If m and n are integers such that $$(m+2n)(2m+n)=27$$, then the maximum possible value of $$2m-3n$$ is

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Sure! We can solve it without explicitly introducing new variables for substitution. We are given ( m + 2 n ) ( 2 m + n ) = 27 (m+2n)(2m+n) = 27 ( m + 2 n ) ( 2 m + n ) = 27 and we want to maximize 2 m − 3 n 2m-3n 2 m − 3 n Because m and n are integers, both (m+2n) and (2m+n) are going to be integer

If m and n are integers such that $(m+2n)(2m+n)=27$ , then the maximum possible value of $2m-3n$ is

If m and n are integers such that $(m+2n)(2m+n)=27$ , then the maximum possible value of $2m-3n$ is

If m and n are integers such that (m+2n)(2m+n)=27(m+2n)(2m+n)=27(m+2n)(2m+n)=27, then the maximum possible value of 2m−3n2m-3n2m−3n is

If m and n are integers such that (m+2n)(2m+n)=27(m+2n)(2m+n)=27(m+2n)(2m+n)=27, then the maximum possible value of 2m−3n2m-3n2m−3n is

If m and n are integers such that $(m+2n)(2m+n)=27$ , then the maximum possible value of $2m-3n$ is

If m and n are integers such that $(m+2n)(2m+n)=27$ , then the maximum possible value of $2m-3n$ is