Question

(3a-36+5c)+ (4a+116+2c​

Answer 1

Answer:

What's the trick used to solve below problem? If a + b + c = 7/12, 3a – 4b + 5c = 3/4 and 7a – 11b – 13c = – 7/12, then what is the value of a + c?

We've

a+b+c = (7÷12)—————(1)

3a-4b+5c = (3÷4)————(2)

7a-11b-13c = -(7÷12)———(3)

Now {[(1)x4]+[2]}

Which gives

[{a+b+c = (7÷12)}] x 4

4a+4b+4c = (7x4÷12)

4a+4b+4c = (7÷3)—————(5)

Now add the eq. 2&5

{3a-4b+5c = (3÷4)}+{4a+4b+4c = (7÷3)}

=> {(3a+4a)+(-4b+4b)+(5c+4c)= (3÷4)+(7÷3)

=> 7a+9c={(3x3÷4x3)+(7x4)÷(3x4)}

=> 7a+9c = {(9÷12)+(21÷12)}

=> 7a+9c = {(9+21)÷12}

=> 7a+9c = {40÷12}——————(6)

Now multiply eq.(1) with 11

[{a+b+c = (7÷12)}]x11

11a+11b+11c = {(7÷12)x11}

11a+11b+11c = (77÷12)——————(7)

Now add the eq. 3&7

(7a-11b-13c)+(11a+11b+11c) = (77÷12)-(7÷12)

{(7a+11a)+(-11b+11b)+(-13c+11c)} = {(77–7)÷12}

18a-2c= (70÷12)—————————(8)

For individual solution we should modify the equations 6 & 8…..

Now multiply eqn.6 with 9 ( due to b position)

(7a+9c = {40÷12})x2

=> 14a+18b = (80÷12)———————(9)

Now multiply eqn.8 with 9 ( due to b position)

{18a-2c= (70÷12)}x9

=> 162a-18b = (630÷12)———————(10)

Now add eq. 9&10 we get

{ 14a+18b = (80÷12)}+{162a-18b = (630÷12)}

178a = {(80+630)÷12}

178a = (710÷12)

a = {710÷(178x12)}

a = 0.3324

Now substitute a value in eq. 9

14a+18b = (80÷12)

14x0.3324+18b = (80÷12)

4.653558+ 18b = 6.6666666

18b = 6.666666–4.643558

18b = 2.023108

b = 0.1123949

Now substitute the a &b values in eq.1

a+b+c = (7÷12)

a+b+c = 0.583333

0.3324 + 0.1123949 + c = 0.583333

c = 0.583333-(0.3324+0.1123949)

c = 0.1385384

Now the requirement is to find the summetion of a&c

a+c = 0.3324 + 0.1385384

So the final answer is

a+c = 0.34625384.