Найдите решение системы неравенств с наибольшим значением x. |3x+5y-8|

Відповідь:

Покрокове пояснення:

$\left \{ {{3x+5y-8\leq4 } \atop {2x+y-3\leq2 }} \right.\\\\\left \{ {{3x+5y\leq4+8 } \atop {y\leq2+3-2x }} \right.\\\\\left \{ {{3x+5y\leq12 } \atop {y\leq5-2x }} \right.\\\\\left \{ {{3x+5(5-2x)\leq12 } \atop {y\leq5-2x }} \right.\\\\\left \{ {{3x+5*5-5*2x\leq12 } \atop {y\leq5-2x }} \right.\\\\\left \{ {{3x+25-10x\leq12 } \atop {y\leq5-2x }} \right.\\\\\left \{ {{3x-10x\leq12-25 } \atop {y\leq5-2x }} \right.\\\\\left \{ {{-7x\leq-13 } \atop {y\leq5-2x }} \right.$

$\left \{ {{x\leq-13:(-7) } \atop {y\leq5-2x }} \right.\\\\\left \{ {{x\geq \frac{13}{7} } \atop {y\leq5-2x }} \right.\\\\\left \{ {{x\geq1\frac{6}{7} } \atop {y\leq5-2x }} \right.$

$y\leq5-2x\\y\leq 5-2x\\y\leq 5-2*1\frac{6}{7} \\y\leq 5-2*\frac{13}{7} \\y\leq 5-\frac{26}{7} \\y\leq \frac{35}{7} -\frac{26}{7} \\y\leq \frac{9}{7} \\y\leq 1\frac{2}{7}$

x=1 - наибольшее значение