Answer:
Q1: p = - 33
Q2: d = - 99
Q3: t = - 13
Step-by-step explanation:
Q1: [tex]$ \textbf{-} \frac{\textbf{p}}{\textbf{3}} \hspace{1mm} \textbf{-} \hspace{1mm} \textbf{8} \hspace{1mm} \textbf{=} \hspace{1mm} \textbf{3}[/tex]
We solve this taking LCM.
We get: [tex]$ \frac{-p - 24}{3} = 3 $[/tex]
[tex]$ \implies - p - 24 = 9 $[/tex]
[tex]$ \implies \textbf{p} \hspace{1mm} \textbf{=} \hspace{1mm} \textbf{- 33} $[/tex]
Q4: [tex]$ \frac{\textbf{d}}{\textbf{11}} \hspace{1mm} \textbf{-} \hspace{1mm} \textbf{4} \hspace{1mm} \textbf{=} \hspace{1mm} \textbf{- 13} $[/tex]
Again we proceed like Q1 by taking LCM.
We get: [tex]$ \frac{d - 44}{11} = - 13 $[/tex]
[tex]$ \implies d - 44 = - 13 \times 11 = - 143 $[/tex]
[tex]$ \implies d = - 143 + 44 $[/tex]
[tex]$ \implies \textbf{d} \hspace{1mm} \textbf{=} \hspace{1mm} \textbf{- 99} $[/tex]
Q7: 5t + 12 = 4t - 1
We club the like terms on either side.
[tex]$ \implies 5t - 4t = - 1 - 12 $[/tex]
[tex]$ \implies (5 -4)t = - 13 $[/tex]
[tex]$ \implies \textbf{t} \hspace{1mm} \textbf{=} \hspace{1mm} \textbf{- 13} $[/tex]
Hence, the answer.
