The answer to your question is -3 !!
The last 2 options I belive
So first we distribute
1/3(9a-18)
a three can be factored out of (9a-18)=3(3a-6)
so 1/3 times 3(3a-6)=3a-6
so the equation end up as 6-a=3a-6
add 6 to both sides
12-a=3a
add a to both sides
12=4a
divide both sides by 4
3=a
so there is one solution or A
440 gallons of 80 % antifreeze solution must be mixed with 80 gallons of 15% antifreeze to get a mixture that is 70% antifreeze
<em><u>Solution:</u></em>
Let "x" be the gallons of 80 % antifreeze added
Therefore, "x" gallons of 80 % antifreeze solution must be mixed with 80 gallons of 15% antifreeze
Final mixture is x + 80
Therefore, we can frame a equation as:
"x" gallons of 80 % antifreeze solution must be mixed with 80 gallons of 15% antifreeze to get (x + 80) gallons of 70 % antifreeze
Thus, we get,
x gallons of 80 % + 80 gallons of 15 % = (x + 80) gallons of 70 %
![x \times \frac{80}{100} + 80 \times \frac{15}{100} = (x+80) \times \frac{70}{100}\\\\0.8x + 80 \times 0.15 = (x+80) \times 0.7\\\\0.8x+12 = 0.7x+56\\\\0.8x-0.7x=56-12\\\\0.1x = 44\\\\x = \frac{44}{0.1}\\\\x = 440](https://tex.z-dn.net/?f=x%20%5Ctimes%20%5Cfrac%7B80%7D%7B100%7D%20%2B%2080%20%5Ctimes%20%5Cfrac%7B15%7D%7B100%7D%20%3D%20%28x%2B80%29%20%5Ctimes%20%5Cfrac%7B70%7D%7B100%7D%5C%5C%5C%5C0.8x%20%2B%2080%20%5Ctimes%200.15%20%3D%20%28x%2B80%29%20%5Ctimes%200.7%5C%5C%5C%5C0.8x%2B12%20%3D%200.7x%2B56%5C%5C%5C%5C0.8x-0.7x%3D56-12%5C%5C%5C%5C0.1x%20%3D%2044%5C%5C%5C%5Cx%20%3D%20%5Cfrac%7B44%7D%7B0.1%7D%5C%5C%5C%5Cx%20%3D%20440)
Thus 440 gallons of 80 % antifreeze solution must be mixed
5x 7 tens = 35 tens is your answer