Answer :
In the word problem above, we are asked to write two or three algebra equations then solve them. This can be seen below;
Explanation
Let the time taken for the speaker to lecture in classrooms be x and the time taken to write the papers to be y.
Therefore, we can transform the given activities of the lecturer into two equations below;
[tex]\begin{gathered} \text{Monday}\rightarrow4x+3y=7hours \\ \text{Tuesday}\rightarrow3x+7y=8hours\text{ 25 mins} \end{gathered}[/tex]For ease of calculation, we would make 8 hours 25 mins, to just hours. This gives 25mins as 5/12 hours and altogether 101/12 hours. We can then solve the above equations simultaneously.
[tex]\begin{gathered} \begin{bmatrix}4x+3y=7 \\ 3x+7y=\frac{101}{12}\end{bmatrix} \\ \text{Isolate for x for 4x+3y=7:} \\ 4x=7-3y \\ x=\frac{7-3y}{4} \\ \end{gathered}[/tex]We substitute the value of x in the second equation.
[tex]\begin{gathered} \begin{bmatrix}3\cdot\frac{7-3y}{4}+7y=\frac{101}{12}\end{bmatrix} \\ \text{Next we simplify} \\ \begin{bmatrix}\frac{21+9y}{4}+7y=\frac{101}{12}\end{bmatrix} \\ \begin{bmatrix}\frac{21+19y}{4}=\frac{101}{12}\end{bmatrix} \\ We\text{ can break down the denominator;} \\ \begin{bmatrix}21+19y=\frac{101}{3}\end{bmatrix} \\ Cross\text{ multiply} \\ 3\times(21+19y)=101 \\ 63+57y=101 \\ 57y=101-63 \\ 57y=38 \\ y=\frac{38}{57} \\ y=\frac{2}{3} \end{gathered}[/tex]We can then substitute the value of y below
[tex]\begin{gathered} x=\frac{7-3y}{4} \\ \text{since y=}\frac{2}{3} \\ We\text{ can have } \\ x=\frac{7-3\cdot\frac{2}{3}}{4} \\ x=\frac{7-2}{4} \\ x=\frac{5}{4} \end{gathered}[/tex]The values of x and y imply that
[tex]\begin{gathered} It\text{ would take the speaker}\frac{5}{4}\text{hours to speak in one classroom } \\ It\text{ would also take the speaker }\frac{2}{3}\text{hours to write one paper} \end{gathered}[/tex]With this, we can find how long it would take the speaker to lecture once classroom and write four papers. This interprets as
[tex]\begin{gathered} x+4y=\text{?} \\ We\text{ can then substitute the values of x and y} \\ =\frac{5}{4}+\frac{2}{3}\times4 \\ =\frac{5}{4}+\frac{8}{3} \\ =\frac{15+32}{12} \\ =\frac{47}{12} \\ =3\frac{11}{12} \end{gathered}[/tex]From the derived answer, we can convert to the final answer in hours and minutes
[tex]\begin{gathered} 3hours+\frac{11}{12}\text{hours} \\ Since\text{ 60 mins = 1hour } \\ \therefore\frac{11}{12}\text{hours = }\frac{11}{12}\times60\text{ mins} \\ =11\times5\min s \\ =55\min s \end{gathered}[/tex]Therefore the time taken by the speaker to lecture one classroom and write four papers is;
Answer
[tex]3\text{hours 55 mins}[/tex]
