# Elementary system of inequalities - set of solutions 3 x &#x2212;<!-- − --> 1 &#x2264;<!

Elementary system of inequalities - set of solutions
$3x-1\le 1+4x$
$\frac{5}{6}x+\frac{2}{3}\ge 1+\frac{1}{2}x$
$\frac{x-1}{6}<\frac{1}{4}$
$⇒$
$x\ge -2$
$x\ge 1$
$x<\frac{5}{2}$
I compared the above reduced inequalities which gave me as answer that the solutions are from [$\frac{5}{2}$]. I think it is correct but I'm not sure. Can someone affirm this or debunk this please and what's the reason?
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Louis Lawrence
The $x\ge 1$ trumps the $x\ge -2$ so your solution set is $\left[1,\frac{5}{2}\right)$.
To find the solution set, you take the intersection of each solution set.
$x\ge -2$ has solution set $\left[-2,\mathrm{\infty }\right)$
$x\ge 1$ has solution set $\left[1,\mathrm{\infty }\right)$
$x<\frac{5}{2}$ has solution set $\left(-\mathrm{\infty },\frac{5}{2}\right)$
The intersection of $\left[-2,\mathrm{\infty }\right)$ and $\left[1,\mathrm{\infty }\right)$ is $\left[1,\mathrm{\infty }\right)$ because $-2<1$. Furthermore, the intersection of $\left[1,\mathrm{\infty }\right)$ and $\left(-\mathrm{\infty },\frac{5}{2}\right)$ is $\left[1,\frac{5}{2}\right)$.