A chemist is mixing 45% salt solution with a 30% salt solution to make 60 L of a new solution that will contain 40% salt. How much of each of the original solutions should the chemist use?

30 ml of 45% salt solution and 20 ml of 30% salt solution.

Step-by-step explanation:

Let x be the amount of 45% salt solution and y be the amount of 30% salt solution

the two equation will be:

x+y=60 -- A

0.45x+0.3y=0.4(60) -- B

Solve by substitution

Rewrite Equation A

x+y=60

x=60-y --- C

Substitute C into B

0.45(60-y)+0.3y=24

 27-0.45y+0.3y=24

                -0.15y=-3

                        y=20

Substitute the y=20 into equation C

x=60-20

x=30

Therefore, there was 30 ml of 45% salt solution and 20 ml of 30% salt solution.

{x = -1/2 ,   y = -1

step-by-step explanation:

solve the following system:

{2 x + y = -2

{10 x - 3 y = -2

in the first equation, look to solve for y:

{2 x + y = -2

{10 x - 3 y = -2

subtract 2 x from both sides:

{y = -2 x - 2

{10 x - 3 y = -2

substitute y = -2 x - 2 into the second equation:

{y = -2 x - 2

{10 x - 3 (-2 x - 2) = -2

10 x - 3 (-2 x - 2) = (6 x + 6) + 10 x = 16 x + 6:

{y = -2 x - 2

{16 x + 6 = -2

in the second equation, look to solve for x:

{y = -2 x - 2

{16 x + 6 = -2

subtract 6 from both sides:

{y = -2 x - 2

{16 x = -8

divide both sides by 16:

{y = -2 x - 2

{x = -1/2

substitute x = -1/2 into the first equation:

{y = -1

{x = -1/2

collect results in alphabetical order:

answer:   {x = -1/2 ,   y = -1

x = -5

step-by-step explanation: