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258 Two-by-Two Linear Systems
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The terms eby and bey are additive inverses, so they disappear from the sum Next, we can apply the distributive law backward in the left side, obtaining (ea bd )x = ec bf We can solve for x if we divide through by (ea bd ), assuming that (ea bd ) 0: x = (ec bf )/(ea bd )
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Have you noticed that the taboos in the above derivations are actually two different ways of saying the same thing They re stated like this: (db ae) 0
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(ea bd ) 0 Both of these inequalities are equivalent to the statement ae bd Can you see why What do you think will happen if a linear system has coefficients in the above form such that ae = bd You ll get a chance to explore a situation like that in exercises 5, 6, and 7 at the end of this chapter!
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A two-by-two linear system can be unraveled by renaming one variable in terms of the other, and then creating a single-variable, first-degree equation from the result We solve that equation, and then plug the number into a strategic spot to solve for the other variable This process is usually called the substitution method I like to call it rename and replace
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Rename one variable When we want to solve a two-by-two linear system by rename-and-replace, we begin by morphing one of the equations into SI form Consider this pair of equations:
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7v + w + 10 = 0 and 4v + 8w = 40 We can take the first equation and add 7v to each side, getting w + 10 = 7v
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Then we can subtract 10 from each side to obtain this SI equation with w playing the role of the dependent variable: w = 7v 10
Make a first-degree equation The second step involves substituting our new name for w into the equation we haven t touched yet, which in this case is the second original That gives us
4v + 8(7v 10) = 40 The distributive law of multiplication over subtraction can be applied to the second addend on the left side of the equals sign to get 4v + 56v 80 = 40 Summing the first two addends in the left side of this equation gives us 60v 80 = 40 Adding 80 to each side, we obtain 60v = 40 This tells us that v = 40/60 = 2/3
Plug the number into the best place Now that we have solved for one of the variables, we can replace the resolved unknown with its solution in any relevant equation containing both variables The simplest approach is to use is the SI equation we derived in the first step:
w = 7v 10 When we replace v by 2/3 here, we get w = 7 2/3 10 Taking the product on the right side of the equals sign, and changing 10 into 30/3 to obtain a common denominator, we come up with w = 14/3 30/3 Now it s a matter of mere arithmetic: w = (14 30)/3 = 16/3 We ve derived the solution to this system: v = 2/3 and w = 16/3
260 Two-by-Two Linear Systems
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As always, we had better check our work to be sure the solutions we obtained satisfy both of the original equations Here s the first check: 7v + w + 10 = 0 7 (2/3) + ( 16/3) + 10 = 0 14/3 16/3 + 10 = 0 30/3 + 10 = 0 10 + 10 = 0 0=0 All right! Here s the second check: 4v + 8w = 40 4 2/3 + 8 ( 16/3) = 40 8/3 128/3 = 40 (8 128)/3 = 40 120/3 = 40 40 = 40 All right again! Our solutions are correct
Here s a challenge!
Solve the following pair of equations as a two-by-two linear system using the substitution method: 3x y = 1 and 8x + 2y = 4 This process is going to be messy! If we remember that is a plain old real number, it won t be too bad The signs will be tricky, though
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