I assume you want it using the other method? It's much easier to substitute in cases like this because otherwise you'd end up taking away numbers with different powers (which is a lot harder to solve).
I assume you want it using the other method? It's much easier to substitute in cases like this because otherwise you'd end up taking away numbers with different powers (which is a lot harder to solve).
please explain what your issue is - if you feel that you can answer the question using a different one to the one suggested then your question is answered - but the suggestion made by Robbie is the sensible approach
Are you trying to solve the equations suggested by Robbie, or are you actually trying to solve the equations 2x+2y=1 and 32x+y=27, which are much simpler to solve?
Are you trying to solve the equations suggested by Robbie, or are you actually trying to solve the equations 2x+2y=1 and 32x+y=27, which are much simpler to solve?
Are you trying to solve the equations suggested by Robbie, or are you actually trying to solve the equations 2x+2y=1 and 32x+y=27, which are much simpler to solve?
please explain what your issue is - if you feel that you can answer the question using a different one to the one suggested then your question is answered - but the suggestion made by Robbie is the sensible approach
2^x+2y=2^0 x+2y=0----i then 3^2x+y=3^3 so 2x+y=3----ii so x+2y=0--*2 2x+y=3 --*1
therefore 2x+4y=0---ii - 2x+y=3 0+3y=-3 so y=-1 then plug it bck into any equations to find x