A problem solvable by a first-degree equation.

Question

In a two-digit number, the tens digit exceeds the units digit by $3$. The difference between twice the number with the digits reversed and the number itself is $42$. What is the number?

My solution: Let be $\boxed{x}\boxed y$ the number with two digits. $\boxed{x}\boxed y=10x+y$. The tens digit exceeds the units digit by $3$. Hence:

$$x=10\cdot (y+3)+y.$$

But $2\boxed{y}\boxed x-\boxed{x}\boxed y=42 \iff 20\cdot (y+3)+2y \ldots $.

I don't know how to set it up from the beginning. How should it be done?

Answer 1

You could call the digit of tens $a$ and the digit of units $b$. From the first condition you get $a=b+3$ and from the second $2(10b+a)-(10a+b)=42$. Solving you should get $a=9, b=6$.

Answer 2

In a two-digit number, the tens digit exceeds the units digit by 3 . The difference between twice the number with the digits reversed and the number itself is 42 . What is the number?

Notice that to solve this you need to use first-degree "system of equations composed of 2 equation in 2 variables" as will be shown below.

$$N=xy=x+10y$$

The tens digit exceeds the units digit by 3 $$y-x=3 \tag 1 $$

The difference between twice the number with the digits reversed and the number itself is $42$

$$2(10x+y)-x-10y=42$$ $$20x+2y-x-10y=42$$ $$19x-8y=42$$ Using (1), substitute the value of y in the last equation to get: $$19x-8*(3+x)=19x-24-8x=42$$ $$11x=66$$ $$x=6$$

To get the value of $y$ Use (1) with the obtained value of $x=6$: $$y-6=3$$

$$y=9$$

Now verify the answer:

$N=6+9*10$

$$9-6=3...true$$

$$2*69-96=42...true$$