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Start your free trialJosh O.
2,370 PointsI do not understand what I am doing wrong. I have added 5 to integerOne even printed the result (echo).
I Added 5 to $integerOne and subtracted 1 from $integerTwo. I even displayed the correct answer by using echo. Then I started over and just added and subtracted the appropriate numbers. still it returned an warning "Bummer: you haven't changed $integerOne.
Here was my original code: $integerOne =1; $integerTwo = 2; $floatOne = 1.5; $answerOne = $integerOne + 5; $answerTwo = $integerTwo -1; echo $answerOne; echo " and " echo $answerTwo;
<?php
//Place your code below this comment
$integerOne = 1;
$integerTwo = 2;
$floatOne = 1.5;
$integerOne + 5;
$integerTwo - 1;
?>
1 Answer
joelearner
54,915 PointsHi Josh,
Since you are adding a value to a variable that has been initialized, you will need to specify the variable name when you add or subtract a number. It looks kind of confusing when you're first starting off. You can do something like this:
variable1 = variable1 + 5;
Or this is also correct:
variable1 += 5;
Cheers!
Josh O.
2,370 PointsJosh O.
2,370 PointsThank you very much!!!