<span>1/root(5-x) = 1/(root5 * root 1- x/5)
1/root5 * (1-x/5)^-1/2
Using the 1st 2 terms of the binomial expansion 1 +nx:
1/root5 ( 1 + x/10)
= X/10root5 + 1/root5</span>
We label each equation
3x + 4y = 5 (1)
2x – 3y = 9 (2)
We now want to get rid of one of the variables (x or y). Lets get rid of x:
We need (1) and (2) to have the same number of x's so we multiple (1) by 2 and (2) by 3 so they both have 6 x's
6x + 8y = 10 (1)*2
6x - 9y = 27 (2)*3
Now to get rid of the x's we take one of the equations from the other. It is easier to do (2)*3 -(1)*2
6x - 9y = 27 (2)*3
6x + 8y = 10 (1)*2
-17y = 17
If we devide this equation by -17 we get
y = -1
We can plug this value of y into (1) to get
3x -4 = 5
add 4 to both sides of the equation
3x = 9
Divide by 3
x = 3
Now to check put x and y into (2)
6 - (-3) = 9
As this is true, we have the solution
x = 3, y = -1
Answer:
38.3 hours
Step-by-step explanation:
Given data
We have been given the function that describes his pay, which is
y=6.50h
Where h is the number of hours
now we have the amount y= 249
substitute
249=6.50*h
make the subject of the formula
249/6.5=h
h=38.3
Hence he needs to work for at least 38.3 hours
