MaxWong

\(\dfrac{\sqrt{6y + 2}}{\sqrt{2y + 2}} = \dfrac52\\ \dfrac{6y + 2}{2y + 2} = \dfrac{25}4\\ \dfrac{3y + 1}{y + 1} = \dfrac{25}4\\ 4(3y + 1) = 25(y + 1)\\ 12y + 4 = 25y + 25\\ 13y = -21\\ y = -\dfrac{13}{21}\text{ (rej.)}\)

There are no solutions.

Melody gave an answer. Are you having difficulties understanding her solution?

We factorize it first.

\(11! + 12! + 13! = 11! (1 + 12 + 12 \times 13) = 13^2 \times 11!\)

All prime factors of \(11!\) are smaller than or equal to 11 since \(11! = 11 \times 10 \times 9 \times \cdots \times 2\times 1\).

The largest prime factor is 13.

Since you didn't reply, I will assume that it denotes the symmetric difference \(A \oplus B = (A \cup B) - (A \cap B)\).

First, \(C - B\) is just the collection of elements in C that is not in B, i.e., \(C - B = \{2, 4\}\).

Now, we calculate \(A \oplus (C - B) = A \oplus \{2 ,4\}\).

Note that \(A \cap \{2, 4\} = \varnothing\) since the two sets do not have common elements. Then \(A \oplus \{2, 4\} = A \cup \{2, 4\} = \{1,2,4,5,6,8\}\).

Hence, we have \(A \oplus (C - B) = \{1,2,4,5,6,8\}\).

Multiply \(x^2 - 7x + 10 = (x -2)(x - 5)\) on both sides:

\(Bx - 11 = A(x - 5) - 3(x - 2)\)

Since this is true for any real number x, we can substitute whatever real number we want.

Substitute x = 5.

\(5B - 11 = -3(5 - 2)\\ 5B - 11 = -9\\ 5B = 2\\ B = \dfrac25\)

Substitute B = 2/5 and x = 2.

\(\dfrac45 - 11 = A(2 - 5)\\ A = \dfrac{17}5\)

Are you sure that ACE is a triangle?

We have \(\lfloor r \rfloor = 12.2 - 2r\). Then 12.2 - 2r must be an integer. Then \(r = k + 0.1\text{ or }r = k + 0.6\) for some integer k. Now you can substitute that and try to solve it like a linear equation.

Summing A, B, C gives the interior angle sum as 15x.

What is the interior angle sum of a triangle?

Note that \(\dfrac13 = 0.333\dots\) and \(\dfrac23 = 0.666\dots\).

Then \(\dfrac1{300} = 0.00333\dots\) and \(\dfrac1{150} = 0.00666\dots\).

\(0.72666\dots + 0.02333\dots = 0.72 + 0.02 + 0.00666\dots + 0.00333\dots = 0.74 + \dfrac1{300} + \dfrac1{150}\)

I believe you can continue here.

Just multiply them together.

\(\text{Result} = Ba=\begin{bmatrix}7&-6\\3&-4\end{bmatrix}\begin{bmatrix}5\\-3\end{bmatrix} = \begin{bmatrix}53\\27\end{bmatrix}\)