If \(2(2x+3)-10\le 6(x-1)\) and \(\displaystyle\frac{2x-3}{4}+6\ge 3+\frac{4x}{3},\ x\in N\) then number of solutions for 'x' are
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 2
x must satisfy both inequalities and be a natural number.
First inequality:
\(4x+6-10\le 6x-6\)
\(4x-4\le 6x-6\)
\(2\le 2x\)
\(x\ge 1\)
Second inequality, multiplied by 12:
\(3(2x-3)+72\ge 36+16x\)
\(6x+63\ge 36+16x\)
\(27\ge 10x\)
\(x\le 2.7\)
So \(1\le x\le 2.7\) with x a natural number: x = 1 or x = 2, that is 2 solutions.