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El triplo de la suma de dos números es 63, y el número mayor es 6 veces el menor. Entonces, el número mayor es?

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Final answer:

To solve this problem, we can use algebra and solve a system of equations to find the values of the two numbers. The larger number is 18.

Step-by-step explanation:

To solve this problem, let's use algebra. Let's call the smaller number x and the larger number y. We are given two pieces of information: the triple sum of the numbers is 63, and the larger number is 6 times the smaller number.

The triple sum of the numbers is 63, so we can write an equation: 3(x + y) = 63

The larger number is 6 times the smaller number, so we can write another equation: y = 6x

We can now solve this system of equations to find the values of x and y. First, let's use the second equation to substitute y in the first equation: 3(x + 6x) = 63

Simplifying this equation gives us: 21x = 63

Dividing both sides of the equation by 21, we find that x = 3. Plugging this value back into the second equation, we find that y = 6(3) = 18.

So, the larger number is 18.

User Doug Paul
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x: número más grande

y: número menor

3 (x + y) = 63

x = 6 años

Sustituimos la x = 6y de la segunda ecuación en la primera

3 (6 años + años) = 63

7 años = 63/3

3 años = 21

y = 21/7

y = 3

Este es el número más pequeño y = 7, por lo que el número más grande es x = 6y = 6 * 3 = 18

El número más grande es 18