ab+cd=34
ac+bd=38
ad+bc=43
what is a+b+c+d=?
multiple choice:
a)15 b)16 c)17 d)18 e)19
First, I tried guessing and testing, but couldn't find the correct answer.
Then I tried isolating a
a=(34-cd)/b
I'm stuck, there are too many variables
Wow, I never thought of combining it, thanks! The problem doesn't ask to solve for a b c & d
To solve these equations, we can use a method called elimination.
From the first equation: ab + cd = 34
From the second equation: ac + bd = 38
From the third equation: ad + bc = 43
Let's solve for a by eliminating the variable b:
Equation 1 (ab + cd = 34) multiplied by c:
ac + c^2d = 34c
Equation 2 (ac + bd = 38) multiplied by d:
ad + bd^2 = 38d
Now, subtract the second equation from the first equation to eliminate b:
(ac + c^2d) - (ad + bd^2) = 34c - 38d
ac + c^2d - ad - bd^2 = 34c - 38d
Now, let's solve for a by factoring:
a(c - d) + d(c - d) = (34c - 38d)
(a + d)(c - d) = 34c - 38d
(a + d) = (34c - 38d)/(c - d)
Let's substitute this value into the third equation:
(ad + bc) = 43
[(34c - 38d)/(c - d)]d + bc = 43
(34cd - 38d^2)/(c - d) + bc = 43
(34cd - 38d^2) + bc(c - d) = 43(c - d)
Now, let's simplify the equation:
34cd - 38d^2 + bc^2 - bcd = 43(c - d)
34cd - 38d^2 + b(c^2 - cd) = 43(c - d)
34cd - 38d^2 + b[(a + d)(c - d) - cd] = 43(c - d)
At this point, we have eliminated variables a and b, but we still have two variables, c and d. To continue solving for a + b + c + d, we will need additional information or more equations.
To solve this system of equations, we can use a technique called elimination or substitution. Let's start by eliminating variable a.
From the equation ac + bd = 38, we can express a in terms of b and d:
a = (38 - bd) / c
Now, substitute this expression for a in the equation ad + bc = 43:
[(38 - bd) / c]d + bc = 43
To eliminate denominators, we can multiply the whole equation by c:
(38 - bd)d + b(c^2) = 43c
Expanding and rearranging the terms:
38d - bdd + bc^2 = 43c
Now, let's isolate b in terms of d and c:
38d + bc^2 = 43c + bdd
b = (43c + bdd - 38d) / c^2
We can substitute this expression for b back into our equation ab + cd = 34:
[(34 - (43c + bdd - 38d) / c^2)b] + cd = 34
Simplifying and rearranging the terms:
(34c^2 - 43c - bdd + 38d)b + c^3d = 34c^2
At this point, we can substitute the values of b and a back into the equation ad + bc = 43 to solve for c and d. However, this further algebraic manipulation becomes quite complex.
So, let's use another approach called matrix elimination, which is a systematic method for solving systems of equations. Representing the equations in matrix form:
| a b | | c | | 34 |
| | x | | = | |
| c d | | d | | 38 |
| | | | | |
| a d | | b | | 43 |
| | | | | |
| a c | | a | | 0 |
| | | | | |
Using matrix elimination, let's solve this system:
1. Multiply the first equation by c, subtracting the third equation multiplied by a:
ac + bc = 34c
ac + ad = 43a
--------------
bc - ad = 34c - 43a (Equation 4)
2. Multiply the second equation by a, subtracting the third equation multiplied by b:
ac + bd = 38a
ab + bd = 43b
--------------
ac - ab = 38a - 43b (Equation 5)
Now we have obtained two equations (Equation 4 and Equation 5) with only two variables, b and c. We can solve this system by substitution or elimination.
Substituting Equation 5 into Equation 4:
bc - ad = 34c - 43a
bc - (38a - 43b) = 34c - 43a
bc - 38a + 43b = 34c - 43a
bc + 43b = 34c - 5a
43b + bc = 34c - 5a
Now, we have a single equation with only b and c as variables. Unfortunately, at this point, this equation is still difficult to solve without additional information or further simplification. Therefore, it seems there may have been a mistake in the given system of equations or a missing constraint.
Without additional information or constraints, there is not enough information to determine the values of a, b, c, and d. Therefore, we cannot find the sum a + b + c + d.
In general, you are correct. There are not enough equations to solve for four variable. However they may be looking only for answers a,b,c,d that are integers. That restricts your options. Note that:
ab + cd + ac + bd = (a+d)(b+c) = 72
ab + ad + cd + bc = (a+c)(b+d) = 77
ac + bd + ad + bc = (a+b)(c+d) = 81
IF the solutions must be single-digit integers, the last equation tells you that
a + b = 9 and
c + d = 9
Therefore
a + b + c + d = 18
To solve for all four, you could say that
a + c = 11
b + d = 7
Which is consistent with a+b+c+d = 18
I can't find a way of satisfying
(a+d)(b+c) = 72
such that the sum of the two factors is 18.
I give up.