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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=-1
Found 2 solutions by Edwin McCravy, bucky:
Answer by Edwin McCravy(20064)   (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:
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(I) 2a+3b-4c+6d=6
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(II) 7a-5b-c=-7
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(III) 13a-9b=6
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(IV) d^2-2d=-1
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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.
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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:
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d^2 - 2d + 1 = 0
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The left side is a perfect square as follows:
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(d - 1)^2 = 0
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To make the left side equal to the zero on the right side we need to have:
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d - 1 = 0
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Solve for d by adding 1 to both sides to get:
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d = 1
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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:
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2a + 3b - 4c + 6(1) = 6
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subtract 6 from both sides and equation I reduces to:
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(I) 2a + 3b - 4c = 0
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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:
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(II) -28a + 20b + 4c = 28
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We now have the new equations (I) and (II) as:
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( I ) 2a + 3b - 4c = 0
(II)-28a +20b + 4c = 28
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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:
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(I&II) -26a + 23b = 28
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This looks a lot like equation (III) and the pair of equations is now:
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(I&II)-26a + 23b = 28
(III) +13a - 9b = 6
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Let's work to eliminate the variable "a". Multiply equation (III) ... all terms on both
sides by 2 to make the equation pair become:
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(I&II)-26a + 23b = 28
(III) +26a - 18b = 12
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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:
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(I&II&III) 5b = 40
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Divide both sides by 5 and get:
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b = 40/5 = 8
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Two down ... only a and c left to find.
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Now that we know b = 8 we can return to the original equation (III) and substitute
8 for b to get:
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(III) 13a - 9(8) = 6
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Multiply 9 times 8 and the equation changes to:
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(III) 13a - 72 = 6
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Add 72 to both sides and it further becomes:
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(III) 13a = 78
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Solve for a by dividing both sides by 13 to get:
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a = 78/13 = 6
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One more to go ... just c left.
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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:
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(II) 7(6)- 5(8) - c = -7
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Do the multiplications and we get:
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(II) 42 - 40 - c = -7
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Combine the numbers on the left side:
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(II) 2 - c = -7
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Subtract 2 from both sides to reduce this to:
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(II) - c = -9
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Solve for +c by multiplying both sides by -1 and we have:
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c = +9
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That's it ... no more variables to find. In summary: a = 6, b = 8, c = 9, and d = 1.
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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.
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Hope this helps you to understand the problem.
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