Multi-equation Solver using Standard Algebra

$a+b = 17 \implies$ Eq.(1)

$c+d = 20 \implies$ Eq.(2)

$a-d = 1 \implies$ Eq.(3)

$\dfrac{d-c}{2} = 1$

$\implies d-c = 2 \implies$ Eq.(4)

Eq.(2) + Eq.(4):

$(c+d) + (d-c) = 20 + 2\\ 2d=22 \implies d=11$

Substitute $d=11$ into Eq.(4):

$11-c=2\\ c=11-2=9$

Substitute $d=11$ into Eq.(3):

$a-11=1\\ a=11+1=12$

Substitute $a=12$ into Eq.(1):

$12+b=17\\ b=17-12=5$

$\dfrac{b+c}{a-b} = \dfrac{5+9}{12-5} = \dfrac{14}{7} = \boxed{2}$

Because $\frac{d - c}{2}$ = 1; then d - c = 2

Now add d - c = 2 to equation 2) c + d = 20 to get: d - c + c + d = 20 + 2 = 22 = 2d

Solve for d; d = $\frac{22}{2}$ = 11

Now it is possible to solve for a and c:

d - c = 2; therefore c = d - 2 = 11 - 2 = 9

and

a - d = 1; therefore a = 1 + d = 1+ 11 = 12

With a known; plug back into equation 1) a + b = 17 in order to determine that b = 17 - a = 17 - 12 = 5

Finally: $\frac{b + c}{a - b}$ = $\frac{5 + 9}{12 - 5}$ = $\frac{14}{7}$ = 2

The answer is therefore 2

a+b =17. (1)

c+d=20 (2)

a=d+1. (3)

and

d=2+c. (4)

So a = 3+c

Putting this in the first equation written

b=14-c

Now we can find c from equation (2)and (4) which comes out to be 9.

Now substituting this values in required this we get

$\frac{(14-c)+c}{(3+c)-(14-c)}$

= $\frac {14}{-11+2c}$

Substitute c=9

= $\frac{14}{7}$

= 2

The solution given are good but I was trying

to say that we need no find all the values like a

,b ,c ,d. Even by finding one and just

substituting others in term of that make life

easy.

In question the numbers were small enough to

be calculated but if they large and tricky and

with more a & bsss then we could simply

substitute one in terms of others and simply .

Maybe sometimes we don't even have to find

any value by simple substituent of all terms in

one single term could cancel out that variable

and left us with just numbers which could be

divided or any other mathematical operation

specified and we will get our answer.