The sum of digits of a 2 digit number is 7, If digits are interchanged the new number formed is 3 less than four times of original number. ( use 2 variables)
t=tens digit
u=units digit
t+u = 7
10u+t = 4(10t+u)-3
42
To solve this problem, let's assume that the tens digit of the 2-digit number is 'x' and the ones digit is 'y'.
According to the given information, the sum of the digits is 7. This can be represented as:
x + y = 7 ----(equation 1)
Also, when the digits are interchanged, the new number formed is 3 less than four times the original number. Mathematically, we can write this as:
10y + x = 4(10x + y) - 3 ----(equation 2)
Now we have a system of two equations (equation 1 and equation 2) with two variables (x and y). We can solve this system to find the values of x and y.
To solve the system, we can use the method of substitution or elimination. Let's use substitution:
From equation 1, we can modify it to express x in terms of y:
x = 7 - y
Now substitute this expression for x in equation 2:
10y + (7 - y) = 4(10(7 - y) + y) - 3
Simplify the equation:
10y + 7 - y = 4(70 - 3y + y) - 3
9y + 7 = 4(70 - 2y) - 3
9y + 7 = 280 - 8y - 3
Combine like terms:
9y + 7 = 277 - 8y
Move all the terms to one side:
9y + 8y = 277 - 7
17y = 270
Divide both sides by 17:
y = 270 / 17
y = 15
Now substitute this value for y back into equation 1 to find x:
x + 15 = 7
x = 7 - 15
x = -8
Therefore, the tens digit is -8 and the ones digit is 15, which is not possible for a 2-digit number with positive digits. Hence, there is no solution to this problem.