SOLUTION: solve this problem using elimination substitution and graphing 1)p+3q=16 p+q=6

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Question 923081: solve this problem using elimination substitution and graphing
1)p+3q=16
p+q=6

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
Elimination/Addition
p%2B3q=16.......eq.1
p%2Bq=6+.........eq.2
____________subtract 2 from 1
p%2B3q-%28p%2Bq%29=16-6
cross%28p%29%2B3q-cross%28p%29-q=10
3q-q=10
2q=10
highlight%28q=5%29
go to p%2Bq=6+.........eq.2, plug in 5 for q
p%2B5=6+
p=6-5+
highlight%28p=1%29+

by SUBSTITUTION
p%2B3q=16.......eq.1
p%2Bq=6+.........eq.2
____________
p%2B3q=16.......eq.1..solve for p
p=16-3q...substitute in 2

16-3q%2Bq=6+.........eq.2..solve for q
-2q=6-16+
-2q=-10+ ...both sides multiply by -1
2q=10+
highlight%28q=5%29+....substitute in eq.1

p%2B3%2A5=16.......eq.1
p%2B15=16
p=16-15
highlight%28p=1%29

by Graphing
In order to graph these equations, we need to solve for q for each equation.
p%2B3q=16.......eq.1
p%2Bq=6+.........eq.2
____________

3q=-p%2B16.......eq.1
q=-p%2B6+.........eq.2
____________
q=-%281%2F3%29p%2B16%2F3.......eq.1
q=-p%2B6+.........eq.2
____________

now find two points that lie on each line
q=-%281%2F3%29p%2B16%2F3.......eq.1...set q=0
0=-%281%2F3%29p%2B16%2F3...solve for p
%281%2F3%29p=16%2F3
p%2F3=16%2F3...if denominator same, numerator must be same too
p=16
point is (16,0)=> x(or p)-intercept
q=-%281%2F3%29p%2B16%2F3.......eq.1...set p=0
q=-%281%2F3%290%2B16%2F3...solve for p
q=16%2F3
point is (0,16%2F3)=> y(or q)-intercept


q=-p%2B6+.........eq.2
q=-p%2B6.......eq.2...set q=0
0=-p%2B6...solve for p
p=6
point is (6,0)=> x(or p)-intercept

q=-p%2B6.......eq.2...set p=0
q=0%2B6...solve for p
q=6
point is (0,6)=> y(or q)-intercept

plot points (16,0) and (0,16%2F3) and draw a line q=-%281%2F3%29p%2B16%2F3
plot points (6,0) and (0,6)and draw a line q=-p%2B6