MYSQL/PHP Wont update - php

Question

I have this code which doesn't work for some reason
$con = mysql_connect("$server", "$user", "$pass");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db('$db', $con);
// BLOCK 1
$supplier=$_POST["supplier"];
$supplierNowBalance =$_POST["supplierNowBalance"];
$supplierBalance=$supplierNowBalance - $supplier;
mysql_query("UPDATE inv SET balance='".$supplierBalance."' WHERE username='supplier' AND name='bingo'");
echo $supplierBalance;
// END OF BLOCK 1
Everything before and after the UPDATE statement works. I copied the update statement to PHPMyAdmin and it updates just fine! I don't get it. for some reason php is not running the mysql_query
Few tips: I have few of this block of code in the program so it goes through all of them posts and calculates, and updates
I have tried every possible combination for the mysql query (with and without 's )
###############ANSWER__________
Thanks for the answer but I figured it out. Apparently PHP is desperate to have double quote "" around the connection string.
On line 6, when I descrbed the database variable
mysql_select_db('$db', $con);
I had single qoute, I changed it to double qoute and bam, it worked, Thanks
mysql_select_db("$db", $con);

Answer 1

balance is numeric right? you shouldn't wrap it with single quotes.
mysql_query("UPDATE inv SET balance= " . $supplierBalance . " WHERE username='supplier' AND name='bingo'");
using PDO makes it invulnerable with SQL INJECTION
<?php
// other codes
$stmt = $dbh->prepare("UPDATE inv SET balance = ? WHERE username = ? AND name = ?");
$stmt->bindParam(1, $supplierBalance);
$stmt->bindParam(2, 'supplier');
$stmt->bindParam(3, 'bingo');
$stmt->execute();
// other codes
?>

Answer 2

Thanks for the answer but I figured it out. Apparently PHP is desperate to have double quote "" around the connection string.
On line 6, when I descrbed the database variable
mysql_select_db('$db', $con);
I had single qoute, I changed it to double qoute and bam, it worked, Thanks
mysql_select_db("$db", $con);