a + b = 3
-b + c = 3
a + 2c = 10
and
p + 4r = -7
p - 3q = -8
q + r = 1
any help is appreciated!!Can you give me some advice for solving these math problems (solving systems of equations in three variables)?
Let's start with the first set of equations.
I. a + b = 3
II. -b + c = 3
III. a + 2c = 10
Take equation I. and find a:
I. a + b = 3
** a = 3 - b **
Take equation II. and find c:
II. -b + c = 3
** c = 3 + b **
Plug the two ** equations above into equation III:
III. a + 2c = 10
(3 - b) + 2(3 + b) = 10
3 - b + 6 + 2b = 10
9 + b = 10
b = 1
Plug b = 1 into the equations I. and II. to find a and c:
I. a + b = 3
a + 1 = 3
a = 2
II. -b + c = 3
-1 + c = 3
c = 4
So, a = 2, b = 1, c = 4.
Now let's do the second set of equations.
I. p + 4r = -7
II. p - 3q = -8
III. q + r = 1
Take equation II. and find p:
II. p - 3q = -8
** p = -8 + 3q **
Take equation III. and find r:
III. q + r = 1
** r = 1 - q **
Plug the two ** equations above into equation I:
I. p + 4r = -7
(-8 + 3q) + 4(1 - q) = -7
-8 + 3q + 4 - 4q = -7
-4 - q = -7
3 = q
Plug 3 = q into the equations II. and III. to find p and r:
II. p - 3q = -8
p - 3(3) = -8
p - 9 = -8
p = 1
III. q + r = 1
3 + r = 1
r = -2
So, p = 1, q = 3, r = -2.
No comments:
Post a Comment