
How do you solve \[\dfrac{{3r + 9}}{6} = \dfrac{{5r - 10}}{5}\] ?
Answer
537.9k+ views
Hint: Here we need to solve the equation to find the value of \[r\] . So this problem can be easily solved by isolating the \[r\] term on one side and constant on the other side. Now here firstly we will use the cross multiplication method and then we will expand the equation. Lastly by using elimination method and combining like terms we will solve for \[r\] .
Formula: For solving the above problem addition or subtraction is used for isolating variable terms on one side of the equation while to solve the variable cross multiplication or subtraction is used.
Complete step by step solution:
The question that we need to solve is \[\dfrac{{3r + 9}}{6} = \dfrac{{5r - 10}}{5}\]
So here we will firstly simplify the equation by cross multiplication and we get
\[(3r + 9) \times 5 = (5r - 10) \times 6\]
Now we will expand the term and the equation we get is
\[15r + 45 = 30r - 60\]
Then while keeping the equation balanced we will isolate the term \[r\] on one side and other terms on the other side by combining the like terms. So we get
\[15r = 105\]
Lastly we will simplify and after simplification we get
\[r = 7\]
So the value of \[r\] comes out to be \[7\]
So, the correct answer is “ \[7\] ”.
Note: In order for solving such types of problems in a better way we need to simplify each side of the equation by removing parentheses and combining like terms.
We can solve such problems by making everything to zero. Remember that to remove the number we need to decide how to remove it and should do opposite of what is currently done. It means if the number is added we will remove the number and if the number is multiplied then by keeping the equation balanced we will divide to remove the number. Simplify the equation till the value of \[r\] is obtained.
Formula: For solving the above problem addition or subtraction is used for isolating variable terms on one side of the equation while to solve the variable cross multiplication or subtraction is used.
Complete step by step solution:
The question that we need to solve is \[\dfrac{{3r + 9}}{6} = \dfrac{{5r - 10}}{5}\]
So here we will firstly simplify the equation by cross multiplication and we get
\[(3r + 9) \times 5 = (5r - 10) \times 6\]
Now we will expand the term and the equation we get is
\[15r + 45 = 30r - 60\]
Then while keeping the equation balanced we will isolate the term \[r\] on one side and other terms on the other side by combining the like terms. So we get
\[15r = 105\]
Lastly we will simplify and after simplification we get
\[r = 7\]
So the value of \[r\] comes out to be \[7\]
So, the correct answer is “ \[7\] ”.
Note: In order for solving such types of problems in a better way we need to simplify each side of the equation by removing parentheses and combining like terms.
We can solve such problems by making everything to zero. Remember that to remove the number we need to decide how to remove it and should do opposite of what is currently done. It means if the number is added we will remove the number and if the number is multiplied then by keeping the equation balanced we will divide to remove the number. Simplify the equation till the value of \[r\] is obtained.
Recently Updated Pages
Complete reduction of benzene diazonium chloride with class 12 chemistry CBSE

How can you identify optical isomers class 12 chemistry CBSE

The coating formed on the metals such as iron silver class 12 chemistry CBSE

Metals are refined by using different methods Which class 12 chemistry CBSE

What do you understand by denaturation of proteins class 12 chemistry CBSE

Assertion Nitrobenzene is used as a solvent in FriedelCrafts class 12 chemistry CBSE

Trending doubts
What is BLO What is the full form of BLO class 8 social science CBSE

What are the 12 elements of nature class 8 chemistry CBSE

Citizens of India can vote at the age of A 18 years class 8 social science CBSE

Full form of STD, ISD and PCO

Convert 40circ C to Fahrenheit A 104circ F B 107circ class 8 maths CBSE

Advantages and disadvantages of science

