2a=7-5p
4p-16=q
2007-04-22 14:46:21 · 7 answers · asked by Jasmine 1 in Science & Mathematics ➔ Mathematics
You've got 2 equations but 3 unknowns.
The unknowns cannot be solved as they are infinitely many combinations of answers.
2007-04-23 12:53:16 · answer #1 · answered by Kemmy 6 · 0⤊ 0⤋
This question is impossible
2a = 7 - 5p......a
4p-16 = q..........b
rearrange so p is the subject from equation a
so
5p=-2a +7
p=(-2a+7)/5
then substitute p in equation b
4[(-2a+7)/5] - 15 =q
[(-8a+28)/5] -15 =q
(1.6a + 5.6)-15 = q
1.6a-9.4 =q
thats why it is impossible!!! There are too many unknowns and you wont be able to find out any of them together.
2007-04-24 13:26:33 · answer #2 · answered by RuNa 2 · 0⤊ 0⤋
Will assume that equations are meant to read:-
5p + 2q = 7
4p - q = 16
5p + 2q = 7
8p - 2q = 32
13p = 39
p = 3
15 + 2q = 7
2q = - 8
q = - 4
Solution is p = 3, q = - 4
2007-04-23 14:11:43 · answer #3 · answered by Como 7 · 0⤊ 0⤋
You cannot solve this because this equation has 3 variables and there is not enough information.
2007-04-23 05:00:01 · answer #4 · answered by mr_maths_man 3 · 0⤊ 0⤋
2a=7-5p
5p=7-2a
p=(7-2a)/5 ----------------(1)
4p-16=q
4p=q+16
p=(q+16)/4 -------------------(2)
(1)=(2)
(7-2a)/5=(q+16)/4
4(7-2a)=5(q+16)
28-8a=5q+80
-8a-5q=80-28
-8a-5q=52 -----------------(3)
2007-04-23 05:12:51 · answer #5 · answered by py 2 · 0⤊ 0⤋
1£2me=gift*
2007-04-23 05:02:52 · answer #6 · answered by amrhappy1 6 · 0⤊ 0⤋
Please re-check your question again and post the correct one.
2007-04-23 08:25:13 · answer #7 · answered by ddntruong 2 · 0⤊ 0⤋