Question 23450
SEE THE FOLLOWING EXAMPLE AND FOLLOW THE METHOD TO SOLVE YOUR PROBLEM.COME BACK IF NEEDED
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Solve the system of linear equations by graphing: x+2y=5..
OR 2Y=5-X..OR Y=(5-X)/2
                                                  x-y=2...
OR Y=X-2
that is the first one 
 the second one is:     4x-y=6
                        3x+y=1

SOLVE BY GRAPHING MEANS DRAW THE GRAPHS AND FIND THE 2 GRAPHS OF 2 EQUATIONS CUT EACH OTHER .THAT IS THE SOLUTION.LET ME SHOW YOU THE FIRST PROBLEM. THEN YOU CAN DO SECOND PROBLEM THE SAME WAY 
LET X=.........................0......1......2........3.........4......5....ETC.
Y=(5-X)/2..IN FIRST EQN........5/2....2......3/2......1.........1/2....0...ETC
Y=(X-2).....IN SECOND EQN.......-2....-1......0........1.........2......3...ETC
THE GRAPHS WILL LOOK LIKE THIS 
{{{ graph( 500, 500, -10, 10, -10, 10, (5-x)/2, x-2) }}}
WE FIND THAT THE TWO GRAPHS ARE CUTTING AT X=3 AND Y=1.HENCE THAT IS THE SOLUTION..YOU CAN CHECK BACK
EQN.I.....X+2Y=3+2*1=3+2=5..OK
EQN.II.....X-Y=3-1=2........OK.