If $2x+3y =\frac{1}{10}$ and $3x+2y = \frac{1}{8}$, then how to calculate here $x$ and $y$?
I know it is easy but not for me so please describe your solution in step by step. Thanks for your help.
Answers are $x=\frac{7}{200}$ and $y = \frac{1}{100}$.
On
okay , elementary. $ 2x + 3y = 1/10$ , take that 10 to the LHS and rewrite as
$ 20x + 30y $ = $1$
then similarly , $ 3x+2y = 1/8 $
becomes $24x+16y = 1$
$24x+16y = 1$---------1
$ 20x + 30y $ = $1$ ----------2
multiply 1 by the coeff of x in the next eqn 2 (i.e here 20)
and 2 by the coeff of x in 1 ( here 24) $.
so your eqns become
$ 480x + 320y = 20 $
$480x + 720 y= 24 $
subtract these eqns and we get
$ 400 y = 4 => y = 1/100 $
put the value of y in any one eqn to get $x = 7/{200} $
$$ \left\{ \begin{array}{r} 2x+3y=1/10; \\ 3x+2y=1/8; \\ \end{array} \right. $$
Multiple $1$st equation by $3$, multiple $2$nd equation by $2$ (to get the same coeffficients near $x$):
$$ \left\{ \begin{array}{r} 6x+9y=3/10; \\ 6x+4y=1/4; \\ \end{array} \right. $$
Then we subtract equations:
$$(6x+9y)-(6x+4y)=3/10-1/4;$$ $$5y=6/20-5/20=1/20;$$ $$y=1/100.$$
Then substitute $y$ into $1$st equation (you can substitute into $2$nd equation too):
$$2x+3\cdot 1/100 = 1/10;$$ $$2x=1/10-3/100 = 10/100-3/100=7/100;$$ $$x=7/200.$$
Solution: $x=7/200, y=1/100$.