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## Answers

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