The given system of inequations is
2(2x + 3) − 10 < 6(x − 2), …(i)
And \(\frac{2x-3}{4}\) + 6 ≥ 4 + \(\frac{4x}{2}\) …(ii)
Now, 2(2x + 3) − 10 < 6(x − 2)
⇒ 4x + 6 − 10 < 6x − 12
⇒ 4x + 4 < 6x − 12
⇒ 4x − 6x < −12 + 4
⇒ −2x < −8
⇒ x > 4
⇒ x∈ (4, ∞)
So, the solution set for first inequation is the interval (4, ∞) and \(\frac{2x-3}{4}\) + 6 ≥ 4 + \(\frac{4x}{3}\)
⇒ \(\frac{2x-3+24}{4}\) ≥ \(\frac{12+4x}{3}\)
⇒ \(\frac{2x+21}{4}\) ≥ \(\frac{12+4x}{3}\)
⇒ 3(2x + 21) ≥ 4(12 + 4x)
⇒ 6x − 16x ≥ 48 − 63
⇒ −10x ≥ −15
⇒ x ≥ \(\frac{15}{10}\)
⇒ x ≥ 1.5
⇒ x ∈ (−∞, 1.5)
So, the solution set of in equation (ii) is the internal (−∞, 1.5) The solution set of in equations (i) and (ii) are graphed on real line in fig (i) and fig (ii) respectively.
We observe that there is no common solution of the two inequation, so the given system of equations has no solution.
