1. (5/9 + 2/3) + 1/4 + 5 = ?
2. (6)4/7 + (5/6 + 1/3) = ?
3. if x= 7, solve the equation 6(x+3) / 11(x+7) = ?
4. (6)2 + 4(17+5) - 50 - (36 - 10) = ?
5. 36 divided by 12 + 5 - (4)(6) + (16 - 8) = ?
I'll solve each one in a different comment.
#1
Simplify the expression \(\frac{5}{9}+\frac{2}{3}+\frac{1}{4}+5\).
\(\frac{5}{9}+\frac{2}{3}+\frac{1}{4}+5\) | Get every fraction in a form such that all the fractions have a common denominator. The LCD of 9,3, and 4 is 36. Therefore, out goal is to convert every fraction that it has a denominator of 36. |
\(\frac{5}{9}*\frac{4}{4}=\frac{20}{36}\) | |
\(\frac{2}{3}*\frac{12}{12}=\frac{24}{36}\) | |
\(\frac{1}{4}*\frac{9}{9}=\frac{9}{36}\) | Reinsert all of these fractions into the orginal expression. |
\(\frac{20}{36}+\frac{24}{36}+\frac{9}{36}+5\) | Add the fractions together. |
\(5\frac{53}{36}\) | Now, convert this mixed number into an improper fraction. |
\(5\frac{53}{36}=\frac{36*5+53}{36}=\frac{233}{36}\) | |
Time to do #2. Now, I will simplify the expression, as I interpret it to be, \(\frac{6*4}{7}+\left(\frac{5}{6}+\frac{1}{3}\right)\):
Of course, the parentheses are unnecessary in this expression as any order of adding is allowed by the associative property of addition. There is also one for multiplication.
\(\frac{6*4}{7}+\frac{5}{6}+\frac{1}{3}\) | Simplify the numerator of the leftmost fraction. |
\(\frac{24}{7}+\frac{5}{6}+\frac{1}{3}\) | The LCD in this case is 42, so convert all fractions to have this common denominator. |
\(\frac{24}{7}*\frac{6}{6}=\frac{20*6+4*6}{42}=\frac{144}{42}\) | |
\(\frac{5}{6}*\frac{7}{7}=\frac{35}{42}\) | |
\(\frac{1}{3}*\frac{14}{14}=\frac{14}{42}\) | Now, add the fractions together. |
\(\frac{144}{42}+\frac{35}{42}+\frac{14}{42}=\frac{193}{42}\) | |
For #3, this one contains some ambiguity in which interpretation you mean. You're question is one of the following.
1. \(f(7)=\frac{6(x+3)}{11}*\frac{(x+7)}{1}\)
2. \(f(7)=\frac{6(x+3)}{11(x+7)}\)
While I solve the next 2, please clarify which one you want me to do. In the future, use parentheses (or brackets) to disambiguate any mathematical input.
For example, if the interpretation is #1, use (6(x+3)/11)*(x+7)
For the second interpretation, use 6(x+3)/(11(x+7))
Thanks for responding quickly! I will evaluate it, to your choice, as the latter one of \(f(7)=\frac{6(x+3)}{11(x+7)}\).
\(f(7)=\frac{6(x+3)}{11(x+7)}\) | First, replace every instance of x with 7, as x=7 by the given information. |
\(\frac{6(7+3)}{11(7+7)}\) | Simplify within the parentheses first. |
\(\frac{6*10}{11*14}\) | Now, simplify both the numerator and the denominator. However, to make it easier computationally, notice how 14 and 10 have a common factor of 2, so we can factor it out, so we don't have to do 11*14, which some have not memorized yet. |
\(\frac{6*5}{11*7}\) | Now, simplify the numerator and denominator. |
\(\frac{30}{77}\) | |
#4
I will evaluate \(6^2+4(17+5)-50-(36-10)\) now
\(6^2+4(17+5)-50-(36-10)\) | First, evaluate inside of the parentheses first to adhere to the order of operations. |
\(6^2+4*22-50-26\) | Do any exponents next, in this case 6^2. |
\(36+4*22-50-26\) | Now, do the multiplication since that is prioritized above addition and subtraction. |
\(36+88-50-26\) | Now, compute from left to right. |
\(124-50-26\) | |
\(74-26\) | |
\(48\) | |
#5
I have the same issue as #3 because it can be subject to more than one interpretation, too.
When your use "divided by," do you mean that 36 is the numerator and the quantity of 12 + 5 - (4)(6) + (16 - 8) is the denominator. In other words, do you mean \(\frac{36}{12+5-(4)(6)+(16-8)}\)?
there was no divide sign so I just used that
36 (divide symbol) 12 + 5 - (4)(6) + (16 - 8)
its not a fraction
I understand, I think! I will simplify \(36\div12+5-(4)(6)+(16-8)\).
\(36\div12+5-(4)(6)+(16-8)\) | First, do what is inside of the parentheses. |
\(36\div12+5-4*6+8\) | Let's simplify both the division and multiplication now. |
\(3+5-24+8\) | Now, evaluate from left to right. |
\(8-24+8\) | |
\(-16+8\) | |
\(-8\) | |
It is possible that, after all of this solving, that you have a question. If you do, just ask! I (or any other active member of this forum) will be happy to answer. Don't hesitate.