Problem to solve
Solve \[\frac{6}{5}x+\frac{3}{2}y=5\], \[-\frac{3}{5}x+y=5y+2\]
Solution
STEP 1
Solve the following system:
{(3 y)/2 + (6 x)/5 = 5
y - (3 x)/5 = 5 y + 2
STEP 2
Express the system in standard form:
{(3 y)/2 + (6 x)/5 = 5
-4 y - (3 x)/5 = 2
STEP 3
Express the system in matrix form:
(6/5 | 3/2
-3/5 | -4)(x
y) = (5
2)
STEP 4
Solve the system with Cramer's rule:
x = (5 | 3/2
2 | -4)/(6/5 | 3/2
-3/5 | -4) and y = (6/5 | 5
-3/5 | 2)/(6/5 | 3/2
-3/5 | -4)
STEP 5
Evaluate the determinant 6/5 | 3/2
-3/5 | -4 = (-39)/10:
x = (5 | 3/2
2 | -4)/(-39/10) and y = (6/5 | 5
-3/5 | 2)/(-39/10)
STEP 6
The gcd of 5 | 3/2
2 | -4 and -39/10 is 1/10, so (5 | 3/2
2 | -4)/((-39)/10) = (10 5 | 3/2
2 | -4)/(10 ((-39)/10)) = -10/39 5 | 3/2
2 | -4:
x = -10/39 5 | 3/2
2 | -4 and y = (6/5 | 5
-3/5 | 2)/((-39)/10)
STEP 7
The gcd of 6/5 | 5
-3/5 | 2 and -39/10 is 1/10, so (6/5 | 5
-3/5 | 2)/((-39)/10) = (10 6/5 | 5
-3/5 | 2)/(10 ((-39)/10)) = -10/39 6/5 | 5
-3/5 | 2:
x = (-10 5 | 3/2
2 | -4)/39 and y = -10/39 6/5 | 5
-3/5 | 2
STEP 8
Evaluate the determinant 5 | 3/2
2 | -4 = -23:
x = (-10)/39×-23 and y = (-10 6/5 | 5
-3/5 | 2)/39
STEP 9
(-10)/39 (-23) = 230/39:
x = 230/39 and y = (-10 6/5 | 5
-3/5 | 2)/39
STEP 10
Evaluate the determinant 6/5 | 5
-3/5 | 2 = 27/5:
x = 230/39 and y = (-10)/39×27/5
STEP 11
(-10)/39×27/5 = (-18)/13:
Answer: | 
 | x = 230/39 and y = -18/13