# Thread: On the tip of my fingers but can't get it...

1. ## On the tip of my fingers but can't get it...

I feel really stupid posting this question as I know Im on the verge of finding a way to solve it but I just can't think of it.

Here it is...

Solve for A, B, C, and D in the following equations:

A+C=0
B-D=0
4A+3D+2C=7
A+4B+6C=-6

Thanks again for any and all help

2. Originally Posted by forkball42
I feel really stupid posting this question as I know Im on the verge of finding a way to solve it but I just can't think of it.

Here it is...

Solve for A, B, C, and D in the following equations:

A+C=0
B-D=0
4A+3D+2C=7
A+4B+6C=-6

Thanks again for any and all help
Solve the second equation for D:
$\displaystyle D = B$

Then
$\displaystyle A+C=0$
$\displaystyle 4A+3B+2C=7$
$\displaystyle A+4B+6C=-6$

Solve the first equation for C:
$\displaystyle C = -A$

So
$\displaystyle 2A+3B=7$
$\displaystyle -5A+4B=-6$

Solve the first equation for B:
$\displaystyle B = -\frac{2}{3}A + \frac{7}{3}$

Thus
$\displaystyle -\frac{23}{3}A + \frac{28}{3}=-6$

or
$\displaystyle A = 2$

I leave the rest to you.

-Dan

3. Hello, forkball42!

A slightly different approach . . .

Solve for A, B, C, and D:

. . $\displaystyle \begin{array}{cccc}A+C &=& 0 & [1]\\ B-D & = & 0 & [2]\\4A+2C+3D& =&7 & [3]\\ A+4B+6C&=&\text{-}6 & [4]\end{array}$

$\displaystyle \begin{array}{ccccc} \text{From [1], we have:} & C & = & \text{-}A& [5] \\ \text{From [2], we have:} & D & = & B & [6]\end{array}$

Substitute [5] and [6] into [3] and [4]:

. . $\displaystyle \begin{array}{cccccccc}4A + 2(\text{-}A) + 3(B ) & = & 7 & \Rightarrow & 2A + 3B & = & 7& [7] \\ A + 4B + 6(\text{-}A) & = & \text{-}6 & \Rightarrow & \text{-}5A + 4B & = & \text{-}6& [8]\end{array}$

$\displaystyle \begin{array}{cccc}\text{Multiply [7] by 4:} & 8A + 12B & = & 28 \\ \text{Multiply [8] by -3:} & 15A - 12B & = & 18\end{array}$

Add: .$\displaystyle 23A \:=\:46\quad\Rightarrow\quad\boxed{ A \,=\,2}$

. . Substitute into [5]: .$\displaystyle \boxed{C \,=\,\text{-}2}$

Substitute into [7]: .$\displaystyle 2(2) + 3B \:=\:7\quad\Rightarrow\quad\boxed{ B \,= \,1}$

. . Substitute into [6]: .$\displaystyle \boxed{D \,= \,1}$