To solve this system of equations, we can use either substitution or elimination method.
Substitution method:
From equation (1), we can solve for A in terms of b:
A = 13 - 2b
Now we can substitute this expression for A in equation (2):
2a - 3v = 5
2(13 - 2b) - 3v = 5
26 - 4b - 3v = 5
-4b - 3v = -21
This is a linear equation in two variables (b and v), so we can solve for one variable in terms of the other:
-4b = 3v - 21
b = (-3/4)v + (21/4)
Now we can substitute this expression for b back into equation (1) to solve for A:
A + 2b = 13
A + 2((-3/4)v + (21/4)) = 13
A - (3/2)v + 21 = 13
A - (3/2)v = -8
A = (3/2)v - 8
So the solution to this system of equations is:
A = (3/2)v - 8
b = (-3/4)v + (21/4)
Elimination method:
To use the elimination method, we want to eliminate one variable from the system by multiplying one or both equations by constants so that the coefficients of the variable are opposite (i.e. one positive and one negative).
In this case, we can eliminate b by multiplying equation (1) by 2 and equation (2) by 1:
2A + 4b = 26
2a - 3v = 5
Now we can add the two equations together to eliminate b:
2A + 2a + 4b - 3v = 31
Simplifying:
2(A + a) + 4b - 3v = 31
2(A + a) + (4b - 3v) = 31
2(A + a) = 31 - (4b - 3v)
2(A + a) = 31 - 4b + 3v
This is another linear equation in two variables (A and v), so we can solve for one variable in terms of the other:
2(A + a) = -4b + 3v + 31
A + a = (-2/2)b + (3/2)v + (31/2)
A + a = (-1/2)b + (3/2)v + 15.5
Now we can substitute this expression for A + a back into equation (1) to solve for A in terms of v:
A + 2b = 13
(A + a) + b = 13
(-1/2)b + (3/2)v + 15.5 + b = 13
(-1/2)b + b = (3/2)v - 2.5
(1/2)b = (3/2)v - 2.5
b = 3v/2 - 5
Substituting this expression for b back into equation (1):
A + 2(3v/2 - 5) = 13
A + 3v - 10 = 13
A + 3v = 23
A = -3v + 23
So the solution using the elimination method is:
A = -3v + 23
b = 3v/2 - 5
Find the simultaneous linear of the equation
A+2b=13........(1)
2a-3v=5 ..........(2)
1 answer