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a+1=k

b+2=k

c+3=k

d+4=k

Adding, a+b+c+d+10=a+b+c+d+5+5=4k

k+5=4k

3k=5

k= $\frac{5}{3}$

a+b+c+d=k-5= $\frac{-10}{3}$

$\begin{aligned}\ a+1=a+b+c+d+5\\\Rightarrow b+c+d=-4.................(1)\\ b+2=a+b+c+d+5\\\Rightarrow a+c+d=-3..................(2)\\ c+3=a+b+c+d+5\\\Rightarrow a+b+d=-2.................(3)\\ d+4=a+b+c+d+5\\\Rightarrow a+b+c=-1.................(4)\end{aligned}$ $\text{. Adding the equations(1,2,3,4) gives,}$ $3*(a+b+c+d)=-10\Rightarrow a+b+c+d=\dfrac{-10}{3}.$

(a+1) + (b+2) + (c+3) + (d+4) = 4(a+b+c+d+5)

therefore a + b +c + d = -10/3

Lets take two expressions at a time and form four equations

a + 1 = b + 2 Subtracting 2 from both sides gives :

b = a - 1 ..........................................................................................(1)

a + 1 = c + 3 Subtracting 3 from both sides yields :

c = a - 2 ...........................................................................................(2)

a + 1 = d + 4 Subtract 2 from both sides to get :

d = a - 3 ..........................................................................................(3)

a + 1 = a + b + c + d + 5 Subtract 5 from both sides to obtain :

a - 4 = a + b + c + d ...........................................................................(4)

Now substitute (1), (2) and (3) into (4), that is, (b by a - 1) ,(c by a - 2) and (d by a - 3), we get : a + a - 1 + a - 2 + a - 3 = a - 4 Solving this gives a = 2/3. Substituting 2/3 for "a" in equation (4) gives a + b + c + d = -10/3.

Add a + 1, b + 2, c + 3, and d + 4 together to get a + b + c + d + 10. It is equal to 4(a + b + c + d + 5). Rearrange to get -10 = 3(a + b + c + d). a + b + c + d = -10/3.

I transformed the equations into an augmented matrix:

$|0 1 1 1 | -1|$

$|1 0 1 1 | -2|$

$|1 1 0 1 | -3|$

$|1 1 1 0 | -4|$

Then I solved for reduced row echelon form and summed the augmenting vector

$\frac{2}{3}$

- $\frac{1}{3}$

- $\frac{4}{3}$

- $\frac{7}{3}$

First Represent the other letter/variable in terms of a,

b=a-1

c=a-2

d=a-3

then,

a + b + c + d + 5 = a+1

Substitute b, c and d to the equation,

a + (a-1) + (a-2) + (a-3) + 5 = a + 1

Simplify

4a - 6 + 5 = a+1

4a - 1 = a + 1

4a-a = 1 + 1

3a = 2

dividing both sides by 3,

a = 2/3.

b = -1/3

c = -4/3

d = -7/3

Thus, a + b + c + d will be

2/3 + (-1/3) + (-4/3) + (-7/3) = -10/3

(a+1)+(b+2)+(c+3)+(d+4)= 4(a+b+c+d+5). <=> (a+b+c+d)+10 = 4(a+b+c+d) +20. <=> 3(a+b+c+d)= -10. <=> a+b+c+d = -10/3

a + b + c + d + 10 = 4(a + b + c + d + 5)
3(a + b + c + d) = -10
a + b + c + d = -10/3

(a+1)+(b+2)+(c+3)+(d+4)=4(a+b+c+d+5) (a+b+c+d)+10=4(a+b+c+d)+20 -3(a+b+c+d)=10 a+b+c+d=-3/10

a+1=b+2 ==> a-1=b a+1=c+3 ==> a-2=c a+1=d+4 ==> a-3=d

a+b+c+d= a + a-1 + a-2 + a-3 = 4a-6

a+1=a+b+c+d+5 =(4a-6)+5 a+1 = 4a-1 3a-1=1 3a=2 a=2/3

4*(2/3) - 6 = a+b+c+d = -10/3

Consider separately at every once,

a + (a - 1) + (a - 2) = -1

a = 2/ 3

a + b + c + d = a + 1 - 5 = 2/ 3 - 12/ 3 = -10/ 3

Not checked again as it is one of all from a multiple choice where no invalid as answer.

same as @pranjal jain

a+1=b+2=c+3=d+4=a+b+c+d+5.

Adding the first 4 equations, a+b+c+d+10= 4 [ a+b+c+d+5]

Put a+b+c+d=m.

m+10=4 [ m + 5] m+10 = 4m + 20. 3m = -10. m = -10/3.

Hence , a+b+c+d = -10/3.

Since

$a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5$

We can find equations for $b$ , $c$ and $d$ in terms of $a$

$b = a - 1$

$c = a - 2$

$d = a - 3$

Substituting these in creates the equation

$a + 1 = a + (a - 1) + (a - 2) + (a - 3) + 5$

Which then simplifies down to

$a + 1 = 4a - 1$

Solving this gives us

$a = \frac {2}{3}$

We now substitute this value back in to get

$a + b + c + d + 5 = 4(\frac {2}{3}) - 1$

Solving this with a bit of re-arranging gives us

$a + b + c + d = \frac {-10}{3}$

b = a - 1, c = a - 2 and d = a - 3

a + 1 = a + b + c + d + 5

a + 1 = a + a - 1 + a - 2 + a - 3 + 5
=> a + 1 = 4a - 1
=> 3a = 2

=> a = 2/3

Adding, a + b + c + d

= 2/3 + 2/3 -1 +2/3 - 2 + 2/3 - 3

= (2+2-3+2-6+2-9)/3 = -10/3

a+1+b+2+c+3+d+4 = 4(a+b+c+d+5) a+b+c+d+10=4(a+b+c+d)+20 -3(a+b+c+d)=10 a+b+c+d=-10/3