3-2-39 elimination with fractions
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The easiest way to do this kind of a problem is to actually get
rid of all the fractions.
And so if we look at the entire equation and find the common
denominator for all of it, and then multiply the entire
equation 3 by that common denominator.
So when we look at this top one, the common denominator is going
to be 9.
So if I multiply everything 3 by 9, I'm going to get three X -, y
equaling 45.
Literally, I multiply each and every term by 9, so all three of
those terms three goes into 9 three times.
So 3X nines cancel the next term negative Y 5 * 945.
The next equation, the common denominator is 15, so I'm going
to multiply everything through by 15 and three goes into 15
five times.
So I get five times 2X or 10X5 goes into 15 three times.
So 3 * 1 Y and then 15 * 6 is going to be 90.
So from there, I'm going to take this top equation and I'm going
to multiply it by three so I can get additive inverses.
So nine X -, 3 Y equaling 5 carry 1135.
When I add these, I get 19 X equaling to 25 S To solve for
XI, just divide by 19.
Now let's go back to those original equations, and this
time let's get rid of the X's.
So I'm going to multiply this top equation by -10 so negative
thirty X + 10 Y equal -450 and then I'm going to multiply this
bottom equation by three, so thirty X + 9 Y equal 270.
When I add now the X's will cancel.
We get 19 Y equaling negative from 07 from 15 would give me 8
two from three so -180.
So when I divide I get -180 / 19.
So my solution XY 225 / 19 common -180 / 19.
The fact that there is a solution means it's consistent,
and the fact that it's a single solution means that the two
lines are in dependent of each other.