here,
2x-6y= -12
2(x-3y)= -12
x-3y = -6...........eq1
x+2y=14............eq2
we can subtract the above two given equations
we get,

___________

so,
y=4
now to find the value of x we have to substitute the value of y in any of the above two equations, I choose eq2
we get,

<h3>so,</h3><h3> x=6,y=4</h3>
Without the graph it is impossible the find the value of K
(cube root of 5) * sqrt(5)
--------------------------------- = ?
(cube root of 5^5)
This becomes easier if we switch to fractional exponents:
5^(1/3) * 5^(1/2) 5^(1/3 + 1/2) 5^(5/6)
------------------------ = --------------------- = ------------- = 5^[5/6 - 5/3]
[ 5^5 ]^(1/3) 5^(5/3) 5^(5/3)
Note that 5/6 - 5/3 = 5/6 - 10/6 = -5/6.
1
Thus, 5^[5/6 - 5/3] = 5^(-5/6) = --------------
5^(5/6)
That's the correct answer. But if you want to remove the fractional exponent from the denominator, do this:
1 5^(1/6) 5^(1/6)
---------- * ------------- = -------------- (ANSWER)
5^(5/6) 5^(1/6) 5