Treat the matrices on the right side of each equation like you would a constant.
Let 2<em>X</em> + <em>Y</em> = <em>A</em> and 3<em>X</em> - 4<em>Y</em> = <em>B</em>.
Then you can eliminate <em>Y</em> by taking the sum
4<em>A</em> + <em>B</em> = 4 (2<em>X</em> + <em>Y</em>) + (3<em>X</em> - 4<em>Y</em>) = 11<em>X</em>
==> <em>X</em> = (4<em>A</em> + <em>B</em>)/11
Similarly, you can eliminate <em>X</em> by using
-3<em>A</em> + 2<em>B</em> = -3 (2<em>X</em> + <em>Y</em>) + 2 (3<em>X</em> - 4<em>Y</em>) = -11<em>Y</em>
==> <em>Y</em> = (3<em>A</em> - 2<em>B</em>)/11
It follows that

Similarly, you would find

You can solve the second system in the same fashion. You would end up with

Answer:
ok thx
Step-by-step explanation:
What is the math problem??
Answer:
1) a) yes
b) no
c) a² - 39
(a)² - (sqrt(39))²
(a - sqrt(39))(a + sqrt(39))
This quadratic can be split into real factors, but not rational
sqrt(39) is a real number, but not rational
Conjugate of (a - bi) is (a + bi).
Just change the sign on the imaginary part.
7 + 2i