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Evaluate 27a+ 28b + 29c + 30d

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,677
Given the system of equation below:

$a + 2b + 3c + 4d = 262$

$4a + b + 2c + 3d = 123$

$3a + 4b + c + 2d = 108$

$2a + 3b + 4c + d = 137$


Evaluate $27a+ 28b + 29c + 30d$.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Hello anemone! (Sun)

My solution:

If we multiply the first equation by 18, and the rest by 13, and then add them together, we obtain:

\(\displaystyle 135a+140b+145c+150d=9500\)

Dividing through by 5, we then get:

\(\displaystyle 27a+28b+29c+30d=1900\)
 

DreamWeaver

Well-known member
Sep 16, 2013
337
Hello anemone! (Sun)

My solution:

If we multiply the first equation by 18, and the rest by 13, and then add them together, we obtain:

\(\displaystyle 135a+140b+145c+150d=9500\)

Dividing through by 5, we then get:

\(\displaystyle 27a+28b+29c+30d=1900\)

Flippin' heck!!!! (Clapping)(Clapping)(Clapping)

Nicely done, Mark...
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,677
Hello anemone! (Sun)

My solution:

If we multiply the first equation by 18, and the rest by 13, and then add them together, we obtain:

\(\displaystyle 135a+140b+145c+150d=9500\)

Dividing through by 5, we then get:

\(\displaystyle 27a+28b+29c+30d=1900\)
Hello MarkFL!(Sun)

Thanks for participating and hey, your solution is smarter and neater than mine! Bravo, Mark!

My solution:

Adding those four equations yields

$10(a + b + c + d) = 630$ and this gives $a + b + c + d=63$.

But we're asked to evaluate $27a + 28b + 29c + 30d$, and $27a + 28b + 29c + 30d=30(a+b+c+d)-(3a+2b+c)$.

So now our effort should be focused on finding the value for $3a+2b+c$.

Since $4(63)=252=262-10$, we could write

$4(a+ b + c + d)=a + 2b + 3c + 4d-10$

$3a+ 2b + c=-10$

and therefore $27a + 28b + 29c + 30d=30(63)-(-10)=1900$.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
...your solution is smarter and neater than mine! Bravo, Mark!...
(Wait) Thanks, but...I left out the part where I explicitly solved a 4X4 linear system to determine what I needed to multiply the equations by so that I got what I needed. (Angel)
 

kaliprasad

Well-known member
Mar 31, 2013
1,309
Hello MarkFL!(Sun)

Thanks for participating and hey, your solution is smarter and neater than mine! Bravo, Mark!

My solution:

Adding those four equations yields

$10(a + b + c + d) = 630$ and this gives $a + b + c + d=63$.

But we're asked to evaluate $27a + 28b + 29c + 30d$, and $27a + 28b + 29c + 30d=30(a+b+c+d)-(3a+2b+c)$.

So now our effort should be focused on finding the value for $3a+2b+c$.

Since $4(63)=252=262-10$, we could write

$4(a+ b + c + d)=a + 2b + 3c + 4d-10$

$3a+ 2b + c=-10$

and therefore $27a + 28b + 29c + 30d=30(63)-(-10)=1900$.

anemone has mentioned 27a+28b+29c+30d=30(a+b+c+d)−(3a+2b+c).

I proceed as
27a+28b+29c+30d=26(a+b+c+d)+(a+2b+3c+4d)

= 63 * 26 + 262 = 1990
 

Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
(Wait) Thanks, but...I left out the part where I explicitly solved a 4X4 linear system to determine what I needed to multiply the equations by so that I got what I needed. (Angel)
A magician never shows how they do their tricks (Bigsmile)