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.........
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.........
By the way the solution you posted cannot be correct since 11 + 20 = 31 not 36.
Here is how to solve the problem.
$\displaystyle x + y = 36$
Let's assume $\displaystyle x\ge y$ then
$\displaystyle x = 2y + 3$
Which come from
Now, we can substitute $\displaystyle 2y + 3$ in for $\displaystyle x$ which give
$\displaystyle 2y + 3 + y = 36 \Rightarrow 3y = 33 \Rightarrow y = 11$
Finally, substitute this back in
$\displaystyle x = 2\cdot 11 + 3 = 25$
Answer: 25 and 11
For the second problem it is best to multiply everything out first.
$\displaystyle x-3-2(6-2x) = 2(2x-5)$
$\displaystyle x-3-12+4x = 4x-10$
Then get all the x's on one side and the number on the other.
$\displaystyle x + 4x - 4x = -10 + 3 + 12$
$\displaystyle x + 4x - 4x = 5$
Now you got the correct answer, but you had made some mistakes. Specifically the following
$\displaystyle x-3-2(6-2x) = 4x-10$
$\displaystyle x(6-2x)-4x = -10+5$
I am unsure how you got to that second line.
But, then you made another mistake that corrected the first one
$\displaystyle 6x-2x-4x = -5$ would actually give $\displaystyle 0x = -5$ which obviously doesn't make since. Somehow you have $\displaystyle -1(-x = -5)$.
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
I would suggest creating a new thread if you have any further questions.
The length of th rectangular swimming pool is twice its width.
$\displaystyle L_1 = 2w_1$
The pool is surrounded by by a cement walk 4 feet wide.
$\displaystyle L_2 = L_1 + 8$
$\displaystyle w_2 = w_1 + 8$
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 $\displaystyle L_2\cdot w_2$ to that of the area of the pool ($\displaystyle A_{pool} = L_1\cdot w_1$). So we have
$\displaystyle L_2\cdot w_2 = 784 + A_{pool}$
$\displaystyle L_2\cdot w_2 = 784 + L_1\cdot w_1$
$\displaystyle \left(L_1 + 8\right)\left(w_1 + 8\right) = 784 + L_1\cdot w_1$
Now, you will want to substitute what we know for $\displaystyle L_1$ and then solve.
I will let you complete the problem.
Close. Substitute $\displaystyle 2w_1$ for $\displaystyle L_1$ and you get
$\displaystyle \left(2w_1 + 8\right)\left(w_1 + 8\right) = 784 + 2w_1\cdot w_1$
Now expand
$\displaystyle 2w_1^2 + 8w_1 + 16w_1 + 64 = 784 + 2w_1^2$
Get all $\displaystyle w_1$ to one side and numbers to the other
$\displaystyle 24w_1 = 720 \Rightarrow w_1 = 30$
Substitute the value of $\displaystyle w_1$ into the formula for $\displaystyle L_1$
$\displaystyle L_1 = 2w_1 = 2\cdot 30 = 60$