If 3x-2y=a and 5x-3y=b express the vectors x and y in terms of a and b.
If this is indeed a vector problem with x, y, a, and b all vectors having the same dimension,
then we can simply solve two simultaneous equations.
Rearrange the first equation for x:
x = (1/3) . (a + 2y) ,
And the second:
x = (1/5) . (b + 3y)
Equate these two expressions (both RHS must be equal) and multiply both sides by 15 to get rid of the denominators:
5(a + 2y) = 3(b + 3y)
5a + 10y = 3b + 9y (<--- by expanding the brackets)
y = 3b - 5a (<--- by collecting the y terms on the left)
Now put this back into one of our earlier expressions for x....
(symbol below means 'implies')
x = (1/3) . (a + 2y) ==> x = (1/3) . (a + 2 [3b - 5a] )
x = (1/3) . (a + 6b - 10a ) = (1/3) . (6b - 9a) = - 3a + 2b (by noticing the multiples of 3)
Hope you found this useful!!
Comments
If this is indeed a vector problem with x, y, a, and b all vectors having the same dimension,
then we can simply solve two simultaneous equations.
Rearrange the first equation for x:
x = (1/3) . (a + 2y) ,
And the second:
x = (1/5) . (b + 3y)
Equate these two expressions (both RHS must be equal) and multiply both sides by 15 to get rid of the denominators:
5(a + 2y) = 3(b + 3y)
5a + 10y = 3b + 9y (<--- by expanding the brackets)
y = 3b - 5a (<--- by collecting the y terms on the left)
Now put this back into one of our earlier expressions for x....
(symbol below means 'implies')
x = (1/3) . (a + 2y) ==> x = (1/3) . (a + 2 [3b - 5a] )
x = (1/3) . (a + 6b - 10a ) = (1/3) . (6b - 9a) = - 3a + 2b (by noticing the multiples of 3)
Hope you found this useful!!