How do you solve the system $1/4x+y=3/4$ and $3/2x-y=-5/2$ ?

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Answer 1 · 2015-05-17T12:39:48

From the first equation we can derive a formula for $y$ by subtracting $1/4x$ from both sides to get

$y = 3/4 - 1/4x$

Then substitute that for $y$ in the second equation:

$-5/2 = 3/2x-y$

$= 3/2x-(3/4-1/4x)$

These fractions are a pain, so let's eliminate them by multiplying through by 4 to get:

$-10=6x-(3-x) = 7x-3$

Now add 3 to both sides to get

$7x = -7$

Divide both sides by $7$ to get

$x = -1$

Then $y = 3/4 - 1/4x = 3/4 + 1/4 = 1$

How do you solve the system 1/4x+y=3/41/4x+y=3/4 and 3/2x-y=-5/23/2x-y=-5/2?