# Use the elimination method please?/ Thanks!

Nov 7, 2017

It’s too late for this answer now (sorry) but I can explain the elimination method if that helps ...

#### Explanation:

In essence you are trying to get the equations so that they contain a factor (or variable) that can be removed, leaving a simple linear equation (such as 5a = 2 or whatever.)

Off we go ...

First number the equations for reference:

$x - \frac{y}{2} = \frac{5}{3}$ (I)

$\frac{3}{2} x - \frac{3}{8} y = 1$ (II)

(I) x 6 gives $6 x - 3 y = 10$ (III)

(II) x 8 gives $12 x - 3 y = 8$ (IV)

(III) - (IV) leaves $- 6 x = 2$ so $x = - \frac{1}{3}$ (V) Now y has been “eliminated”

Substitute (V) into (I) to give:

$- \frac{1}{3} - \frac{y}{2} = \frac{5}{3}$ (VI)

So multiplying (VI) by 6 gives:

$- 2 - 3 y = 10$ so $- 3 y = 12$ and $y = - 4$

You should always check you have got it right (I mess up continuously) by resubstituting both values you have got into one of the original equations so choosing (I) it becomes:

$- \frac{1}{3} - - \frac{4}{2} = \frac{5}{3}$ and multiplying through by 6 gives:

$- 2 + 12 = 10$ which sounds about right to me.