How do you solve the system $y=4x; 2y-x=16$ ?

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Answer 1 · 2017-01-09T22:02:54

See full solution process below

Step 1) Because the first equation is already solved for $y$ we can substitute $4x$ for $y$ in the second equation and solve for $x$ :

$(2 xx 4x) - x = 16$

$8x - x = 16$

$8x - 1x = 16$

$(8 - 1)x = 16$

$7x = 16$

$(7x)/color(red)(7) = 16/color(red)(7)$

$(color(red)(cancel(color(black)(7)))x)/cancel(color(red)(7)) = 16/7$

$x = 16/7$

Step 2) We can now substitute $16/7$ for $x$ in the first equation and calculate $y$ :

$y = 4 xx 16/7$

$y = 64/7$

The solution is:

$x = 16/7$ and $y = 64/7$

How do you solve the system y=4x; 2y-x=16y=4x; 2y-x=16?