# SOLUTION: If you could explain to me how to solve this problem, I would appreciate it. I need numerical values for a, b, c, and d. I know this involves substitution, I just don't know where

Algebra ->  Algebra  -> Expressions-with-variables -> SOLUTION: If you could explain to me how to solve this problem, I would appreciate it. I need numerical values for a, b, c, and d. I know this involves substitution, I just don't know where       Log On

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 Algebra: Expressions involving variables, substitution Solvers Lessons Answers archive Quiz In Depth

 Question 116081: If you could explain to me how to solve this problem, I would appreciate it. I need numerical values for a, b, c, and d. I know this involves substitution, I just don't know where to start! Thank you so much. 2a+3b-4c+6d=6 7a-5b-c=-7 13a-9b=6 d^2-2d=-1Found 2 solutions by Edwin McCravy, bucky:Answer by Edwin McCravy(8908)   (Show Source): You can put this solution on YOUR website!``` 2a + 3b - 4c + 6d = 6 7a - 5b - c = -7 13a - 9b = 6 dē - 2d = -1 Solve the last equation first, since that is the only equation that has just one unknown: dē - 2d = -1 dē - 2d + 1 = 0 (d - 1)(d - 1) = 0 d - 1 = 0 or d = 1 Substitute d = 1 into the first equation: 2a + 3b - 4c + 6d = 6 2a + 3b - 4c + 6(1) = 6 2a + 3b - 4c + 6 = 6 2a + 3b - 4c = 0 So now the system of equations is: 2a + 3b - 4c = 0 7a - 5b - c = -7 13a - 9b = 6 Can you solve that system? If not, post again asking how. Answer: a = 6, b = 8, c = 9 and earlier we found d = 1. Edwin``` Answer by bucky(2189)   (Show Source): You can put this solution on YOUR website!Sort of complex. For reference I'm going to use Roman Numerals to Identify the 4 equations as follows: . (I) 2a+3b-4c+6d=6 . (II) 7a-5b-c=-7 . (III) 13a-9b=6 . (IV) d^2-2d=-1 . Cramer's rule might be an interesting way to solve this, but I'm going to assume that you haven't studied that yet. So let's just plod our way through. . The first thing that I noticed was that equation (IV) has just one variable. So we can solve it for d first thing. Add 1 to both sides of equation (IV) and it becomes: . d^2 - 2d + 1 = 0 . The left side is a perfect square as follows: . (d - 1)^2 = 0 . To make the left side equal to the zero on the right side we need to have: . d - 1 = 0 . Solve for d by adding 1 to both sides to get: . d = 1 . One down ... only a, b, and c to find. Before we go further let's go back to equation (I) and substitute 1 for d to make that equation: . 2a + 3b - 4c + 6(1) = 6 . subtract 6 from both sides and equation I reduces to: . (I) 2a + 3b - 4c = 0 . Notice that the form of this equation is now very similar to equation (II). Let's multiply both sides of equation (II) by -4 and it becomes: . (II) -28a + 20b + 4c = 28 . We now have the new equations (I) and (II) as: . ( I ) 2a + 3b - 4c = 0 (II)-28a +20b + 4c = 28 . Notice now what happens if we add these two equations vertically in columns. We eliminate the terms containing the variable c and the equation resulting from this addition (call it equation (I&II) is: . (I&II) -26a + 23b = 28 . This looks a lot like equation (III) and the pair of equations is now: . (I&II)-26a + 23b = 28 (III) +13a - 9b = 6 . Let's work to eliminate the variable "a". Multiply equation (III) ... all terms on both sides by 2 to make the equation pair become: . (I&II)-26a + 23b = 28 (III) +26a - 18b = 12 . If we then add the two equations vertically in columns, the two "a" terms cancel each other and the combined equation that results from the addition (call it equation (I&II&III) becomes: . (I&II&III) 5b = 40 . Divide both sides by 5 and get: . b = 40/5 = 8 . Two down ... only a and c left to find. . Now that we know b = 8 we can return to the original equation (III) and substitute 8 for b to get: . (III) 13a - 9(8) = 6 . Multiply 9 times 8 and the equation changes to: . (III) 13a - 72 = 6 . Add 72 to both sides and it further becomes: . (III) 13a = 78 . Solve for a by dividing both sides by 13 to get: . a = 78/13 = 6 . One more to go ... just c left. . Return to one of the original equations that contains c and substitute the known values for a and b. Let's go way back to equation (II). If we substitute 6 for a and 8 for b that equation becomes: . (II) 7(6)- 5(8) - c = -7 . Do the multiplications and we get: . (II) 42 - 40 - c = -7 . Combine the numbers on the left side: . (II) 2 - c = -7 . Subtract 2 from both sides to reduce this to: . (II) - c = -9 . Solve for +c by multiplying both sides by -1 and we have: . c = +9 . That's it ... no more variables to find. In summary: a = 6, b = 8, c = 9, and d = 1. . You can check these answers by returning to the original 4 equations and substituting the values for a, b, c, and d as needed in each equation. You should (and will) find that with these values the left side of each equation will equal the right side. . Hope this helps you to understand the problem. .