How do you solve the following system of equations?: 13x+13y=16, 3x + 19y=1913x+13y=16,3x+19y=19?

1 Answer
Oct 22, 2017

x = 57/208, y = 199/208x=57208,y=199208

Explanation:

13x + 13y = 1613x+13y=16.....(I)(I)

3x + 19y = 193x+19y=19.....(II)(II)

Rearranging (II)(II) for x for simplicity,

3x = 19 - 19y3x=1919y

impliesx = (19-19y)/3x=1919y3

Substituting for xx in (I)(I)

13( (19-19y)/3) + 13y = 1613(1919y3)+13y=16

implies(247 - 247y)/3 + 13y = 16247247y3+13y=16

implies(247-247y+39y)/3 = 16247247y+39y3=16

implies247-208y = 16*3247208y=163

implies208y = 247 - 48208y=24748

impliesy = 199/208y=199208

Substituting for yy in either equation (we'll choose (II)(II)),

3x + 19(199/208) = 193x+19(199208)=19

implies3x = 19 - 3781/2083x=193781208

impliesx = 1/3*((3952 - 3781)/208) = 1/3 * (171/208) = 171/624 = 57/208x=13(39523781208)=13(171208)=171624=57208