Answers

2014-03-13T14:54:18+01:00
FROM THE QUESTION ASKED WE SAY THAT
2x+3y=1    eq1
5x+7y=3    eq2
USING SIMULTANEOUS METHOD
FROM EQ 1
2x+3y=1
TO MAKE X THE SUBJECT OF THE FORMULA
2x=1-3y
x=1-3y/2
SUBSTITUTE 1-3y/2 FOR x IN eq 2
5(1-3y/2)+7y=3
5-15y/2+7y=3
DIVIDING THROUGH BY L.C.M
5-15y+14y=6
COLLECTING LIKE TERM
-15y+14y=6-5
-y=1
y=-1
SUBSTITUTING -1 FOR Y IN EQ 1
2x+3y=1
2x+3(-1)=1
2x-3=1
2x=1+3
2x=4
x=4/2
x=2
SO THEREFORE X=2 AND Y=-1

ALSO USING ELIMINATION METHOD
WE HAVE
2x+3y=1
5x+7y=3
MULTIPLYING BOTH SIDE BY THE COEFFICIENT OF x IN EACH OF THE EQUATION
2x+3y=1 X5
5x+7y=3 X2

10x+15y=5    EQ 3
10x+14y=6    EQ4
SUBTRACT EQ 4 FROM EQ3
0+y=-1
y=-1
to find x
2x+3(-1)=1
2x-3=1
2x=4
x=2
SO THEREFORE
x=2, y=-1


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