How do you solve the system of equations $4x - 2y = 1$ and $8x + 4y = 6$ ?

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Answer 1 · 2018-06-14T12:36:12

$x=1/2$ and $y=1/2$

$4x-2y=1$
$8x+4y=6$

Eliminate term $x$ to find $y$ by multiplying top equation by $8$ and bottom equation by $4$ :

$32x-16y=8$
$32x+16y=24$

Subtract equations from each other:

$-32y=-16$
$y=(-16)/-32$
$y=1/2$

Substitute $y=1/2$ into one of the original equations and solve for $x$ :

$4x-2y=1$
$4x-2(1/2)=1$
$4x-1=1$
$4x=2$
$x=2/4$
$x=1/2$

How do you solve the system of equations 4x - 2y = 14x - 2y = 1 and 8x + 4y = 68x + 4y = 6?