# How do you find the solution of the system of equations  1/4x+y=3/40 and 3/2x-y=-5/2?

May 12, 2015

Given $\left(\frac{1}{4}\right) x + y = \frac{3}{40}$, subtract $\left(\frac{1}{4}\right) x$ from both sides to get:

$y = \frac{3}{40} - \left(\frac{1}{4}\right) x$

Given $\left(\frac{3}{2}\right) x - y = - \frac{5}{2}$, add $y + \frac{5}{2}$ to both sides to get:

$\left(\frac{3}{2}\right) x + \frac{5}{2} = y$

So we have

$\left(\frac{3}{2}\right) x + \frac{5}{2} = y = \frac{3}{40} - \left(\frac{1}{4}\right) x$

Just keeping the two ends and multiplying both sides by 40 we get:

$60 x + 100 = 3 - 10 x$

Add $10 x$ to both sides to get:

$70 x + 100 = 3$

Subtract 100 from both sides to get:

$70 x = - 97$

Divide both sides by 70 to get:

$x = - \frac{97}{70}$

To calculate the value of $y$, substitute this value of $x$ into one of our previous equations:

$y = \left(\frac{3}{2}\right) x + \frac{5}{2}$

$= \left(\frac{3}{2}\right) \left(- \frac{97}{70}\right) + \frac{5}{2}$

$= - \frac{3 \cdot 97}{2 \cdot 70} + \frac{5 \cdot 70}{2 \cdot 70}$

$= \frac{- 291 + 350}{140}$

$= \frac{59}{140}$

To check our calculation, try plugging these values of $x$ and $y$ back into the first equation:

$\left(\frac{1}{4}\right) x + y = \left(\frac{1}{4}\right) \left(- \frac{97}{70}\right) + \frac{59}{140}$

$= - \frac{97}{280} + \frac{118}{280}$

$= \frac{21}{280}$

$= \frac{3}{40}$

I will leave checking the second one to you.