To solve this system of equations, we can first solve for one of the variables in terms of the other. Let's solve for b in terms of a and c from the first equation:
a + c = 5 c = 5 - a
Now substitute this expression for c into the second and third equations:
ad + b(5-a) = 5 ab + b + b(5 - a) + d = 8
Now we can simplify the equations:
ad + 5b - ab = 5 ab + b - ab + d = 8
Simplifying further:
5b = 5 - ad b + d = 8
Now substitute 5 - ad for b in the second equation:
(5 - ad) + d = 8 5 - ad + d = 8 5 - a = 8 -a = 3 a = -3
Now substitute a = -3 back into c = 5 - a:
c = 5 - (-3) c = 8
Now we can substitute a = -3 and c = 8 into the equations above:
5b = 5 - (-3)d 5b = 5 + 3d b = 1 + 0.6d
Substitute b = 1 + 0.6d and c = 8 into the last equation:
To solve this system of equations, we can first solve for one of the variables in terms of the other. Let's solve for b in terms of a and c from the first equation:
a + c = 5
c = 5 - a
Now substitute this expression for c into the second and third equations:
ad + b(5-a) = 5
ab + b + b(5 - a) + d = 8
Now we can simplify the equations:
ad + 5b - ab = 5
ab + b - ab + d = 8
Simplifying further:
5b = 5 - ad
b + d = 8
Now substitute 5 - ad for b in the second equation:
(5 - ad) + d = 8
5 - ad + d = 8
5 - a = 8
-a = 3
a = -3
Now substitute a = -3 back into c = 5 - a:
c = 5 - (-3)
c = 8
Now we can substitute a = -3 and c = 8 into the equations above:
5b = 5 - (-3)d
5b = 5 + 3d
b = 1 + 0.6d
Substitute b = 1 + 0.6d and c = 8 into the last equation:
(b)(d) = 1
(1 + 0.6d)d = 1
1 + 0.6d^2 = 1
0.6d^2 = 0
d^2 = 0
d = 0
Now that we have found the values of a, b, c, and d, the solution to the system of equations is:
a = -3, b = 1, c = 8, d = 0.