If y>0>x, and (3+5y)/(x−1) < −7, then which of the following must be true?
A. 5y−7x+4 < 0
B. 5y+7x−4 > 0
C. 7x−5y−4 < 0
D. 4+5y+7x > 0
E. 7x−5y+4 > 0
Since given that x is a negative number then x-1 is also some negative number. Multiply both sides of the inequality by x-1. When multiplying by negative flip the sig, so we'll have:
\frac{3+5y}{x-1} < -7 --> 3+5y>-7(x-1) --> 3+5y>7-7x --> 5y+7x-4>0.
Answer: B.
Hope it's clear.
A. 5y−7x+4 < 0
B. 5y+7x−4 > 0
C. 7x−5y−4 < 0
D. 4+5y+7x > 0
E. 7x−5y+4 > 0
Since given that x is a negative number then x-1 is also some negative number. Multiply both sides of the inequality by x-1. When multiplying by negative flip the sig, so we'll have:
\frac{3+5y}{x-1} < -7 --> 3+5y>-7(x-1) --> 3+5y>7-7x --> 5y+7x-4>0.
Answer: B.
Hope it's clear.
