Question 873914
{{{t}}}= tens digit.
{{{u}}}= ones/units digit.
 
One of the statements in the problem says that {{{t+u=10}}}
 
{{{10t+u}}}= value of the two-digit numeral.
{{{10u+t}}}= value of the number named when the digits are reversed.
 
Another statement in the problem says that
{{{10t+u}}} is {{{18}}} more than {{{10u+t}}} .
That can be written as the equation
{{{10t+u=10u+t+18}}} .
Subtracting {{{u}}} and {{{t}}} from both sides we get
{{{10t-t+u-u=10u-u+t-t+18}}} , which simplifies to
{{{9t=9u+18}}}
Then, dividing both sides by {{{9}}} , we get
{{{9t/9=(9u+18)/9}}}--->{{{t=9u/9+18/9}}}--->{{{t=u+2}}}
 
Take the equation above, and {{{t+u=10}}} written at the beginning,
we have the system {{{system(t+u=10,t=u+2)}}} ,
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+2}}} for {{{t}}} in {{{t+u=10}}} , we get
{{{u+2+u=10}}}-->{{{u+u+2=10}}}-->{{{2u+2=10}}}-->{{{2u=10-2}}}-->{{{2u=8}}}-->{{{u=8/2}}}-->{{{highlight(u=4)}}}
 
Now, we substitute the value found for {{{u}}} in {{{t=u+2}}} to find the value of {{{t}}} :
{{{t=u+2}}}-->{{{t=4+2}}}-->{{{highlight(t=6)}}} .
 
So, with {{{t=6}}} and {{{u=4}}} , the two-digit numeral is {{{highlight(64)}}} .