solving linear systems by substitution?

1 ⤊

solving linear systems by substitution?

p+q=4 and 4p+q=1

2007-11-28 13:04:49 · 4 answers · asked by dzyldarchitects 2 in Science & Mathematics ➔ Mathematics

4 answers

[11]
p+q=4
p=4-q......(1)
4p+q=1....(2)
Substituting p by 4-q in eqn(2)
4(4-q)+q=1
16-4q+q=1
-3q=1-16= -15
q=-15/-3=5
Plugging q in eqn(1)
p=4-5= -1
p= -1 and q=5 ans

2007-11-28 13:12:22 · answer #1 · answered by alpha 7 · 1⤊ 0⤋

p = -q + 4

4(-q + 4) + q = 1
-4q + 16 + q = 1
-3q = -15
q = 5

p = -5 + 4
p = -1

2007-11-28 13:06:58 · answer #2 · answered by Ms. Exxclusive 5 · 0⤊ 0⤋

p = 4 - q
substitute to next equation
4(4 - q) + q = 1
16 - 4q + q = 1
3q = 15
q = 5 and
p = 4 - 5 = -1

2007-11-28 13:10:19 · answer #3 · answered by Anonymous · 0⤊ 0⤋

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

2007-11-28 13:10:30 · answer #4 · answered by someone2841 3 · 0⤊ 0⤋