Given {4A + 8b + 3C = 18, 5A + 10B + 3C = 21} then a + 2B + 3C=

Given {4A + 8b + 3C = 18, 5A + 10B + 3C = 21} then a + 2B + 3C=

① 4A + 8b + 3C = 18, ② 5A + 10B + 3C = 21, ② subtracting ① to get a + 2B = 3, then I want to get 3 + 3C ① plus ② to get 9 + 18B + 6C = 39, that is 9 (a + 2b) + 6C = 39, a + 2B = 3, so 6C = 39-27 = 12, C = 2, so 3 + 3C = 9, so a + 2B + 3C = 9
Given a = a ^ 2-2a + 3, B = a ^ 2-7, C = 5A ^ - 4a-6, find the value of a + 2b-3c
A+2B-3C
=a^2-2a+3+2a^2-14-15a^2+12a+18
=-12a^2 + 10a+7
-4（3a~2-3a-1)
A+2B-3C
=a^2-2a+3+2*(a^2-7)-3*(5a^2-4a-6)
=a^2-2a+3+2a^2-14-15a^2+7a+18
=-12a^2 + 5a+7
=-(12a+7)(a-1)
Simplification, (4a-2b + 3C) ^ 2 - (4a + 2b-3c) ^ 2
Help to solve (4a-2b + 3C) - (4a + 2b-3c), and is there a simple method?
(4a+2b-3c)(2b-4a+3c)
Tip: use the square difference formula
(4a+2b-3c)(2b-4a+3c)
=[2b+(4a-3c)][2b-(4a-3c)]
=(2b)²-(4a-3c)²
=4b²-(16a²-24ac+9c²)
=-16a²+4b²-9c²+24ac
, find this [(2a + b) (2a-b) - (2a-b) ^ 2 + 2B (a-b)] / 4b,
The original formula = (4a & sup2; - B & sup2; - 4A & sup2; + 4ab-b & sup2; + 2ab-2b & sup2;) / 4B
=(6ab-4b²)/4b
=6ab/4b-4b²/4b
=1.5a-b
That is, a + B + C = 0
b=-a-c
ax²+(-a-c)x+c=0
ax²-ax-cx+c=0
ax(x-1)-c(x-1)=0
(x-1)(ax-c)=0
x=1,x=c/a
So choose B
ff'(x)=4x³-1
y=3x
Parallel then k = f '(x) = 3
4x³=4
X=1
f(1)=1-1=0
So p (1,0)
Given (2 / a) - (1 / b) = 3, find the value of (2a + 5ab-4b) / (2B + 4ab-a)
(2a+5ab-4b)/(2b+4ab-a)
The numerator and denominator are divided by ab at the same time
=[(2/b-4/a)+5]/[(2/a-1/b)+4]
=[-2*(2/a-1/b)+5]/[(2/a-1/b)+4]
=[(-2)*3+5]/(3+4)
=-1/7
If a-2b = 2, then 9-2a + 4B=
Five
Find this [(2a + b) (2a-b) - (2a-b) ^ 2 + 2B (a-b)] / 4b, and know a = 2, B = 1
Simplification of expansion
[(2a+b)(2a-b)-(2a-b)^2+2b(a-b)]/4b=3a/2-b
Substituting
The result is 3-1 = 2
If a-2b = - 2, then 4-2a = 4B
What's the problem? Let's find the value of ab. how can it become the value of an equation. a=0，b=1
a=0,b=1
This is a problem of binary linear equations. We can get 4B = 2 (a + 2) = 2A + 4 directly from Formula 1, then we can get a = 0 from formula 2, and then we can get b = 1
However, there are some ambiguities in my question. Just like the understanding above, it seems that your second form should not be an equation. Is it your wrong input? He made a good change...... If it's true, it's a problem of finding the values of a and B.
The former formula is reduced to a = - 2 + 2B, and substituted into the latter formula, 4-2 (- 22b) = 4b, B = 1, a = 0