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

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- Jan 28th 2007, 11:35 AMMoneyMakerGrade 7 Algebra
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 PMQuick
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 AMqpmathelp
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

see this page for precedence of operations

order of performing mathematical operations - Feb 4th 2007, 09:29 AMpablohacker
first you collect "like terms", not sure if thats a term you have used but it means adding up whats "like" before subing numbers in so

3A + 4B - 2C - 4A +5B - C becomes

-1A +9B -3C ( as you have added all the A's together and so on.)

then you sub in the numbers so,

-1(2) + 9(-1) -3(4) = -2-9-12 = -23