Given:
One number is 2 less than a second number.
Twice the second number is 2 less than 3 times the first.
Required:
We need to find the smaller number and the larger number,
Explanation:
Let x be the first number and y be the second number.
The first number is 2 less than the second number.
[tex]x=y-2[/tex]
Twice the second number is 2 less than 3 times the first.
[tex]twice\text{ the second number means 2y.}[/tex][tex]three\text{ times the first number means 3x.}[/tex]
Twice the second number is 2 less than 3 times the first.
[tex]2y=3x-2[/tex]
Substitute x =y-2 in the equation.
[tex]2y=3(y-2)-2[/tex]
[tex]2y=3y-3\times2-2[/tex]
[tex]2y=3y-6-2[/tex]
[tex]2y=3y-8[/tex]
Solve for y.
[tex]3y-2y=8[/tex]
[tex]y=8[/tex]
Substitute y =8 in the equation x =y-2, we get
[tex]x=8-2[/tex][tex]x=6[/tex]
Final answer:
[tex]The\text{ smaller number is 6.}[/tex]
[tex]The\text{ larger number is 8.}[/tex]
