Hello from MrBillDoesMath!
Answer:
x = ( -9 + sqrt(33) ) /4
and
x = ( -9 - sqrt(33) ) /4
Discussion:
Using the quadratic formula with a = 2, b = 9 and c = 6 gives
x = ( -b +\- sqrt(b^2-4ac) )/2a
x = ( -9 +\- sqrt(9^2 - 4*2*6)) / (2*2) =>
x = ( -9 +\- sqrt (81- 48)) / 4 =>
x = ( -9 +\- sqrt(33) ) /4
Thank you,
MrB
Hello,
The answer is <u>3x² - x - 2</u>
<h3>Step-by-step explanation:-</h3>
<u>Step 1</u> : Apply distributive property by multiplying each term of " x - 1 " by each term of 3x + 2 :-
→ 3x² + 2x - 3x - 2
<u>Step 2</u> : Combine 2x and -3x to get -x :-
→ <u>3x² - x - 2</u>
✍️ <em>B</em><em>y</em><em> </em><em>Benjemin</em> ☺️
well, let's say that the first solution is 7 liters and has "x" percent of acid, how much acid total in it? well, 7x, <u>keeping in mind that "x" is a decimal form</u>.
likewise, the second solution is 3 liters and has 15% acid, so how much acid is there in that one? (15/100) * 3 = 0.45, so
![\begin{array}{lcccl} &\stackrel{solution}{quantity}&\stackrel{\textit{\% of }}{amount}&\stackrel{\textit{liters of }}{amount}\\ \cline{2-4}&\\ \textit{1st solution}&7&x&7x\\ \textit{2nd solution}&3&0.15&0.45\\ \cline{2-4}&\\ mixture&10&0.29&2.9 \end{array}~\hfill \implies 7x~~ + ~~0.45~~ = ~~2.9 \\\\[-0.35em] ~\dotfill\\\\ 7x=2.45\implies x=\cfrac{2.45}{7}\implies x=0.35~\hfill \stackrel{\textit{converting to \%}}{0.35\cdot 100}\implies \stackrel{\%}{35}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Blcccl%7D%20%26%5Cstackrel%7Bsolution%7D%7Bquantity%7D%26%5Cstackrel%7B%5Ctextit%7B%5C%25%20of%20%7D%7D%7Bamount%7D%26%5Cstackrel%7B%5Ctextit%7Bliters%20of%20%7D%7D%7Bamount%7D%5C%5C%20%5Ccline%7B2-4%7D%26%5C%5C%20%5Ctextit%7B1st%20solution%7D%267%26x%267x%5C%5C%20%5Ctextit%7B2nd%20solution%7D%263%260.15%260.45%5C%5C%20%5Ccline%7B2-4%7D%26%5C%5C%20mixture%2610%260.29%262.9%20%5Cend%7Barray%7D~%5Chfill%20%5Cimplies%207x~~%20%2B%20~~0.45~~%20%3D%20~~2.9%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%207x%3D2.45%5Cimplies%20x%3D%5Ccfrac%7B2.45%7D%7B7%7D%5Cimplies%20x%3D0.35~%5Chfill%20%5Cstackrel%7B%5Ctextit%7Bconverting%20to%20%5C%25%7D%7D%7B0.35%5Ccdot%20100%7D%5Cimplies%20%5Cstackrel%7B%5C%25%7D%7B35%7D)
Elimination is used when there is a common factor between the two equations. Substitution is used in cases where finding a common term just isn’t possible, or too complicated.
