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2 years ago

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the sum of three numbers is 78. the third number is 3 times the first. the first number is 7 more than the second. what are the

numbers?

2 answers:

Lena [83]2 years ago

4 0

Answer:

17

Step-by-step explanation:

x+3x+x-7=78

5x-7=78

5x=78+7

5x=85

x=85/5

x=17

makkiz [27]2 years ago

4 0

<u>Answer:</u>

The three numbers are 17, 10 and 51.

<u>Step-by-step explanation:</u>

Let the first, second, and third numbers be $x$ , $y$ , and $z$ respectively.

• From the question, we know:

sum of the numbers is 78.

∴ $x + y + z = 78$ ----------(1st equation)

Let's express both $x$ and $z$ in terms of $\bf y$ :

• We know that:

the third number is 3 times the first.

∴ $z = 3x$

⇒ $x = \frac{z}{3}$ ----------(2nd equation)

• We also know that:

the first number is 7 more than the second.

∴ $\boxed{x = 7 + y}$

Substituting $x = \frac{z}{3}$ (from 2nd equation)

⇒ $\frac{z}{3} = 7 + y$

⇒ $\boxed{z = 21 + 3y}$

• We can now substitute $x = 7 + y$ and $z = 21 + 3y$ into the first equation:

$x + y + z = 78$

⇒ $(7 + y) + y + (21 + 3y) = 78$

⇒ $5y + 28 = 78$

⇒ $5y = 50$

⇒ $y = \bf 10$

∴ $x = 7 + 10\\$

⇒ $x = \bf 17$

$z = 21 + 3(10)$

⇒ $z = \bf 51$

∴ The three numbers are 17, 10 and 51.

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