Three Linear Equations with three variables

Question

A chemist has three different acid solutions. The first acid solution contains 25% acid, the second contains 45%,

and the third contains 65%. He wants to use all three solutions to obtain a mixture of 238 liters containing 45%

acid, using 3 times as much of the 65% solution as the 45% solution. How many liters of each solution should be used?

How to set up as three linear equations to find the answer?

Answer 1

X = liters of 65%

X=3Y where Y = liters of 45%

Z= liters of 25%

238 = X+Y+Z

=3Y+Y+Z

=4Y + Z

Z=238-4Y

25Z+45Y+65Z=238(45)

Those are the 3 linear equations

25Z+45Y+65X = 10,710 (238x45)

or

.25Z +.45Y+.65X = 107.1 (percent x liters + percent x liters = percent x liters)

Z + Y + X = 238 (liters + liters + liters = total liters

-3Y + X = 0 X = 3Y (65% soluition is 3 times 45% solution)

You could solve it with matrix algebra

or just multiply and equation by a constant and subtract from another equation to eliminate 2 variables and solve for the 3rd

then substitute that value into the equations reducing them to 2 equations 2 unknowns

and solve them either by matrix algebra or the substitution method