Question

[2x+1]÷3+[3x-5]÷4<[4x-3]÷5+[x+1]÷2 ​

Answer 1

Respuesta:

$x<7$

Explicación paso a paso:

$\frac{\left(2x+1\right)}{3}+\frac{\left(3x-5\right)}{4}<\frac{\left(4x-3\right)}{5}+\frac{\left(x+1\right)}{2}\\\\\frac{2x+1}{3}+\frac{3x-5}{4}<\frac{4x-3}{5}+\frac{x+1}{2}\\\\=3\cdot \:2\cdot \:2\cdot \:5\\\\=60\\\\\frac{2x+1}{3}\cdot \:60+\frac{3x-5}{4}\cdot \:60<\frac{4x-3}{5}\cdot \:60+\frac{x+1}{2}\cdot \:60\\\\\frac{2x+1}{3}\cdot \:60\\\\=\frac{\left(2x+1\right)\cdot \:60}{3}\\\\\frac{60}{3}=20\\\\=20\left(2x+1\right)$

$\frac{3x-5}{4}\cdot \:60\\\\=\frac{\left(3x-5\right)\cdot \:60}{4}\\\\\frac{60}{4}=15\\\\=15\left(3x-5\right)\\\\\frac{4x-3}{5}\cdot \:60\\\\=\frac{\left(4x-3\right)\cdot \:60}{5}\\\\\frac{60}{5}=12\\\\=12\left(4x-3\right)\\\\\frac{x+1}{2}\cdot \:60\\\\=\frac{\left(x+1\right)\cdot \:60}{2}\\\\\frac{60}{2}=30\\\\=30\left(x+1\right)\\\\20\left(2x+1\right)+15\left(3x-5\right)<12\left(4x-3\right)+30\left(x+1\right)\\\\20\left(2x+1\right)+15\left(3x-5\right)$

$=40x+20+15\left(3x-5\right)\\\\=40x+20+45x-75\\\\=85x-55\\\\85x-55<78x-6\\\\85x-55+55<78x-6+55\\\\85x<78x+49\\\\85x-78x<78x+49-78x\\\\7x<49\\\\\frac{7x}{7}<\frac{49}{7}\\\\x<7$