Let f:R5→R3f(x1,x2,x3,x4,x5)=(x1,x2,x3,0,0) 
One can check that f is a homomorphism .
Moreover it is onto.
Kernel(f)={(x1,x2,x3,x3,x4,x5)∣f(x1,x2,x3,x4,x5)=(0,0,0,0,0)} 
                ={(0,0,0,x4,x5)∣x4,x5∈R}  ≅R2  
Therefore by 1st isomorphism theorem  R5/R2≅R3 
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