If one number is thrice the other and their sum is $16$, find the numbers If one number is thrice the other and their sum is $16$, find the numbers.
I tried, 
Let the first number be $x$ and the second number be $y$
Acc. to question  
$$
\begin{align}
x&=3y &\iff x-3y=0 &&(1)\\
x&=16-3y&&&(2)
\end{align}
$$
 A: I assume you have $$x=\color{red}{3y}, ~~x+y=16$$ Then $3(x+\color{red}{y})=3\times 16=48$ and so $3x+\color{red}{3y}=48$ and so $3x+x=48$ and so $4x=48$...
A: The problem statement says
$$x+3x=16,$$
hence
$$x=4,\\3x=12.$$
A: $$x=3y$$
$$y+x=16 \Rightarrow y+3y=16 \Rightarrow 4y=16 \Rightarrow y=4$$
So, $x=12$.
A: Let the first number be $x$. 
Let the second number be $y$. 
According to question
$$ \tag{1}
x+y=16 
$$
$$
\tag{2}   
x=3y
$$
So, $x-3y=0 \tag{2}$
Multiply equation $(1)$ by $3$. 
Solve both equations:
$$\tag{1} 3x+3y=48$$
$$\tag{2} x-3y=0$$
$$\tag{1) + (2}4x=48$$
$$\tag{3}x=12$$
Putting in equation $(1)$: 
$$\tag{1} x+y=16$$
$$\tag{1),(3} 12+y=16$$
$$\tag{4}y=16-12$$
$$\tag{5}y=4$$
A: Let the 1st number be $x$ and the 2nd number be $3x$. Since $x + 3x = 16$, $4x = 16$, so $x=4$. Therefore, 1st number is $4$ and the other is $12$.   
A: Let the first number be $x$
And second which is thrice be $3y$
Acc to ques.. 
$x=3y\ldots \textrm{equation 1}$
$x+y=16\ldots\textrm{equation 2}$
By Elimination method 
       x-3y  =    0
     -(x+ y) =  -16
     ______________
       0- 4y = - 16

Dividing the equation by $(-4)$. The result is $y=4$
Put $y=4$ into the first equation :$x=3\cdot 4\Rightarrow x=12$
I hope this will help you.. Thank you!
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