The sum of the digits of a certain two-digit number is 7. Reversing its digits increase the number by 9. What is the number. Use x and y. Please help me!!!!
in the original number, let the unit digit be x, then tens digiti be y
so the original number is 10y + x
where x+y = 7
the number reversed will be 10x + y
new number - old number = 9
10x + y - (10y + x) = 9
9x - 9y = 9
or x-y = 1
x+y=7
x-y=1
add them
2x = 8
x = 4, then y = 3
the number is 10y+x = 34
check:
original number is 34
reversed number is 43
is the number increased by 9 ??? YES
is the sum of the digits 7 ??? YES
YEAHHH
thank you!! this really helped me sincemy parents are busy.
The sum of the digits of a two-digit number is 7. The value of the number is 2 less than 12 times the tens digit. Find the number.
To solve this problem, let's use the variables "x" and "y" to represent the tens and ones digits of the two-digit number, respectively.
According to the problem, the sum of the digits is 7, so we can write the equation:
x + y = 7 ----(Equation 1)
Also, it is given that reversing the digits increases the number by 9. This means that the original number minus the reversed number equals 9. We can express this as:
(10x + y) - (10y + x) = 9
Simplifying this equation gives:
9x - 9y = 9
Dividing both sides by 9, we get:
x - y = 1 ----(Equation 2)
Now we have a system of linear equations with two unknowns. We can solve this system by substitution or elimination method. I will use the elimination method to find the values of x and y.
Adding Equation 1 and Equation 2, we get:
(x + y) + (x - y) = 7 + 1
2x = 8
Dividing both sides by 2, we find:
x = 4
Substituting the value of x back into Equation 1, we can solve for y:
4 + y = 7
y = 7 - 4
y = 3
So the tens digit (x) is 4, and the ones digit (y) is 3. Therefore, the number is 43.