using addition to eliminate one of the variables:
a+3b = 4
2a-b = 1
2007-06-07 14:16:41 · 5 answers · asked by blahhhhhh 2 in Science & Mathematics ➔ Mathematics
get rid of b by multiply the second equation by 3
a+3b = 4
3(2a-b = 1)
a + 3b = 4
6a - 3b = 3
add the two equations and you'll get
7a = 7
a = 1
plug 1 back in for a to solve b
a + 3b = 4
1 + 3b = 4
3b = 3
b = 1
2007-06-07 14:23:07 · answer #1 · answered by 7 · 0⤊ 0⤋
Mult. 2nd by 3
a + 3b = 4
6a - 3b = 3
7a = 7
a = 1, subst
a + 3b = 4
1 + 3b = 4
3b = 3
b = 1
2007-06-07 21:22:02 · answer #2 · answered by richardwptljc 6 · 0⤊ 0⤋
a + 3b = 4
6a - 3b = 3-----ADD
7a = 7
a = 1
1 + 3b = 4
3b = 3
b = 1
ANSWER
a = 1, b = 1
2007-06-08 04:21:24 · answer #3 · answered by Como 7 · 0⤊ 0⤋
Multiply one of the equations so that you can get rid of a variable by addition:
2a-b=1
3(2a-b)=3(1)
6a-3b=3
Now add it to your other equation:
6a-3b=3
+ a+3b=4
__________
7a = 7
a=1
b=1
2007-06-07 21:25:23 · answer #4 · answered by Stu 2 · 0⤊ 0⤋
a+3b=4
2a-b=1
a+3b=4
3(2a-b=1)
a+3b=4
6a-3b=3
7a=7
a=1
1+3b=4
1-1+3b=4-1
3b=3
b=1
2007-06-07 21:37:03 · answer #5 · answered by Dave aka Spider Monkey 7 · 0⤊ 0⤋