• Jan 28th 2007, 11:35 AM
MoneyMaker
I need help here is the question i need to find out how to do this step to step
3A+ 4b-2c-4A+5B-1C
A=2
B=-1
c=4
• Jan 28th 2007, 12:17 PM
Quick
Quote:

Originally Posted by MoneyMaker
I need help here is the question i need to find out how to do this step to step
3A+ 4b-2c-4A+5B-1C
A=2
B=-1
c=4

I have spread out the expression to make it easier for you.

You have: $\displaystyle 3A\quad+\quad 4B\quad - \quad 2C \quad -\quad 4A \quad + \quad 5B \quad - \quad 1C$

Substitute: $\displaystyle 3(2)\quad+\quad 4(-1)\quad - \quad 2(4) \quad -\quad 4(2) \quad + \quad 5(-1) \quad - \quad 1(4)$

Solve: $\displaystyle 6\quad+\quad -4\quad - \quad 8 \quad -\quad 8 \quad + \quad -5 \quad - \quad 4$

Adding a negative number means you subtract: $\displaystyle 6\quad-\quad 4\quad - \quad 8 \quad -\quad 8 \quad - \quad 5 \quad - \quad 4$

And I'm sure you can do it from there.
• Jan 29th 2007, 05:07 AM
qpmathelp
Quote:

Originally Posted by MoneyMaker
I need help here is the question i need to find out how to do this step to step
3A+ 4b-2c-4A+5B-1C
A=2
B=-1
c=4

I guess you mean all upper quotes

3A+ 4B-2C-4A+5B-1C = (3*2) + [4*(-1) ] - (2*4) - (4*2) + [ 5*(-1) ] - (1*4)

= 6 + (-4) - 8 -8 + (-5) - 4

= 6 - 4 - 8 - 8 - 5 - 4

= 6 - 29

= -23

remember that
+ve * +ve = +ve

+ve * -ve = -ve

-ve * -ve = +ve