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Simultaneous equations with 3/4 variables
By Sarah Hildreth on March 27, 1998:

Having trouble with the following systems of equations:

A + B + C = 180 <---we have to solve for A, B, and C
B = 2 + 3A
C = 8 + A


w + x + y + z = 2 <----need to solve for w, x, y, and z
w + 2x + 2y + 4z = 1
w - x + y + z = 6
w - 3x - y + z = 2


Sarah.

By Sujata Krishna on March 27, 1998:

Dear Sarah,

Let us take the first question first and solve for
a, b and c. Let us label the equations to help refer to them later on:


a + b + c = 180 ...........................(1)

b = 2 + 3a ...........................(2)

c = 8 + a ...........................(3)


Notice that equations (2) and (3) give you the values of b and c in terms of a; so if you substituted these in (1) you would get an equation where the only known would be a. If you do this equation (1) becomes:

a + (2 + 3a) + (8 + a) = 180 ...............(4)

or, collecting the terms in a together we have:

5a + 10 = 180

Subtract 10 from both sides to get:

5a = 170,

or,

a = 170 ÷ 5 = 34.


Thats great, we now have a and we know b and c in terms of a from equations (2) and (3)! So all you need to do is to substitute the value of a in those equations to get b = 104 and c = 42.


Now you have your answer, and you could stop here. But you can check to see if your answer is correct. To do this feed into the equation (1) the values of a, b and c and see if they add up to 180.

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You should stop and do all this for yourself before you move on to the next question of calculating w,x,y and z. This is because the two questions are similar, and the second one is a touch more tricky as there are now 4 equations and 4 unknown quantities.

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I am going to proceed to the second question now. Again label all the equations so that you can refer to them:


w + x + y + z = 2 ...................................(1)

w + 2x + 2y + 4z = 1 .................................(2)

w - x + y + z = 6 ....................................(3)

w - 3x - y + z = 2 ....................................(4)



You now need to carefully observe the equations and see whether they bear any relation to each other. In particular, think about whether you can simplify your work by trying to eliminate one of w, x, y and z to reduce the system to 3 equations and 3 unknowns.


I am going to eliminate x from the above equations, but you should try later by eliminating one of the other terms, say w.


To eliminate x I notice that x occurs as x, 2x, -x and -3x in the 4 equations. So, I am going to:

multiply (1) by 2 and then substract (2):


2w + 2x + 2y + 2z = 4

-w - 2x - 2y - 4z = -1

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w + 0 + 0 - 2z = 3



or,


w = 3 + 2z ...................................(5)



Similarly, add (1) and (3) to eliminate x and get:


2w + 2y + 2z = 8



or, dividing through by 2 we get:


w + y + z = 8 ...................................(6)



Again, multiply (1) by 3 and take away (4) to get:


2w + y + 2z = 4 .................................(7)



So, we have now reduced our system to 3 equations (5, 6 and 7) with 3 unknowns (w,y and z). Equation (5) gives you w in terms of z, you can substitute this in (6) and (7) to get:


3 + 2z + y + z = 8 .............................(8)


and


6 + 4z + y + 2z = 4 ..............................(9)



Subtract (9) from (8) to get: y = 4.



Place this value of y in (8) to get: z = -1


Use (5) to give you: w = 1



and finally substitute these values of w, y and z in (1) to get

x = -2.


Again, check your answer by seeing that these values fit the 4 equations you started off with.

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I hope this detailed working out helps you to understand how to plan your approach to such a question (try eliminating knowns) and how to work out the actual values.


all the best,

Sujata.