Achemical company makes two brands of antifreeze. the first brand is 60% pure antifreeze, and the second brand is 85% pure antifreeze. in order to obtain 110 gallons of a mixture that contains 75% pure antifreeze, how many gallons of each brand of antifreeze must be used

Let’s say we have x gallons of brand 1 antifreeze and y gallons of brand 2 antifreeze. We know we need a total of 140 gallons, so one equation to relate the variables would be x+y=140. We also know that our final mixture should have 60% antifreeze, so another equation would be the weighted average of the two brands: .55x+.80y140=.60. (The weighted average is basically multiplying each % (55 and 80) by the number of gallons you have (x and y

) and adding those together, and dividing that by the total number of gallons (140). This equals the % of the final mixture (60).)

Now we have a system of two equations that we can solve.

x+y=140

.55x+.80y140=.60

*rearrange first equation to solve for y

:

y=140−x

*substitute this value in for y

in the second equation:

.55x+.80(140−x)140=.60

*use algebra and solve for x

:

The algebra shouldn’t be too complex, and I’m hoping you’re asking this about the setup rather than the actual algebra, and I’m lazy, and I used a calculator to solve this, and this is probably a long run-on sentence, and I got x=112

.

*plug this in to first equation and solve for y

:

y=140−112=28

x=112,y=28

112 gallons of brand 1 antifreeze, 28 gallons of brand 2 antifreeze. That’s a whole lot of antifreeze.

thanx hey

a. (14,375+570)(5.3) b. 158,417$

step-by-step explanation:

a. you are basically just setting up the equation. f= amount of debt hence 14,375. g= how much she is borrowing from her parents weekly hence 570 and t= the interest on her debt. creating the equation.

b.   14,375+570= 14,945

      14,945 x 5.3 = 79,208.5

      79,208.5 x 2(years) = 158,417(answer)