Question

for what values of a does the following system have at least 1 solution

3(a-5x)<1+x, 2-x/2>3+5(x-a)

Answer 1

Answer:

• for the first inequality:

$\dashrightarrow \: { \tt{3(a - 5x) < 1 + x}} \\ \\ { \tt{3a - 15x < 1 + x}} \\ \\ { \tt{3a < 1 + 16x}} \\ \\ { \boxed{ \tt{a = \frac{1 + 16x}{3} \: \: { \red{}} }}}$

• for the second inequality:

$\dashrightarrow \: { \tt{ \frac{2 - x}{2} > 3 + 5(x - a) }}$

substitute for a:

${ \tt{ \frac{2 - x}{2} > 3 + 5(x - \frac{1 + 16x}{3} ) }} \\ \\ { \tt{ \frac{2 - x}{2} > 3 - \frac{5 + 80x}{3} }} \\ \\ { \tt{2 - x > 6 - \frac{10 + 160x}{3} }} \\ \\ { \tt{6 - 3x > 18 - 10 - 160x}} \\ \\ { \tt{157x > 2}} \\ \\ { \tt{x > \frac{2}{157} }}$

• substitute to get value of a: [ first inequality ]

${ \tt{a < \frac{1 + 16( \frac{2}{157}) }{3} }} \\ \\{ \boxed{ \tt{a < \frac{63}{157} }}}$

[ second inequality ]

${ \boxed{ \tt{a > \frac{65}{157} }}}$