Prepare 350g of a 2% lidocaine cream using a 1% lidocaine cream a 5% lidocaine cream

A)    1% lidocaine -x g,      0.01x pure lidocaine 
        5% lidocaine -y g,      0.05y pure lidocaine
(1)0.01x+0.05y=350*0.02
0.01x  g we get from1% lidocoine and 0.05y g we get from 5% lidocoine

(2) x+y=350 
x g is mass 1% lidocoine, we combine it with y g 5%,
so we get x+y, and we need 350g,
so x+y should be equal 350, x+y=350

we get system of 2 equations that we need to solve to find x and y

(1)0.01x+0.05y=350*0.02
(2) x+y=350

we can solve it by substitution, find x from (2) equation
x=350-y, substitute it into  (1)
instead of x we put (350-y)
0.01(350-y)+0.05y=7
3.5-0.01y+0.05y=7
0.04y =3.5
y=87.5g 5% lidocaine

b)x=350-87.5=262.5 g 1% lidocaine
c) 350g =350000 mg

Define x and y:

Let x be the amount of 1% lidocaine needed.
Let y be the amount of 5% lidocaine needed.


Construct Equations:

x + y = 350

0.01x + 0.05y = 0.02 x 350
0.01x + 0.05y = 7


Solve x and y:

x + y = 350              (Equation 1)
0.01x + 0.05y = 7    (Equation 2)


From equation 1:

x + y = 350
x = 350 - y 


Substitute x = 350 - y into equation 2:

0.01(350 - y) + 0.05y = 7
3.5 - 0.01y + 0.05y = 7
3.5 + 0.04y = 7
0.04y = 7 - 3.5
0.04y = 3.5
y = 87.5


Substitute y = 87.5 into equation 1:

x + y = 350
x + 87.5 = 350
x = 350 - 87.5
x = 262.5


Find x and y:

x = 262.6g (5% lidocaine cream)
y = 87.5g   (1% lidocaine cream)


Convert 350g to milligram:

350g = 350 x 1000 mg
350g = 350 000mg



(a) Amount of 5% lidocaine needed = 87.5g
(b) Amount of 1% lidocaine needed = 262.g
(c) 350g = 350 000mg

The 8th term is 512