Question

A lab assistant needs to create a 500 mL mixture that is 5% hydrochloric acid. The assistance has solutions of 3.5% and 6% in supply at the lab. Using the variables x and y to represent the number of milliliters of the 3.5% solution and the number of milliliters of the 6% solution respectively, determine a system of equations that describes the situation. Enter the equations below separated by a comma.How many milliliters of the 3.5% solution should be used? How many milliliters of the 6% solution should be used?

Answer 1

Let

x -----> the number of milliliters of the 3.5% solution

y ----> the number of milliliters of the 6% solution

we have that

3.5%=3.5/100=0.035

6%=6/100=0.06

5%=5/100=0.05

so

0.035x+0.06y=0.05(500) -------> equation A

and

x+y=500 -----> x=500-y ------> equation B

solve the system of equations

substitute equation B in equation A

0.035(500-y)+0.06y=0.05(500)

solve for y

17.5-0.035y+0.06y=25

0.025y=25-17.5

0.025y=7.5

y=300

Find out the value of x

x=500-300=200

therefore

the number of milliliters of the 3.5% solution is 200 ml

the number of milliliters of the 6% solution is 300 ml