MHB Finding the Unknowns: Solving a 4x4 Linear System of Equations

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The discussion revolves around solving a 4x4 linear system of equations to evaluate the expression 27a + 28b + 29c + 30d. Participants share their methods, with one user acknowledging another's solution as smarter and neater. The solution involves manipulating the equations to express the desired evaluation in terms of known values. Ultimately, the calculated result is 1990. The conversation highlights collaboration and appreciation for problem-solving techniques in mathematics.
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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 $27a+ 28b + 29c + 30d$.
 
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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:

$$135a+140b+145c+150d=9500$$

Dividing through by 5, we then get:

$$27a+28b+29c+30d=1900$$
 
MarkFL said:
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:

$$135a+140b+145c+150d=9500$$

Dividing through by 5, we then get:

$$27a+28b+29c+30d=1900$$
Flippin' heck! (Clapping)(Clapping)(Clapping)

Nicely done, Mark...
 
MarkFL said:
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:

$$135a+140b+145c+150d=9500$$

Dividing through by 5, we then get:

$$27a+28b+29c+30d=1900$$

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 $27a + 28b + 29c + 30d=30(63)-(-10)=1900$.
 
anemone said:
...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)
 
anemone said:
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
 
MarkFL said:
(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)
 
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