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.
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