SOLUTION: The sum of the digits of a two-digit numeral is 10. The number named by the numeral is 18 more than the number named when the digits are reversed. What is the original numeral? Eq

Algebra ->  Average -> SOLUTION: The sum of the digits of a two-digit numeral is 10. The number named by the numeral is 18 more than the number named when the digits are reversed. What is the original numeral? Eq      Log On


   

Question 873914: The sum of the digits of a two-digit numeral is 10. The number named by the numeral is 18 more than the number named when the digits are reversed. What is the original numeral?
Equation & answer please.. Thank You!
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
t= tens digit.
u= ones/units digit.

One of the statements in the problem says that t%2Bu=10

10t%2Bu= value of the two-digit numeral.
10u%2Bt= value of the number named when the digits are reversed.

Another statement in the problem says that
10t%2Bu is 18 more than 10u%2Bt .
That can be written as the equation
10t%2Bu=10u%2Bt%2B18 .
Subtracting u and t from both sides we get
10t-t%2Bu-u=10u-u%2Bt-t%2B18 , which simplifies to
9t=9u%2B18
Then, dividing both sides by 9 , we get
9t%2F9=%289u%2B18%29%2F9--->t=9u%2F9%2B18%2F9--->t=u%2B2

Take the equation above, and t%2Bu=10 written at the beginning,
we have the system system%28t%2Bu=10%2Ct=u%2B2%29 ,
which we call "a system of linear equations."
We can solve it "by substitution,"
meaning that we take an expression that is equal to one variable from one equation,
and substitute that expression for the variable in the other equation.

Substituting u%2B2 for t in t%2Bu=10 , we get
u%2B2%2Bu=10-->u%2Bu%2B2=10-->2u%2B2=10-->2u=10-2-->2u=8-->u=8%2F2-->highlight%28u=4%29

Now, we substitute the value found for u in t=u%2B2 to find the value of t :
t=u%2B2-->t=4%2B2-->highlight%28t=6%29 .

So, with t=6 and u=4 , the two-digit numeral is highlight%2864%29 .