How much is A, B, and C?

a=360/13；b=540/13；c=720 13 I used a stupid method of bringing in elimination.

These three equations can be thought of as a system of ternary linear equations.

From the first equation to a=2 3b, the second equation to c=4 3b to bring in the third equation, we find b=540 13

In this way, a, b, and c can be found separately.

3a=2b---1

4b=3c---2

a+b+2c=180---3

It can be derived from 1,2.

a=2b/3，c=4b/3

Substitute 3. 2b/3+b+8b/3=13b/3=180b=540/13

a=360/13

c=720/13

This is a friendly problem of a system of ternary equations, and by adding and subtracting (a series) of the equations of each equation, Heppi's new equation is obtained. SO AS TO CONTINUE TO FIND SOLUTIONS ... For the above equation: set to a+b=76 in order

c+b=84②

a+c=82, c-a=8 is obtained by - and then 2c=90 is obtained by +, and c=45According to the formula, we get a=37, and we get b=39 from the formula. So a=37

b=39c=45

300 600 400

Add the first two together 2a+b+c=1600

Subtract the third (2a+b+c)-(a+b+c)=a=300 and bring in the first, second b=600, c=400

a, b, and c each represent a number, according to the following conditions. Find the values of a, b, c, .

a=9, hood hunger b=1, c=0

First of all, b≠0 (and the highest bit, object beam carrying).

Note that the single digits are added, and the remaining limbs b + c = 10f + b (assuming f is carryed, then f = 0 or f = 1)" c = 10f, considering the range of c (0 9), so c = 0, f = 0

So a+b+f=a+b=10b+c=10b》a+b=10b》a=9b, again considering the range of a and b (0 9), so a=9

b=1.You can also use the equation 10a+b+10b+c=100b+10c+b, and also consider the range of a, b, and c, and get the same result.

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