Question

solve it please with explanation ​

Answer 1

★ Question 1 ::

If x ∈ Z, Find the Solution set of the inequation ::

$\sf{10 - 2(1 + 4x) < 20}$

Also represent the Solution set on number line,

★ Answer ::

$\sf{ \implies \: 10 - 2(1 + 4x) < 20} \\ \\ \\ \sf{ \implies \: 10 - 2 - 8x < 20} \\ \\ \\ \sf{ \implies \: - 8 - 8x < 20} \\ \\ \\ \sf{ \implies \: - 8x < 20 - 8} \\ \\ \\ \sf{ \implies \: - 8x < 12} \\ \\ \\ \sf{ \implies \: x < \: \dfrac{ - 12}{8}} \\ \\ \\ \sf{ \implies \: x < \: \cancel{ \dfrac{ - 12}{8}}} \: \\ \\ \\ \sf{ \implies \: x < \: \dfrac{ - 3}{2} \: or \: - 1.5}$

Now, number line,

$\setlength{\unitlength}{2.5mm}\begin{picture}(10,10)\linethickness{0.45mm}\put(20,20){\vector(2,0){18}}\put(20,20){\vector(-2,0){18}}\multiput(6,19.35)(2,0){13}{\line(0,1){1.3}}\put(20,20){\circle*{1}}\put(19.3,17){ \sf \: 0 }\put(21.1,17){ \sf \: 1 }\put(23.3,17){ \sf \: 2 }\put(25.2,17){ \sf \: 3 }\put(27.3,17){ \sf \: 4 }\put(29.3,17){ \sf \: 5 }\put(16.5,17){ \sf \: -1 }\put(14.3,17){ \sf \: -2 }\put(12.2,17){ \sf \: -3 }\put(10.3,17){ \sf \: -4 }\put(8.3,17){ \sf \: -5 }\put(20,22){ \sf \: x \: > \: -1.5 }\put(20,20.6){\vector(2,0){12}}\end{picture}$

★ Question 2 ::

Solve the inequation ::

$\sf{ \dfrac{2y + 1}{3} + 1 \leqslant 3, \: y \in \: W}$

★ Answer ::

$\sf{ \implies \dfrac{2y + 1}{3} + 1 \leqslant 3} \: \\ \\ \\ \sf{ \implies \dfrac{2y + 1}{3} \leqslant 3 - 1} \\ \\ \\ \sf{ \implies \dfrac{2y + 1}{3} \leqslant 2} \\ \\ \\ \sf{ \implies \dfrac{2y + 1}{3} \leqslant 2 \times 3} \\ \\ \\ \sf{ \implies 2y + 1 \leqslant 6} \\ \\ \\ \sf{ \implies 2y \: \leqslant 6 - 1} \\ \\ \\ \sf{ \implies 2y \leqslant 5} \\ \\ \\ \sf{ \implies y \leqslant \dfrac{5}{2} \: or \: 2.5} \\ \\ \\ \sf{As \: \: y \in \: W} \\ \\ \\ \sf{ \: \therefore \: y = {\lgroup \: 0,1,2 \rgroup}}$

Therefore, y = [ 0, 1, 2 ]

Answer 2

Answer:

Hey! army....

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