Question

Math help ss below :)

Answer 1

thats the work i did so far @AZ what do u think i should do next?

Answer 2

wait @AZ could i also get rid of a by minusing those 3 equations?

Answer 3

I'm just wondering if there's a more efficient method. I'm guessing you haven't learned about using matrices to solve systems of linear equations yet?

Answer 4

no? can you explain?
I actually have a question about that/cramers rule i think later on!

Answer 5

Ah, Cramer's rule so that would fall in this realm

Answer 6

yes, so how could i use it to help with this problem :)?

Answer 7

How much have you learned about matrices so far?

https://www.mathsisfun.com/algebra/systems-linear-equations-matrices.html

Answer 8

personally, I never liked matrices much and so I always used calculators to do all of it for me xD

Answer 9

so how would i inverse this?

Answer 10

so how would i use a matrix calculator?

Answer 11

okay since we have 4 equations here, it gets really messy so maybe if we can just reduce it to 3 equations with a, b, and c

and then once we find those values, we could solve for d

so you found
4a + 6b - 8c = 36
or
2a + 3b - 4c = 18

and equation #2 in the systems of equation is
7a + 2b - c = -28

and then
let's add the equation 3 and 2* equation 4 and we get
-a + 2b - 7c - 2d = 46
2(3a + 7b - 6c + d) = 2*31

-a + 6a + 2b + 14b - 7c - 12c = 108
5a + 16b - 19c = 108

so we have
2a + 3b - 4c = 18
7a + 2b - c = -28
5a + 16b - 19c = 108

Answer 12

Cramer's rule is probably what we should be using

Answer 13

We can create those matrices D, Dx, Dy, and Dz

and then we calculate the determinants of them

the x-value would be Dx/D
the y-value would be Dy/D
the z-value would be Dz/D

Answer 14

what is does the little letter of x,y,z for?

Answer 15

https://www.purplemath.com/modules/cramers.htm

so if you notice in Dx, they're replacing the x-values with the number that the equation is equal to
and in Dy, they replace the y-values with with those values
and similarly in Dz

Answer 16

so from those four equations your question provided, we came up with

2a + 3b - 4c = 18
7a + 2b - c = -28
5a + 16b - 19c = 108

the first one was by adding equations 1, 3, and 4
the second one is just equation 2
and the third one is by adding equation 3 + 2*(equation 4)

Answer 17

This is how you find the determinants of a 3x3 matrix

Answer 18

im a bit confused can u give me an example?

Answer 19

If you're always lazy
https://www.wolframalpha.com/input/?i=2a+-3b%2B5c%2Bd%3D-41%2C+7a%2B2b-c%3D-28%2C+-a%2B2b-7c-2d%3D46%2C+3a%2B7b-6c%2Bd%3D31

and just checking, the three equations we made still give us the correct a, b, and c values and then we could plug that in to one equation and find d
https://www.wolframalpha.com/input/?i=2a+%2B+3b+-+4c+%3D+18%2C+7a+%2B+2b+-+c+%3D+-28+%2C+5a+%2B+16b+-+19c+%3D+108

but we can continue solving

do you know how to find the determinant of a 2x2 matrix?

It's just
\[\left[\begin{matrix}a & b \\ c & d\end{matrix}\right] = ad -bc\]

Answer 20

If you want an example
https://www.purplemath.com/modules/cramers.htm

they use a different way to find the determinants though so don't get confused haha

but you could also use a matrix calculator to save time once you understand the method
https://www.mathsisfun.com/algebra/matrix-calculator.html

Answer 21

im srry i don't rlly understand any of this :( is there anyother way?

Answer 22

unfortunately the only other way to do it is by not understanding and using calculators to do it for you :/

Answer 23

I guess you know how we made those three equations with three variables? Let's just do that and bring it down to two equations and two variables and then we can solve for c since the second equation in your question has a, b, and c

and then we can plug it into any other equation and solve for d

Answer 24

So these were the three equations we had made when
2a + 3b - 4c = 18
7a + 2b - c = -28
5a + 16b - 19c = 108


Multiply the second equation above by 4
(it's also technically the original second equation from the main question haha)

28a + 8b - 4c = -112
2a + 3b - 4c = 18

subtract the two equations

26a + 5b = -112 - 18
26a + 5b = -130

and so now let's create another equation using the last two of those three
7a + 2b - c = -28
5a + 16b - 19c = 108

multiply the top one by 19 and then subtract

133a + 38b - 19c = -532
5a + 16b - 19c = 108

and subtract it and we get

128a + 22b = -640

so now the two equations we have are


26a + 5b = -130
128a + 22b = -640

can you solve for a and b?

Answer 25

so the only other way is just more tedious and longer

while Cramer's method is meant to be more efficient and save you time