Question

Please solve these linear in equations.........​

Answer 1

(i)

$\quad$

$\displaystyle\longrightarrow\sf {\dfrac {x}{2}+5\leq\dfrac {x}{3}+6}$

$\quad$

$\displaystyle\longrightarrow\sf {\dfrac {x}{2}+5-\dfrac {x}{3}-6\leq0}$

$\quad$

$\displaystyle\longrightarrow\sf {\dfrac {x}{2}-\dfrac {x}{3}-1\leq0}$

$\quad$

$\displaystyle\longrightarrow\sf {\dfrac {x}{2}-\dfrac {x}{3}\leq1}$

$\quad$

$\displaystyle\longrightarrow\sf {\dfrac {3x-2x}{2\times 3}\leq1}$

$\quad$

$\displaystyle\longrightarrow\sf {\dfrac {x}{6}\leq1}$

$\quad$

$\displaystyle\longrightarrow\sf {x\leq6}$

$\quad$

Since $\displaystyle\sf {x}$ is a positive odd integer,

$\quad$

$\displaystyle\large\longrightarrow\sf {\underline {\underline {x\in\{1,\ 3,\ 5\}}}}$

$\quad$

(ii)

$\quad$

$\displaystyle\longrightarrow\sf {\dfrac {2x+3}{3}\geq\dfrac {3x-1}{4}}$

$\quad$

$\displaystyle\longrightarrow\sf {\dfrac {2x+3}{3}-\dfrac {3x-1}{4}\geq0}$

$\quad$

$\displaystyle\longrightarrow\sf {\dfrac {4(2x+3)-3(3x-1)}{3\times 4}\geq0}$

$\quad$

$\displaystyle\longrightarrow\sf {\dfrac {8x+12-9x+3}{12}\geq0}$

$\quad$

$\displaystyle\longrightarrow\sf {-\dfrac {x-15}{12}\geq0}$

$\quad$

$\displaystyle\longrightarrow\sf {\dfrac {x-15}{12}\leq0}$

$\quad$

$\displaystyle\longrightarrow\sf {x-15\leq0}$

$\quad$

$\displaystyle\longrightarrow\sf {x\leq15}$

$\quad$

Since $\displaystyle\sf {x}$ is a positive even integer,

$\quad$

$\displaystyle\large\longrightarrow\sf {\underline {\underline {x\in\{2,\ 4,\ 6,\ 8,\ 10,\ 12,\ 14\}}}}$