Answer:
The three numbers are 17, 10 and 51.
Step-by-step explanation:
Let the first, second, and third numbers be [tex]x[/tex], [tex]y[/tex], and [tex]z[/tex] respectively.
• From the question, we know:
sum of the numbers is 78.
∴ [tex]x + y + z = 78[/tex] ----------(1st equation)
Let's express both [tex]x[/tex] and [tex]z[/tex] in terms of [tex]\bf y[/tex] :
• We know that:
the third number is 3 times the first.
∴ [tex]z = 3x[/tex]
⇒ [tex]x = \frac{z}{3}[/tex] ----------(2nd equation)
• We also know that:
the first number is 7 more than the second.
∴ [tex]\boxed{x = 7 + y}[/tex]
Substituting [tex]x = \frac{z}{3}[/tex] (from 2nd equation)
⇒ [tex]\frac{z}{3} = 7 + y[/tex]
⇒ [tex]\boxed{z = 21 + 3y}[/tex]
• We can now substitute [tex]x = 7 + y[/tex] and [tex]z = 21 + 3y[/tex] into the first equation:
[tex]x + y + z = 78[/tex]
⇒ [tex](7 + y) + y + (21 + 3y) = 78[/tex]
⇒ [tex]5y + 28 = 78[/tex]
⇒ [tex]5y = 50[/tex]
⇒ [tex]y = \bf 10[/tex]
∴ [tex]x = 7 + 10\\[/tex]
⇒ [tex]x = \bf 17[/tex]
[tex]z = 21 + 3(10)[/tex]
⇒ [tex]z = \bf 51[/tex]
∴ The three numbers are 17, 10 and 51.
