Please help me with the solution for this problem

**vendonz** · July 24th 2010, 04:48 PM

The sum of 2 numbers is 36. If the larger number is divided by the smaller number, the
qoutient is 2 and the remainder is 3. Find the numbers.

Anwers: 20 and 11

Can you show me the solution for this problem.........

**lvleph** · July 24th 2010, 05:54 PM

By the way the solution you posted cannot be correct since 11 + 20 = 31 not 36.
Here is how to solve the problem.
$x + y = 36$

Let's assume $x\ge y$ then
$x = 2y + 3$
Which come from

Originally Posted by vendonz

If the larger number is divided by the smaller number, the
qoutient is 2 and the remainder is 3.

Now, we can substitute $2y + 3$ in for $x$ which give
$2y + 3 + y = 36 \Rightarrow 3y = 33 \Rightarrow y = 11$
Finally, substitute this back in
$x = 2\cdot 11 + 3 = 25$

Answer: 25 and 11

**vendonz** · July 25th 2010, 12:38 AM

Yeah sir your right its 25 and 11 sorry my fault ,, can you explain it to me more simple im a slow learner ,, can you use x as our substitution

**vendonz** · July 25th 2010, 12:49 AM

hehehe ,, i got it now ^_^ ,, how about this sir i confuse to this, Solve for x:
x-3-2(6-2x) = 2(2x-5)

I Got this ;
x-3-2(6-2x) = 2(2x-5)
x-3-2(6-2x) = 4x-10
x(6-2x)-4x = -10+5
6x-2x-4x = -5
-1(-x = -5)
x = 5
is this right?? the answer for this is 5

**lvleph** · July 25th 2010, 06:18 AM

For the second problem it is best to multiply everything out first.
$x-3-2(6-2x) = 2(2x-5)$
$x-3-12+4x = 4x-10$
Then get all the x's on one side and the number on the other.
$x + 4x - 4x = -10 + 3 + 12$
$x + 4x - 4x = 5$

Now you got the correct answer, but you had made some mistakes. Specifically the following
$x-3-2(6-2x) = 4x-10$
$x(6-2x)-4x = -10+5$
I am unsure how you got to that second line.
But, then you made another mistake that corrected the first one
$6x-2x-4x = -5$ would actually give $0x = -5$ which obviously doesn't make since. Somehow you have $-1(-x = -5)$ .

**vendonz** · July 25th 2010, 06:41 AM

I solved that a while ago,, but thanks for the effort this is the problem that im solving for now can you help me again???

The length of th rectangular swimming pool is twice its width. The pool is surrounded by by a cement walk 4 feet wide. If the area of the walk is 784 square feet. Find the dimension of the pool.

This is asking for width and length

**lvleph** · July 25th 2010, 07:39 AM

I would suggest creating a new thread if you have any further questions.

The length of th rectangular swimming pool is twice its width.
$L_1 = 2w_1$

The pool is surrounded by by a cement walk 4 feet wide.
$L_2 = L_1 + 8$
$w_2 = w_1 + 8$
Since there is an extra four feet on all sides of the pool.

The area of the walk is 784 square feet. Here The best thing to do is compare the area made by $L_2\cdot w_2$ to that of the area of the pool ( $A_{pool} = L_1\cdot w_1$ ). So we have
$L_2\cdot w_2 = 784 + A_{pool}$
$L_2\cdot w_2 = 784 + L_1\cdot w_1$
$\left(L_1 + 8\right)\left(w_1 + 8\right) = 784 + L_1\cdot w_1$
Now, you will want to substitute what we know for $L_1$ and then solve.
I will let you complete the problem.

**vendonz** · July 25th 2010, 08:03 PM

Thanks for the help ....
I got Width: 60
Length: 120
Is this correct???

**lvleph** · July 25th 2010, 08:12 PM

Close. Substitute $2w_1$ for $L_1$ and you get
$\left(2w_1 + 8\right)\left(w_1 + 8\right) = 784 + 2w_1\cdot w_1$
Now expand
$2w_1^2 + 8w_1 + 16w_1 + 64 = 784 + 2w_1^2$
Get all $w_1$ to one side and numbers to the other
$24w_1 = 720 \Rightarrow w_1 = 30$
Substitute the value of $w_1$ into the formula for $L_1$
$L_1 = 2w_1 = 2\cdot 30 = 60$

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