the units digit of a 2-digit number exceeds twice the tens digit by 1. find the number if the sum of its digits is 10.
Please help i am completley lost!!
x y
y = 2x+1
y = 10-x
so
10-x = 2x+1
3 x = 9
etc
im still pretty confused, but i think u r suppossed to solve the 2 equations u gave and got (3,7) but i think im way off
Check it by using 37 in the problem statement.
the units digit (7) of a 2-digit number (37)exceeds twice the tens (2*3=6) digit by 1.(6+1=7 sure enough)) find the number if the sum of its digits is 10. (3+7=10 sure enough)
thank u
To solve this problem, we can use algebra to represent the given information and find the answer. Let's break down the problem step by step:
Step 1: Let's assume that the tens digit of the 2-digit number is represented by the variable 'x,' and the units digit is represented by the variable 'y.'
Step 2: According to the problem, the units digit exceeds twice the tens digit by 1. So, we can write the equation: y = 2x + 1.
Step 3: It is also given that the sum of the digits is 10. Therefore, we can write another equation: x + y = 10.
Now we have a system of two equations: y = 2x + 1 and x + y = 10.
To solve this system, we can substitute the value of y in the second equation with the value of y from the first equation.
Substituting y = 2x + 1 into x + y = 10, we get:
x + (2x + 1) = 10.
Simplifying this equation: 3x + 1 = 10.
Subtracting 1 from both sides: 3x = 9.
Dividing both sides by 3: x = 3.
Now, substitute x = 3 into y = 2x + 1, we get:
y = 2(3) + 1 = 6 + 1 = 7.
Therefore, the tens digit is 3, and the units digit is 7. Thus, the number is 37.