A collector bought 47 coins at auction, some bronze and some silver. The bronze coins cost him $ 14 each and the silver coins, $ 18 each. If in total he spent $ 750, how many bronze coins did he buy and how many silver? (with the steps)

please help me, it's for an exam

24 bronze coins and 23 silver coins

B= bronze coins S = silver coins
$14xB + $18 x S =$750
B + S = 47 total coins
Get in terms of S=47-B by subtracting B from each side, then replace the S in the first equation with this so that:
14xB+18x(47-B)=750
14B+846-18B=750
Subtract the 846 and combine the abs
-4B=-96 then divide the -4 on both sides
To get B=24 then 47-24=23=S