Re:
mmm4444bot said:
jlneedshelp said:
0.3x - 0.2y = 4 (100)
0.2x + 0.5y = -21/23 (230)
I understand your intent above, but it's better to show the multiplication on both sides.
jlneedshelp,
actually, you \(\displaystyle must\) show each side being multiplied by their respective "compensating
multipliers," because what you show done to one side must be shown being done to another.
Or you might choose not to bother showing the multiplications on each side at all. Or you might
state what you are doing to each side for the next step, and then write the new equations from
that, for example.
\(\displaystyle 100(0.3x - 0.2y) = 100(4)\)
\(\displaystyle 230(0.2x + 0.5y) = 230(-21/23)\)
-----------------------------------------------
\(\displaystyle 30x \ -\ 20y \ = \ 400\)
\(\displaystyle 46x + 115y = -210\)
-----------------------------
If you decide to eliminate the y terms by adding these two equations, note that the
least common multiple of their coefficients is 460.
\(\displaystyle 23(30x - 20y) = 23(400)\)
\(\displaystyle 4(46x + 115y) = 4(-210)\)
-------------------------------------
\(\displaystyle 690x - 460y = \ 9200\)
\(\displaystyle 184x + 460y = -840\)
------------------------------
\(\displaystyle 874x \ \ \ \ \ \ \ \ \ = \ \ 8360\)
\(\displaystyle x = \frac{8360}{874}\)
\(\displaystyle x = \frac{220}{23} . . . . . .exact \ (in \ lowest \ terms)\)
\(\displaystyle or \ x = \ \bigg{9}\frac{13}{23} . . . . . .exact\)
Then you can substitute a convenient form of the x-value into one
of the equations to solve for y. And then the exact solution to the
system of equations will be:
\(\displaystyle \bigg(\frac{220}{23}, \ \frac{-130}{23} \bigg)\)