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

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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(20060)   (Show Source): You can put this solution on YOUR website!

 2a + 3b - 4c + 6d =  6 
 7a - 5b - c       = -7 
13a - 9b           =  6 
           d² - 2d = -1


Solve the last equation first, since
that is the only equation that has 
just one unknown:

       d² - 2d = -1
   d² - 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.
.
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