Answer: it;s D..........................
Let <em>n</em> be the unknown number. We can write it as
<em>n</em> = 10<em>a</em> + <em>b</em>
with <em>a</em> and <em>b</em> integers between 1 and 9 (either with positive or negative sign).
Reversing the digits gives another number
<em>m</em> = 10<em>b</em> + <em>a</em>
The first number is increased by 54 when the digits are reversed, which means
<em>m</em> = <em>n</em> + 54 → 10<em>b</em> + <em>a</em> = 10<em>a</em> + <em>b</em> + 54 → 9<em>b</em> - 9<em>a</em> = 54 → <em>b</em> - <em>a</em> = 6
The digit in the tens place of <em>n</em> is 3 times the digit in the ones place, so
<em>a</em> = 3<em>b</em>
Substitute this into the previous equation and solve for <em>b</em> :
<em>b</em> - <em>a</em> = <em>b</em> - 3<em>b</em> = -2<em>b</em> = 6 → <em>b</em> = -3
Solve for <em>a</em> :
<em>a</em> = 3<em>b</em> = 3(-3) = -9
Then the original number is <em>n</em> = 10<em>a</em> + <em>b</em> = 10(-9) + (-3) = -93
The only system that has infinitely many solutions is NUMBER 2. Where you solve a system, and both sides end up being 0=0, then the system will have infinitely many solutions. Well, in these four choices, Number 2 is the only one that ends up being 0=0 after I solved it.
I’m pretty sure it’s the 2nd one
