Part 1 :-
[tex]{:\implies \quad \sf \dfrac{2x}{3}=\dfrac84}[/tex]
[tex]{:\implies \quad \sf 2x=\dfrac84 \times 3}[/tex]
[tex]{:\implies \quad \sf 2x=6}[/tex]
[tex]{:\implies \quad \sf x=\dfrac62 =\boxed{\bf{3}}}[/tex]
Part 2 :-
Given that, f(x) = x² + x + 7, for f(-2), put x = -2
[tex]{:\implies \quad \sf f(-2)=(-2)^{2}+(-2)+7}[/tex]
[tex]{:\implies \quad \sf f(-2)=4-2+7=\boxed{\bf{9}}}[/tex]
Part 3 :-
We will use • between small numbers and +, - in big numbers of the above given problem as:
[tex]{:\implies \quad \sf 4\underline{\boxed{\bf +}}\sf 5\underline{\boxed{\bf \cdot}}\sf 2\underline{\boxed{\bf -}}\sf 6=8}[/tex]
Part 4 :-
7, is a natural number, and every natural number is whole number, and every whole number is an integer, so (a),(b), and (c) are correct. Also, 7 is not an irrational number. So, option (d) is incorrect
Part 5 :-
[tex]{:\implies \quad \sf 2+3+x=1+8-2}[/tex]
[tex]{:\implies \quad \sf 5+x=7}[/tex]
[tex]{:\implies \quad \sf x=7-5=\boxed{\bf 2}}[/tex]
Part 6 :-
[tex]{:\implies \quad \sf 12b-5b+6b-20b}[/tex]
[tex]{:\implies \quad \sf 7b-14b}[/tex]
[tex]{:\implies \quad \boxed{\bf{-7b}}}[/tex]
