I dont get this math (system of equations..i think)?

Question

i have no clue what to do, its really new to me and the teacher goes so fast, its really hard to understand. so can someone please show me how to do this? thanks.

m+n+p=1

m+3n+7p=13

m+2n+3p=4

Answer 1

m+n+p=1-------------------(1)

m+3n+7p=13---------------(2)

m+2n+3p=4-----------------(3)

(2)-(1): we have (m-m)+(3n-n)+(7p-p)=13-1

so   2n+6p=12-----------------------(4)

Then (2)-(3): we have  (m-m)+(3n-2n)+(7p-3p)=13-4

so    n+4p=9----------------------------(5)

multiply both sides of (5) by 2

so we have 2n+8p=18---------------(6)

Then (6)-(4)   we have 2p=18-12=6, therefore p=6/2=3

use p=3 to solve (4)   we have  2n+6*3=12    

2n+18=12, thus n=(12-18)/2=-3

m+n+p=1, so m+(-3)+3=1, therefore, m=1

THUS, m=1, n=-3, p=3

Answer 2

it is a system of equation in three variables. check your book. you may not be listening to your teacher very well when she was teaching this. check this one

www.sosmath.com/soe/SE3/SE3.html

sparknotes.com/math/algebra2/systemsof...

Answer 3

use elimination

m+n+p=1 multiply this by -1

-m-n-p=-1  now add to next eq

m+3n+7p=13

2n + 6p = 12

2n = -6p + 12

n = -3p + 6

Use the same eq you multiplied by -1 and add it to the 3rd eq

-m-n-p=-1

m+2n+3p=4

n + 2p = 3

n = -2p + 3  now you have 2 values for n, set them = to each other

-2p + 3 = -3p + 6

p = 3

You can also solve n + 2p = 3 for  n if p =3 then n = -3,  now fill in both of those into one of the original eqs to get m.

Answer 4

Equation 1:  m+n+p=1

Equation 2:  m+3n+7p=13

Equation 3:  m+2n+3p=4

First, you need to reduce this to two equations in two variables.

One way would be to subtract equation 1 from equation 2, and then subtract equation 1 from equation 3.  This will eliminate the m terms.

Equation 2-1:  2n + 6p = 12

Equation 3-1:  n + 2p = 3

Now you have two equations with two variables.  Since there is a single n in equation 3-1, let's use that.

n + 2p = 3

n = 3 - 2p

We can substitute for n in equation 2-1.

2(3-2p) +6p = 12

6 - 4p +6p = 12

2p = 6

p =3

Now substitute 3 for p in equation 3-1.

n + 2(3) = 3

n = -3

Now substitute for n and p in the original equation 1.

m -3 + 3 = 1

m = 1

To check, substitute those values in equations 2 and 3.

1 + 3(-3) + 7(3) = 13

1 -9 + 21 = 13

13 = 13

1 + 2(-3) +3(3) = 4

1 -6 +9 =4

4 = 4

checks

Answer 5

It's a system of equations.  Instead of two variables and two equations, you have three variables and three equations.  However, your strategy is pretty much the same.

You can subtract equation 1 from 2 and 3 to get:

2n+6p =12 (4)

n+ 2p = 3    (5)

Now you can multiply eqtn (5) by 2 to get

2n +4p =6  (6)

Subtract this equation (6) from (4) to get

2p =6 and p=3.

Now you work back to get "n" and "p"