Need help figuring this out.
A-6=2B
A+B=15
What is A & B
2007-06-29 17:22:55 · 12 answers · asked by JoAnn S 1 in Science & Mathematics ➔ Mathematics
to solve a pair of simultaneous equations you
1. make A or B the subject of either of the two equations
2. substitute this value for A or B into the other equation
it is best demonstrated and will make more sense with an example.
so with your question
A-6=2B ===> equation 1
A+B=15 ===> equation 2
** Step 1. make A the subject of equation 1
A = 2B + 6
**Step 2. substitute A= 2B + 6 into equation 2
equation 2 is A + B = 15. so replacing A with 2B + 6 you get
2B + 6 + B = 15
3B + 6 = 15
3B = 9
.: B = 3
now that you know that B = 3 you substitute this value into either of the equations 1 or 2 to find A.
so substituting B = 3 into equation 2 you get
A + 3 = 15
.: A = 12
--------------
therefore the solution is A = 12, B = 3
2007-06-29 17:55:38 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Write each equation in standard form:
A - 2B = 6
A + B = 15
multiply each term of the second equation by -1 so that the coefficients of the A terms are opposites
A - 2B = 6
-A - B = -15
---------------- add the equations to eliminate A
0 - 3B = -9
-3B = -9
B = 3
If B = 3 and A + B = 15, then
A + 3 = 15
A = 12
Answer: A = 12 and B = 3
2007-06-30 00:37:58 · answer #2 · answered by mathjoe 3 · 0⤊ 0⤋
A-6=2B can be rewritten as-
A-2B = 6
A+B=15
To solve a pair of equations you need to eliminate one variable. If you multiply the 2nd equation by 2 you will get-
2A+2B=30
Now you can add to two equations to eliminate the 2B terms-
A -2B = 6
2A+2B = 30 yields
3A = 36
so A =12
Substitute 12 for A in one of the original equations-
12 - 6 = 2B
6=2B
B=3
2007-06-30 00:33:22 · answer #3 · answered by skipper 7 · 0⤊ 0⤋
Apply the formulas for a 2-system simultaneous equations where the first equation and the second equation is as follows.
a1x + b1y = c1
a2x + b2y = c2
To solve for x and y,
x = (c1b2 - b1c2)/(a1b2 - b1a2)
y = (a1c2 - c1a2)/(a1b2 - b1a2)
In this equation,
A - 6 = 2B --- (1)
A + B = 15 --- (2)
From (1) A - 2B = 6 -- (3)
x = [(15)(-2) - (1)(6)]/[(1)(-2) - (1)(1)]
x = (-36)/(-3)
x = 12
y = [(1)(6) - (15)(1)]/[(1)(-2) - (1)(1)]
y = (-9)/(-3)
y = 3
2007-06-30 00:46:16 · answer #4 · answered by Anonymous · 0⤊ 0⤋
A=12
B=3
2007-06-30 00:34:55 · answer #5 · answered by aeidensmommy 3 · 0⤊ 0⤋
The easiest approach is to sub in equation 2 into equation 1 for "A" in terms of "B" , A= 15-B
Then, we have 15-B-6=2B
We can put the B-term on the right of the "="
9 = 3B and B=3
So A=12
Is that so hard?
2007-06-30 00:29:11 · answer #6 · answered by cattbarf 7 · 0⤊ 0⤋
A = 2B + 6
2B + 6 + B = 15
3B + 6 = 15
3B = 9
B = 3
A = 2*3 + 6
A = 12
2007-06-30 00:26:10 · answer #7 · answered by therealchuckbales 5 · 0⤊ 0⤋
1st equation (finding A):
A - 6 = 2B
A = 2B + 6
2nd equation (finding B, substitute A):
(2B + 6) + B = 15
2B + 6 + B = 15
3B = 15 - 6
3B = 9
B = 9 / 3
B = 3
1st equation (substitute B):
A = (2 * 3) + 6
A = 6 + 6
A = 12
2007-07-03 11:29:06 · answer #8 · answered by Jun Agruda 7 · 2⤊ 0⤋
A + B = 15 or
2 (A + B) = 2(15)
2A + 2B = 30 ..........eqn 1
A - 6 = 2B
A - 2B = 6........................eqn 2
Add eqn 1 and eqn 2
2A + 2B = 30
A - 2B = 6
2A + A +2B - 2B = 30 + 6
3A = 36
A = 12
A + B = 15
B = 15 - A
B = 15 - 12
B = 3
Therefore A = 12 and B = 3
2007-06-30 00:30:07 · answer #9 · answered by Sam 3 · 0⤊ 0⤋
2B = A- 6
therefore A = 2B+6
By substitution
2B+6+B = 15
3B = 9
B = 3
And, A = 15 - 3 = 12
2007-06-30 00:27:36 · answer #10 · answered by Sgcray 2 · 0⤊ 0⤋