Answer:
D) -2
Step-by-step explanation:
I thinks its the bottom one
If 'a' is a rational number and c is rational, then
a = p/q
c = r/s
where p,q,r,s are integers (q and s can't be zero)
Subtracting c-a gives
b = c-a
b = (p/q) - (r/s)
b = (ps/qs) - (qr/qs)
b = (ps - qr)/(qs)
The quantity pq - qr is an integer. The reason why is because ps and qr are both integers (multiplying any two integers leads to another integer). Subtracting any two integers results in another integer.
So we have (ps - qr)/(qs) in the form (integer)/(integer) = rational number
Therefore, b is a rational number, but this contradicts the given info that b is irrational. If b is irrational, then we CANNOT write it as a ratio of integers.
This contradiction proves the assumption "a+b = c and c is rational" is incorrect
The sum is irrational.
Therefore, if a+b = c, where 'a' is rational and b is irrational, then c is irrational.
<span>Nope. A consistent system is either dependent (both equations represent the same line) or independent. So in an independent system, the lines intersect in exactly one point, which means there is exactly one solution. A dependent system has an infinite number of solutions. An inconsistent system has no solution, so no system of linear equations can have exactly two solutions.</span>