Question 275440
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You have a problem common among students first learning to solve problems using algebra: turning the given information into a correct algebraic equation.<br>
You do, however, have a good start on another part of setting the problem up -- using the information that the sum of the two number is 10 to call the two numbers "x" and "10-x".<br>
Now for converting "Three times one integer is 3 less than 8 times the other integer" into a correct equation....<br>
"Three times one integer..."
you have that part okay:
Your equation to here is "3x..."<br>
"... is"
That is your equals sign; think of it as saying "is equal to" instead of just "is"
Now your equation is "3x =..."<br>
"...3 less than 8 times the other integer"
Here is where beginning algebra students make a very common error.  They see the "3 less..." and they write "3-...".  But "3 less than A" is "A - 3", not "3 - A".  So<br>
"x" is your "one number", so the other number is "10-x".  Then "8 times the other number" is "8(10-x)", and finally "3 less than 8 times the other number" is "8(10-x)-3".<br>
And now you have the complete equation:<br>
"3x = 8(10-x)-3"<br>
Now you can solve using basic algebraic operations.<br>
{{{3x=8(10-x)-3}}}
{{{3x=80-8x-3}}}
{{{3x=77-8x}}}
{{{11x=77}}}
{{{x=7}}}<br>
And the numbers are x=7 and (10-x)=3<br>
ANSWERS: 7 and 3<br>