Given that m and N are opposite to each other, C and D are reciprocal to each other, and the distance from a to the origin is 1, find the value of 3M + 3N + 2CD + a

Given that m and N are opposite to each other, C and D are reciprocal to each other, and the distance from a to the origin is 1, find the value of 3M + 3N + 2CD + a

∵ m, n are opposite numbers to each other, ∵ m + n = 0, ∵ C, D are reciprocal numbers to each other, ∵ CD = 1, ∵ A's distance to the origin is 1, ∵ a = ± 1, a = 1, 3M + 3N + 2CD + a = 3 (m + n) + 2CD + a = 0 + 2 + 1 = 3, a = - 1, 3M + 3N + 2CD + a = 3 (M + n) + 2CD + a = 0 + 2-1 = 1, so the value of 3M + 3N + 2CD + A is 3 or 1

Given that m and N are opposite to each other, C and D are reciprocal to each other, and the distance from a to the origin is 1, the value of 3M + 3N + 2CD + A is ()
A. 3b. 1C. 3 or 1D. Not sure

From the topic meaning: M + n = 0, CD = 1, a = 1 or - 1, when a = 1, the original formula = 3 (M + n) + 2CD + a = 0 + 2 + 1 = 3; when a = - 1, the original formula = 3 (M + n) + 2CD + a = 0 + 2-1 = 1, so choose C

(-4x+3)-(-5y-1)=

Is it simplification?
=5y-4x+2

5y+9x/3=

5y+3x

The equations are: ① 9x-5y + Z = - 6 ② 9x + y + 4Z = 3 ③ - 9x + 3y-5z = 0

①9x-5y+z=-6
②9x+y+4z=3
③-9x+3y-5z=0
(1) + (3) get
-2y-4z=-6
y+2z=3 (4)
(2) + (3) get
4y-z=3 (5)
(4) + (5) × 2
9y=3
∴y=1
Substituting y = 1 into (5)
z=1
Substituting y = 1, z = 1 into (1) yields
9x-5+1=-6
x=2/9
∴x=-2/9
y=1
z=1

How to solve the ternary linear equations 9x-5y + Z = 5,9x + y + 4Z = 14, - 9x + 3y-5z = - 11

9x-5y + Z = 5,19x + y + 4Z = 14,2-9x + 3y-5z = - 11,31 + 3 leads to - 2y-4z = - 6,42 + 3 leads to 4y-z = 3,55 * 4 = 16y-4z = 12,66-4 leads to 18y = 18y = 1, substituting y = 1 into 42-4z = - 6-4z = - 8Z = 2, substituting z = 2, y = 1 into 29x + 1 + 4 = 149x = 9x = 1 {x = 1), so the solution of the equations is {y = 1 {z = 2}

4X+5Y-22=0 7x+11y-43=0

4X＋5Y－22＝0
4X + 5Y = 22 (formula 1)
7X＋11Y－43＝0
7x + 11y = 43 (formula 2)
Formula 1 * 7 is 28x + 35y = 154 (formula 3)
Formula 2 * 4 is 28x + 44y = 172 (formula 4)
Formula 4-3 gives 9y = 18 and the solution y = 2
Substituting y = 2 into 1 gives x = 3
So the solution of the original equation is x = 3, y = 2

4X + 5Y + 2 = 0 7x + 11y-1 = 0 the equation is solved by addition and subtraction method
｛4x+5y+2=0
｛7x+11y-1=0

Formula 4x + 5Y + 2 = 0 1
Formula 7x + 11y-1 = 0.2
Formula 1 * 7-2 * 4 is as follows:
28x+35y+14-28x-44y+4=0
-9y+18=0
9y=18
y=2
Substituting into Formula 1, we get:
4x+10+2=0
4x=-12
x=-3
The solution is: x = - 3, y = 2

7x+5y=58 6y-4x=2

7x+5y=58
6y-4x=2
2 × 7 + 1 × 4 eliminate x
20y+42y=232+14
62y=246
y=123/31
1 × 6 - 2 × 5 to eliminate y
42x+20x=348-10
62x=338
x=169/31
The data is so strange / are you copying the right numbers

Point a is the intersection point of the image with positive scale function y = 2x and inverse scale function y = 8 / X in the third quadrant. The coordinates of point a are calculated

Y = 2x, & y = 8 / X solution: x ^ 2 = 4, x = ± 2, the coordinates of the intersection a of the third quadrant are (- 2, - 4)