Question

If the replacement set is {0, 1, 2, 3, 4, 5, 6, 7}, find the solution for each inequation
$\bf \: i) \: \: \: \: \: \: \: \: \frac{15}{2} < 2x + \frac{5}{2}$
$\bf \: ii) \: \: \: \: \: \: \: \: 4(x - 2) < 3x - 1$
$\bf \: iii) \: \: \: \: \: \: \: n + 2 > \frac{3n - 1}{2}$
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Answer 1

$\huge\pink{\boxed{\blue{\boxed{ \purple{ \boxed{{\pink{Answer}}}}}}}} \\ \large\pink{\boxed{\blue{\boxed{ \purple{ \boxed{{\pink{Your~answer↓}}}}}}}}$

$\bf \: i) \: \: \: \: \: \: \: \: \frac{15}{2} < 2x + \frac{5}{2}$

$\bf \: ii) \: \: \: \: \: \: \: \: 4(x - 2) < 3x - 1$

$\bf \: iii) \: \: \: \: \: \: \: n + 2 > \frac{3n - 1}{2}$

$\small\bold\red{Solution \: of \: (i)}$

$\frac{15}{2} < 2x + \frac{5}{2} \\ \frac{15}{2} - \frac{5}{2} < 2x \\ \frac{15 - 5}{2} < 2x \\ 5 < 2x \\ = > \: x > \frac{5}{2} \\ so \: x = {3, 4, 5, 6, 7}$

$\small\bold\red{Solution \: of \: (ii)}$

$4(x - 2) < 3x - 2 \\ 4x - 8 < 3x - 2 \\ 4x - 3x < - 2 + 8 \\ x < 6 \\ so \: x = 0, 1, 2, 3, 4, 5$

$\small\bold\red{Solution \: of \: (iii)}$

$n + 2 > \frac{3n - 1}{2} \\ 2n + 4 > 3n - 1 \\ 2n - 3n > - 1 - 4 \\ - n > - 5 \\ = > \: n < 5 \\ so \: n \: = 0, 1, 2, 3, 4$

$\huge \fcolorbox{black}{cyan}{♛hope \: it \: helps \: you♛}$

Answer 2

★ǫᴜᴇsᴛɪᴏɴ★

If the replacement set is {0, 1, 2, 3, 4, 5, 6, 7}, find the solution for each inequation

$\bf \: i) \: \: \: \: \: \: \: \: \frac{15}{2} < 2x + \frac{5}{2}$

$\bf \: ii) \: \: \: \: \: \: \: \: 4(x - 2) < 3x - 1$

$\bf \: iii) \: \: \: \: \: \: \: n + 2 > \frac{3n - 1}{2}$

Answer:-

$\: \boxed { \bf \: Solution \: for \: (i)}$

$\bf \frac{15}{2} < 2x + \frac{5}{2}$

$\bf \to \frac{15}{2} - \frac{5}{2} < 2x$

$\bf \to \frac{10}{2} < 2x$

$\bf\to \frac{10}{2 \times 2} < x$

$\bf \to \frac{5}{2} < x$

$\small \bf∴The \: solution \: set \: is \: \{3,4,5,6,7 \}$

$\: \boxed { \bf \: Solution \: for \: (ii)}$

$\bf4(x - 2) < 3x - 2$

$\bf \to4x - 8 < 3x - 2$

$\bf \to4x - 3x < 8 - 2$

$\bf \to \: x < 6$

$\small \bf∴The \: solution \: set \: is \: \{0,1,2,3,4,5 \}$

$\: \boxed { \bf \: Solution \: for \: (iii)}$

$\bf \: n + 2 > \frac{3n - 1}{2}$

$\bf \to2(n + 2) > 3n - 1$

$\bf \to2n + 4 > 3n - 1$

$\bf \to2n - 3n > - 1 - 4$

$\bf \to - n > - 5$

$\bf \to \: n < 5$

$\small \bf∴The \: solution \: set \: is \: \{0,1,2,3,4 \}$

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