Let the first(smaller) number = x
Let the other(larger) number= y
One number is four times the other number.
• y=4x
If 6 is added to the smaller number, we have: x+6
If 4 is added to the larger number, we have: y+4
Then the later number becomes twice the other number
• The later number = y+4
Therefore:
y+4=2(x+6)
![\begin{gathered} y+4=2\left(x+6\right) \\ \text{Recall: y=4x} \\ \text{Therefore we have:} \\ 4x+4=2(x+6) \\ 4x+4=2x+12 \\ 4x-2x=12-4 \\ 2x=8 \\ \text{Divide both sides by 2} \\ x=4 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/3u3lwj7bsdjwjrb5h9xnyatpe7wq4y3c8a.png)
Next, we substitute x=4 to find the value of y
![\begin{gathered} y=4x \\ y=4*4 \\ y=16 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/32c81i6dt3guyh1dyx1l7ensi3pxm3na2g.png)
The numbers are 4 and 16.