How do you solve (1/2)x - (1/3)y = -3(12)x−(13)y=−3 and (1/8)x + (1/6)y = 0(18)x+(16)y=0?

1 Answer
José F.
Jan 18, 2016

(x,y)=(-4,3)(x,y)=(−4,3)

Explanation:

(1/2)x-(1/3)y=-3(12)x−(13)y=−3 and (1/8)x+(1/6)y=0(18)x+(16)y=0

The minimum common multiple between 2,3,6 and 8 is 24. So
let's convert all these fractions in something/24:

(12/24)x-(8/24)y=-(72/24)(1224)x−(824)y=−(7224) and (3/24)x+(4/24)y=0(324)x+(424)y=0

Then multiply all the equalities by 24:

12 x - 8y =-7212x−8y=−72 and 3 x +4 y =0 3x+4y=0

If 3 x + 4 y =0 3x+4y=0 the double 6 x + 8 y6x+8y is also 0, so we can
add it to the other equation without change the result:

6 x + 8 y=06x+8y=0 plus 12 x - 8y =-7212x−8y=−72 gives 18x=-7218x=−72

So, x=-72/18=-4x=−7218=−4

and applying the value x=-4x=−4 to the expression:

3 x +4 y =0 3x+4y=0

gives

3*(-4)+4y=03⋅(−4)+4y=0
4y=124y=12
y=3y=3

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