Thread: Grade 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

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:


Substitute:


Solve:


Adding a negative number means you subtract:

And I'm sure you can do it from there.

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



see this page for precedence of operations

order of performing mathematical operations

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