How do you solve equations that contain multiplication or division

2 answers:

6 0

You multiply or divide, depending on what is contained in the equation. But if there is addition or subtraction in the equation as well, then you do that too. But the order you do it in all depends on how the equation is written, what you're solving for, whether there are parentheses or not, and if there are, where they are placed.
If there's a particular problem you're stuck on that involves multiplication/division, feel free to post it or pm me about it and I'll try to help out.

kodGreya [7K]3 years ago

5 0

Always perform the same operation to both sides of the equation. When you multiply or divide, you have to multiply and divide by the entire side of the equation. Try to perform addition and subtraction first to get some multiple of x by itself on one side.

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what are two equivalent expressions to the following radical expression: (√10 to the 4th power x √12 to the 4th power) to the 8t

brilliants [131]

Answer:

(14400x^4)^8

Step-by-step explanation:

sqrt10 to the 4th is 10^2 or 100, and xsqrt12 to the 4th is 144x^4. Multiply these to get (14400x^4), and take the 8th to get (14400x^4)^8.

3 0

3 years ago

How many integers in the set {n ∈ Z | 1 ≤ n ≤ 700} are divisible by 2 or 7?

sukhopar [10]

$\large\begin{array}{l}\\\\ \textsf{This question gives us a set}\\\\ \mathsf{S=\{n \in\mathbb{Z}:~1\le n\le 700\}}\\\\ \mathsf{S=\{1,\,2,\,3,\,\ldots,\,699,\,700\}}\\\\\\ \bullet~~\textsf{Set of integers that are divible by 2 (even integers):}\\\\ \mathsf{A=\{n\in \mathbb{Z}:~n=2k,\,k\in\mathbb{Z}\}}\\\\ \mathsf{A=\{\ldots,\,-4,\,-2,\,0,\,2,\,4,\,\ldots\}}\\\\\\ \bullet~~\textsf{Set of integers that are divible by 7:}\\\\ \mathsf{B=\{n\in \mathbb{Z}:~n=7k,\,k\in\mathbb{Z}\}}\\\\ \mathsf{A=\{\ldots,\,-14,\,-7,\,0,\,7,\,14,\,\ldots\}} \end{array}$

___________

$\large\begin{array}{l}\\\\ \textsf{We want to know how many elements there are in the}\\\textsf{following set:}\\\\ \mathsf{S\cap (A\cup B)=(S\cap A)\cup(S\cap B)\qquad(i)} \end{array}$

____________

$\large\begin{array}{l}\\\\ \bullet~~\mathsf{S\cap A=\{n\in\mathbb{N}:~n=2k~~and~~1\le n\le 700,\,k\in\mathbb{Z}\}}\\\\ \mathsf{S\cap A=\{2,\,4,\,6,\,\ldots,\,698,\,700\}}\\\\ \mathsf{S\cap A=\{1\cdot 2,\,2\cdot 2,\,3\cdot 2,\,\ldots,\,349\cdot 2,\,350\cdot 2\}}\\\\\\ \textsf{So, there are 350 elements in }\mathsf{S\cap A:}\\\\ \mathsf{\#(S\cap A)=350.} \\\\\\ \bullet~~\mathsf{S\cap B=\{n\in\mathbb{N}:~n=7k~~and~~1\le n\le 700,\,k\in\mathbb{Z}\}}\\\\ \mathsf{S\cap B=\{7,\,14,\,21,\,\ldots,\,693,\,700\}}\\\\ \mathsf{S\cap B=\{1\cdot 7,\,2\cdot 7,\,3\cdot 7,\,\ldots,\,99\cdot 7,\,100\cdot 7\}} \\\\\\ \textsf{So, there are 100 elements in }\mathsf{S\cap B:}\\\\ \mathsf{\#(S\cap B)=100.} \end{array}$

____________

$\large\begin{array}{l}\\\\ \textsf{Therefore,}\\\\ \mathsf{\#\big[S\cap (A\cup B)\big]}\\\\ =\mathsf{\#\big[(S\cap A)\cup(S\cap B)\big]}\\\\ =\mathsf{\#(S\cap A)+\#(S\cap B)-\#\big[(S\cap A)\cap(S\cap B)\big]}\\\\ =\mathsf{350+100-50}\\\\ =\mathsf{450-50}\\\\ =\mathsf{400~elements.}\\\\\\ \textsf{There are 400 integers in S that are divisible by 2 or 7.} \end{array}$

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$\large\begin{array}{l}\\\\ \textsf{Any doubts? Please, comment below.}\\\\\\ \textsf{Best wishes! :-)} \end{array}$

Tags: set theory divibilility divisible integers union intersection

3 0

3 years ago

Pls help me

cricket20 [7]

Answer:

No solution because it is minus 9

6 0

2 years ago

Read 2 more answers

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WARRIOR [948]

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