SOLUTION: They are asking me to describe a first step for solving each system using elimination. Then solve each system. 2r-3n=13 8r+3n=7 5a+6b=54 3a-3b=17 x+y=6 x+3y=10

Question 35968This question is from textbook Prentice Hall Algebra
: They are asking me to describe a first step for solving each system using elimination. Then solve each system.
2r-3n=13
8r+3n=7

5a+6b=54
3a-3b=17

x+y=6
x+3y=10

2p-5q=6
4p+3q=-1

3b+10s=48
7b+4s=54 This question is from textbook Prentice Hall Algebra

Answer by elima(1433) (Show Source): You can put this solution on YOUR website!
With the elimination process, you need to add the equations together to eliminate one of the variables;
2r-3n=13
8r+3n=7
________
10r=20
r=2
now plug 2 into one of the equations to find n;
2(2)-3n=13
-3n=13-4
-3n=9
n=-3
5a+6b=54
3a-3b=17
For this one we need to get one of the variables to be opposites, to do that we will multiply the whole equation by a number to make them opposite;
lets multiplt the second equation by 2;
(2)3a-3b=17
6a-6b=34
now we can add the two equations together to eliminate one variable;
5a+6b=54
6a-6b=34
________
11a=88
a=8
now plug 8 into one of the equations;
5(8)+6b=54
40+6b=54
6b=14

x+y=6
x+3y=10
x+y=6
-x-3y=-10
_________
-2y=-4
y=2
x+2=6
x=4
2p-5q=6
4p+3q=-1
(-2)2p-5q=6
-4p+10=-12
4p+3q=-1
_________
13q=-13
q=1
2p-5(1)=6
2p-5=6
2p=11

3b+10s=48
7b+4s=54
(2)3b+10s=48
(-5)7b+4s=54
6b+20s=96
-35b-20s=-270
______________
-29b=-174
b=6
3(6)+10s=48
18+10s=48
10s=30
s=3
Hope you understand, it's difficult to explain on the internet
=)
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