Does YACC (LALR) parses left recursive grammars - parsing

Question

I was doing a simple C parser project.
This problem occurred when I was writing the grammar for if-else construct.
the grammar that I have written is as following:
iexp: IF OP exp CP block eixp END{
printf("Valid if-else ladder\n");
};
eixp:
|ELSE iexp
|ELSE block;
exp: id
|NUMBER
|exp EQU exp
|exp LESS exp
|exp GRT exp
|OP exp CP;
block: statement
|iexp
|OCP statement CCP
|OCP iexp CCP;
statement: id ASSIGN NUMBER SEMICOL;
id: VAR;
where the lex part looks something like this
"if" {return IF;}
"else" {return ELSE;}
[0-9]+ {return NUMBER;}
">" {return GRT;}
"<" {return LESS;}
"==" {return EQU;}
"{" {return OCP;}
"}" {return CCP;}
"(" {return OP;}
")" {return CP;}
"$" {return END;}
";" {return SEMICOL;}
"=" {return ASSIGN;}
[a-zA-Z]+ {return VAR;}
. {;}
I am getting o/p as
yacc: 9 shift/reduce conflicts, 1 reduce/reduce conflict.
When I eliminate the left recursion on exp derivation the conflicts vanish but why it's so ?
the revised grammar after eliminating left recursion was :
exp: id
|NUMBER
|id EQU exp
|id LESS exp
|id GRT exp
|OP exp CP;
I was able to parse successfully the grammar for the evaluation of arithmetic expressions. Is it so that %right, %left made it successful
%token ID
%left '+' '-'
%left '*' '/'
%right NEGATIVE
%%
S:E {
printf("\nexpression : %s\nResult=%d\n", buf, $$);
buf[0] = '\0';
};
E: E '+' E {
printf("+");
($$ = $1 + $3);
} |
E '-' E {
printf("-");
($$ = $1 - $3);
} |
E '*' E {
printf("*");
($$ = $1 * $3);
} |
E '/' E {
printf("/");
($$ = $1 / $3);
} |
'(' E ')' {
($$ = $2);
} |
ID {
/*do nothing done by lex*/
};