The two digit number needed is valued 58.
Given that:
The tens of a two digit number = unit's digit - 3
Number reversed + The number itself = 143
Let the ones digit of that two digit number be x and the tens digit be y. Then by place value concept, we have:
[tex]10(y) + x[/tex]
Since the tens digit = unit digit - 3
Thus we have:
[tex]y = x - 3[/tex]
or the number is :
[tex]10(x-3) + x = 11x - 30[/tex]
Now sum of the number and the number reversed is 143, which means:
[tex]143 = 10(x) + (y) + 10(y) + x = 10x + (x-3) + 11x - 30\\143 = 22x - 3 - 30\\x = 8[/tex]
Thus, the number is [tex]10(x-3) + x = 50 + 8 = 58[/tex]
Thus, the two digit number needed is valued 58.
Learn more about the sum of digits of a number:
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