# There are two numbers whose sum is 47. Three times the smaller number is equal to 9 more than the...

## Question:

There are two numbers whose sum is 47. Three times the smaller number is equal to 9 more than the larger number. What are the numbers?

## Elimination Method:

In the elimination method, we solve a system of two equations of two variables by adding or subtracting the equations. By doing so, we get a linear equation in one variable, which we can solve easily.

Let us assume the two numbers to be {eq}x \text{ (smaller)} {/eq} and {eq}y \text{ (larger)} {/eq}.

The problem says, "The sum of two numbers is 47".

So we get:

$$x+y=47 \,\,\,\,\,\,\,\rightarrow (1)$$

The problem also says, "Three times the smaller number is equal to 9 more than the larger number".

So we get:

$$3x =y+9 \\[0.3cm] \text{Subtracting y from both sides}, \\[0.3cm] 3x-y=9 \,\,\,\,\,\,\,\rightarrow (2)$$

Adding the equations (1) and (2):

$$(x+y)+(3x-y)= 47+9 \\[0.3cm] \text{Combining the like terms}, \\[0.3cm] 4x= 56 \\[0.3cm] \text{Dividing both sides by 4}, \\[0.3cm] x= 14$$

Substitute this in (1):

$$14+y= 47 \\[0.3cm] \text{Subtracting 14 from both sides}, \\[0.3cm] y=33$$

Therefore, the two numbers are {eq}\boxed{\mathbf{14}} {/eq} and {eq}\boxed{\mathbf{33}} {/eq}.