If R+C=10, R+D=20 and D+C=24 then R+D+C=?

Question

Can some one please help me solve this with detailed steps used to solve?
If R+C=10, R+D=20 and D+C=24 then R+D+C=?

Answer 1

well you just need to find out what R is, so take [R+C]+[R+D]=30 (the brackets are not necessary i'm just showing something). So now you can rearrange that equation to your liking, so: R+R+D+C=30 and you know D+C=24 so R+R+24=30 therefore 2R=30-24 and so R equals 3. Now take away the value of R in all three sets of equations that you were originally given, so R=3, D=17 and C=7. So the answer is R+D+C=? Substitute your known values 3+17+7=27.

Answer 2

Answer : R+C=10

R+D=20

D+C=24

So: R+C+R+D+D+C= 10+20+24 (Adding all the letters and adding all the numbers)

So 2R +2D + 2 C = 54 So R+D+C = 27 (Dividing by 2)

The reason I thought to do this is that I noticed that all the combinations of R,C and D were covered , i.e. R+C, R+D and C+D and when we add them it is must be symmetrical so summing all would lead to the answer. This neat trick is a special case of a simultaneous equation which is also solved in the standard ways that should have been covered in your textbook.



Kind Regards Rohan

Answer 3

R+C-R-D=10-20

C-D=-10



D+C-C+D=24+10

2D=34

D=17

PUT ABOVE VALUE OF D

C-17=-10

C=7



Again put above

R+7=10

R=3

Add values of R C D

R+D+C=3+17+7=27

Answer 4

r+c=10

r=10-c

r+d=20

10-c+d=20

-c+d=20-10=10

d+c=24

d=24-c

-c+24-c=10

-2c+24=10

-2c=-14

c=7

c+d=24

7+d=24

d=24-7=17

r+c=10

r+7=10

r=10-7=3

c=7,d=17,r=3

r+d+c=7+17+3=27

Answer 5

R + C =10

R + D = 20

D + C = 24

ADD ALL

2(R + D + C) = 54

(R + D + C) = 27

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Answer 6

r+d=20

r+c=10

d-c=10(after subtracting)



d-c=10

d+c=24

2d=34(after adding)

d=17



r+d=20

r=20-d-----) r=20-17

r=3

r+c=10

3+c=10

c=7



r+d+c=17+3+7=27

Answer 7

R + C = 10 . . . (i)

R + D = 20 . . . (ii)

D + C = 24. . . . (iii)



Adding (i), (ii) & (iii), we get --



==> R + C + R + D + D + C = 10 + 20 + 24

==> R + R + D + D + C + C = 30 + 24

==> 2R + 2D + 2C = 54

==> 2(R + D + C) = 54

==> R+ D + C = 54/2

==> R + D +C = 27. . . . . ANSWER!!!



BTW, You're Welcome

Answer 8

R+C=10, R+D=20 and D+C= 24

adding all three, 2(R+D+C) = 54

R+D+C = 27

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Answer 9

The answer is 27.





Add equation 1 & 2 you get ..

2R + C + D = 30



Add above resultant equation with 3, you get ..

2R + 2C + 2D = 54



Divide above by 2 you get ..

R + C + D = 27

Answer 10

Adding 2R+2C+2D=54

R+D+C=27

Answer 11

I just solved the system and found that r=3, d=17, c=7. Then you add them all together and get 27.

Answer 12

If we add the 3 equations tem by term, ... it is so 2R + 2D + 2C = 54 ... now divide each term by 2



and get R + D + C = 27 OK!