Question 1111018
you have 2 equations that need to be solved simultaneously.


they are:


1/A + 1/B = 7


1/A - 1/B = 3


add the 2 equations together to get:


2/A = 10


solve for A to get A = 2/10 = 1/5


when A = 1/5, 1/A + 1/B = 7 becomes:


1/(1/5) + 1/B = 7


simplify to get 5 + 1/B = 7


subtract 5 from both sides to get 1/B = 2


solve for B to get B = 1/2


you have A = 1/5 and B = 1/2.


your first equation is 1/A + 1/B = 7


replace A with 1/5 and B with 1/2 to get:


1/(1/5) + 1/(1/2) = 7


simplify to get 5 + 2 = 7, which is true.


your second equaton is 1/A - 1/B = 3


replace A with 1/5 and B with 1/2 to get:


1/(1/5) - 1/(1/2) = 3


simplify to get 5 - 2 = 3, which is true.


both equations are true when A = 1/5 and B = 1/2, therefore the solution can be assumed to be good.


how do you get 5 out of 1/(1/5)?


it's a simple matter of multiplying both numerator and denominator by 5.


1/(1/5) * 5/5 = (5*1) / (5*1/5) = 5/1 = 5


you can do this without changing the fraction because multiplying by 5/5 is the same as multiplying by 1 which keeps the value of the original fraction the same.


to confirm, used your calculator to divide 1 by (1/5) and you will see that the answer is 5.


the same procedure to solve these simultaneous equations is used to solve simultaneous equations that do not include fractions.


since the first equation had + 1/B and the second equation had - 1/B, adding the equations together eliminated B from the equation and allowed you to solve for A.


once you solved for A, it was a matter of replacing A with it's value to solve for B in either of the original equations.


here's a reference on how to solve simultaneous equations if you think you might need it.


<a href = "http://www.purplemath.com/modules/systlin1.htm" target = "_blank">http://www.purplemath.com/modules/systlin1.htm</a>


here's a reference on how to add fractions with different denominators if you think you need it.


<a href= "http://www.purplemath.com/modules/fraction4.htm" target = "_blank">http://www.purplemath.com/modules/fraction4.htm</a>