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

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Answer 1 · 2015-05-17T12:39:48

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

$y = \frac{3}{4} - \frac{1}{4} x$ $y = 3/4 - 1/4x$

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

$- \frac{5}{2} = \frac{3}{2} x - y$ $-5/2 = 3/2x-y$

$= \frac{3}{2} x - (\frac{3}{4} - \frac{1}{4} x)$ $= 3/2x-(3/4-1/4x)$

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

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

Now add 3 to both sides to get

$7 x = - 7$ $7x = -7$

Divide both sides by $7$ $7$ to get

$x = - 1$ $x = -1$

Then $y = \frac{3}{4} - \frac{1}{4} x = \frac{3}{4} + \frac{1}{4} = 1$ $y = 3/4 - 1/4x = 3/4 + 1/4 = 1$

How do you solve the system 1/4x+y=3/414x+y=341/4x+y=3/4 and 3/2x-y=-5/232x−y=−523/2x-y=-5/2?