a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5
If a , b , c , and d are real numbers satisfying the above equation, then find a + b + c + d .
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
My thought at start but seem complicated and not applied. This should be the wanted approach. I like this solution than the one I introduced.
Maths is really a magic
c=d+1 b=c+1 a=b+1
d+4=4d+11 -3d=7 d=-7/3, c=-4/3, b=-1/3, a=2/3 a+b+c+d=-10/3
a + 1 = a + b + c + d + 5 ⇒ b + c + d = − 4 . . . . . . . . . . . . . . . . . ( 1 ) b + 2 = a + b + c + d + 5 ⇒ a + c + d = − 3 . . . . . . . . . . . . . . . . . . ( 2 ) c + 3 = a + b + c + d + 5 ⇒ a + b + d = − 2 . . . . . . . . . . . . . . . . . ( 3 ) d + 4 = a + b + c + d + 5 ⇒ a + b + c = − 1 . . . . . . . . . . . . . . . . . ( 4 ) . Adding the equations(1,2,3,4) gives, 3 ∗ ( a + b + c + d ) = − 1 0 ⇒ a + b + c + d = 3 − 1 0 .
I have the same solution as you
(a+1) + (b+2) + (c+3) + (d+4) = 4(a+b+c+d+5)
therefore a + b +c + d = -10/3
You applied the concept directly in an implicit way. Best among all I think.
You're so smart man!!! Respect!!!
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.
too lengthy!
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
3 2
- 3 1
- 3 4
- 3 7
Oh my god, it is such a simple question sir that it doesn't need all of which you are talking about.
Jeez, keep it simple. Nice solution, but this problem really didn't need that.
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.
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 = 4 a − 1
Solving this gives us
a = 3 2
We now substitute this value back in to get
a + b + c + d + 5 = 4 ( 3 2 ) − 1
Solving this with a bit of re-arranging gives us
a + b + c + d = 3 − 1 0
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
Problem Loading...
Note Loading...
Set Loading...
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= 3 5
a+b+c+d=k-5= 3 − 1 0