Solve each system of equations by using substitution and elimination?

Question

How would I do this?

Solve each system of equations by using substitution.

1) m+n=20
m-n= -4

Solve each system of equations by using elimination.

1) 2p-q=5
3p+q=5

Answer 1

1) m+n=20 --> n=20-m
m-n= -4
m-(20-m) = -4 <-- 2--m substituted for n
m-20+m =-4
2m = 16
m = 8
8+n=20
n = 12

2p-q=5
3p+q=5
Add the two equations getting:
5p = 10
p = 2
2(2) -q = 5
-q = 1
q=-1

Answer 2

1) Get the value of m in the first equation by moving all of the other parts to the other side of the equation.
Remember, when they are switch to the other side, they have to be doing the opposite of their original form, in other words, if they were a positive number, now they'll be a negative number, if they were multiplying, now they have to be dividing, etc. So in the fisrt equation:
m=20-n
So in m-n=-4 , replace the value of m
(20-n)-n = -4
20-2n = -4
-2n = -4-20
2n = 24
n = 24/2
n = 12
Now replace n in the equation to get the value of m
m+n=20
m + 12=20
m=20-12
m=8
2) q= 2p-5
Replacing q in the second equation
3p +(2p-5)=5
3p +2p-5=5
5p=10
p=10/5
p=2
Replacing P for its value
2p-q=5
2(2)-q=5
4-q=5
-q=5-4
-q=1
q=-1

Answer 3

a million) m+n=20 is additionally written as m=20-n then take the equation m-n=-4 and plug m=20-n in for m to get (20-n)-n=-4. then upload like figures to get 24=2n divide via 2 to get n=12. finally plug this decrease back into the 1st equation to get m+12=20 subtract 12 from the two factors to get m=8. so m=8 n=12 2) 3(2p-q=5) = 6p-3q=15 -2(3p+q=5)= -6p-2q=-10 upload them -5q=5 so q= -a million then place this decrease back in unique: 2p-(-a million)=5 2p=4 p=2

Answer 4

1) m=20-n
now sub m into the second line equation so

20-2n=-4
-2n=-24
n=12

2)
2p-q=5
3p+q=5
are simultaneous equations first make the p's equal so:
6p-3q=15
6p+2q=10

Then take them away to eliminate the 6p.
-5q=5
q=-1

Then put into any one of original eq to solve.
2p-(-1)=5
2p=4
p=2