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Sagot :
Let the number of orders served by Mai, Abdul, and Chris be represented by M, A, and C respectively.
A total of 111 orders were served. Thus,
[tex]M\text{ + A +C = 111 ---- equation 1}[/tex]Abdul served 3 times as many orders as Chris. Thus,
[tex]\begin{gathered} A\text{ = 3}\times C \\ \Rightarrow A\text{ = 3C ------ equation 2} \end{gathered}[/tex]Mai served 6 more orders than Chris. Thus,
[tex]M\text{ = (6 }+\text{ C) ----- equation 3}[/tex]Substitute equations 2 and 3 into equation 1. Thus, we have
[tex]\begin{gathered} M\text{ + A +C = 111} \\ \Rightarrow(6+C)\text{ + 3C+ C=111} \\ \end{gathered}[/tex]Simplify by opening the brackets
[tex]6\text{ + C + 3C + C = 111}[/tex]Collect like terms, we have
[tex]\begin{gathered} C\text{ + 3C + C = 111-6} \\ \Rightarrow5C\text{ = 105} \end{gathered}[/tex]Solve for C dividing both sides of the equation by the coefficient of C.
The coefficient of C is 5. Thus,
[tex]\begin{gathered} \frac{5C}{5}=\frac{105}{5} \\ \Rightarrow C=21 \end{gathered}[/tex]Solve for A by substituting the value of 21 for C in equation 2.
[tex]\begin{gathered} \text{From equation 2,} \\ A\text{ = 3C} \\ \text{= 3}\times21 \\ \Rightarrow A=63 \end{gathered}[/tex]Solve for M by substituting the value of 21 for C in equation 3.
[tex]\begin{gathered} \text{From equation 3,} \\ M\text{ = (6 }+\text{ C)} \\ =6\text{ + 21} \\ \Rightarrow M\text{ = 27} \end{gathered}[/tex]Hence,
number of orders served by Mai = 27
number of orders served by Abdul = 63
number of orders served by Chris = 21
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