How do you solve the following system: $2 x - 5 y = - 19, y + 4 x = 16$ $2x-5y=-19, y + 4x = 16$ ?

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Answer 1 · 2016-04-23T09:21:14

$x = 3 \frac{17}{22}$ $x=3 17/22$ and $y = 4 \frac{10}{11}$ $y=4 10/11$

Putting value of $y$ $y$ from $y + 4 x = 16$ $y+4x=16$ i.e. $y = 16 - 4 x$ $y=16-4x$ in $2 x - 5 y = - 19$ $2x-5y=-19$ , we get

$2 x - 5 (16 - 4 x) = - 19$ $2x-5(16-4x)=-19$ or

$2 x - 80 + 20 x = - 19$ $2x-80+20x=-19$ or

$22 x = 80 - 19 = 61$ $22x=80-19=61$ or $x = \frac{61}{22} = 3 \frac{17}{22}$ $x=61/22=3 17/22$

Now putting this in $y = 16 - 4 x$ $y=16-4x$ we get $y = 16 - \frac{4 \times 61}{22}$ $y=16-(4xx61)/22$ or

$y = 16 - \frac{244}{22} = 16 - 11 \frac{2}{22} = 4 \frac{20}{22} = 4 \frac{10}{11}$ $y=16-244/22=16-11 2/22=4 20/22=4 10/11$

How do you solve the following system: 2x−5y=−19,y+4x=162x-5y=-19, y + 4x = 16 ?