad}c=b \end{aligned}

5. Thank you for expanding on problem #1. Yes, that's what I meant for problem #2. I really appreciate it.