00:01
Okay, so we're asked to solve the linear system, two -thirds x minus one -fifth -y, equals three, and one -half -x plus three -fourths -y equals one.
00:15
So it tells us we can use substitution or elimination.
00:19
I prefer substitution, so that's what i'm going to do here.
00:22
So let's solve for x in this equation.
00:25
So first we'll move, we'll add the one -fifth y to both sides, so we have two -thirds -x.
00:31
Equals 1 5th y plus 3 and then we'll divide by 2 thirds on both sides but dividing by 2 thirds is the same as multiplying by 3 halves so we have x equals 3 over 10 the 3 gets put on top and then 5 gets multiplied by 2 y plus 3 times 3 is 9 over 2 so now we take this x and we plug it in for this x in this second equation here so we have 1 half times 3 over 10 y plus 9 over 2 plus 3 over 4 y equals 1.
01:13
So we want to distribute this 1 half so we get 3 over 20 y plus 9 over 4 plus 3 4's y equals 1.
01:27
So now we want to combine those ys to do that we need a common denominator.
01:32
So we have 3 over 20 y and let's subtract this 9 4 to the other side.
01:38
So we have this equals 1 minus 9 .4s.
01:41
To get a common denominator here, we can multiply this by 5 over 5.
01:46
So we get that's 15 over 20 y.
01:50
So altogether we have 18 over 20 y, which is the same as 9 over 10 y.
01:59
Here let's get a common denominator.
02:02
Four would be our common denominator.
02:03
So we have 4 over 4 minus 9 over 4, which would give us negative 5.
02:08
Force.
02:10
So our last step is to solve for y.
02:12
Again, we can multiply by the reciprocal.
02:15
So we have y equals negative 50 over 36, which equals negative 25 over 18...