A car radiator contains 8 liters of a 32% antifreeze solution. How many liters of this solution should be drained and replaced with pure antifreeze to get a 49% solution?

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A car radiator contains 8 liters of a 32% antifreeze solution. How many liters of this solution should be drained and replaced with pure antifreeze to get a 49% solution?

Mathematics
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OKay so for this... damn it i have to go okay sorry bye!!
8 liters
is 4L 2L 3L or 1.75

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drain 8 liters of 32% and replace 8 liters of 49%
haha thats the lazy way :)
my only choices are 2 3 4 or 1.75 not 8
system of 2 equations Let x be 32% antifreeze, y is amount drained and replaced x+y = 8 .32x + y = .49
oops it should be 8*.49 use elimination subtract row 1 - row 2 x + y = 8 -.32x -y = -(8*.49) __________________ .64x +0 = 4.08 x = 4.08/.64 x = 6.375 y = 8-x y = 8-6.375 y=1.625 1.625L should be drained and replaced
wierd i did something really wrong sorry answer is 2L

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