
How do you solve \[4h + 6 = 30\] ?
Answer
551.7k+ views
Hint: We can first reorder the terms according to the question. After that we can solve the question for the variable \[h\]. We can group the like terms on one side so that it will be easier to solve. Then after forming the easier equation, we can simplify it, and we will get our answer.
Complete step by step solution:
According to the question, our equation is:
\[4h + 6 = 30\]
We will first reorder the terms for easy solving.
\[ \Rightarrow 6 + 4h = 30\]
Then we will solve for the variable \[h\]. We need to add \[ - 6\] in both the parts of the equation. We do this to cancel out the terms from left so that we can solve the equation for the variable \[h\] and get the answer:
\[ \Rightarrow 6 + ( - 6) + 4h = 30 + ( - 6)\]
When we simplify this, we get:
\[ \Rightarrow 6 - 6 + 4h = 30 - 6\]
\[ \Rightarrow 0 + 4h = 24\]
\[ \Rightarrow 4h = 24\]
Now, we have to divide \[4\]to both the terms. This is done to make the variable \[h\] alone. By dividing the terms by \[4\], we get:
\[ \Rightarrow \dfrac{{4h}}{4} = \dfrac{{24}}{4}\]
Here, the numbers which are divisible or are common get cancelled. The \[4\] on the numerator and the denominator gets cancelled from the left side. The number \[24\] gets cancelled from \[4\] leaving a remainder of \[6\], and then we get our answer as:
\[ \therefore h = 6\]
Our final answer here is \[h = 6\].
Note: The above method is very easy and the question is solved very quickly. But we have another method also. We can make the variable \[h\] alone by shifting \[6\]t o the other side. On that side \[6\] gets subtracted from \[30\] and the result is \[24\], and then shifts \[4\] also to the other side. The number \[4\] gets divided with a result that is \[24\] and the final answer is \[6\]. This method is also easy to solve.
Complete step by step solution:
According to the question, our equation is:
\[4h + 6 = 30\]
We will first reorder the terms for easy solving.
\[ \Rightarrow 6 + 4h = 30\]
Then we will solve for the variable \[h\]. We need to add \[ - 6\] in both the parts of the equation. We do this to cancel out the terms from left so that we can solve the equation for the variable \[h\] and get the answer:
\[ \Rightarrow 6 + ( - 6) + 4h = 30 + ( - 6)\]
When we simplify this, we get:
\[ \Rightarrow 6 - 6 + 4h = 30 - 6\]
\[ \Rightarrow 0 + 4h = 24\]
\[ \Rightarrow 4h = 24\]
Now, we have to divide \[4\]to both the terms. This is done to make the variable \[h\] alone. By dividing the terms by \[4\], we get:
\[ \Rightarrow \dfrac{{4h}}{4} = \dfrac{{24}}{4}\]
Here, the numbers which are divisible or are common get cancelled. The \[4\] on the numerator and the denominator gets cancelled from the left side. The number \[24\] gets cancelled from \[4\] leaving a remainder of \[6\], and then we get our answer as:
\[ \therefore h = 6\]
Our final answer here is \[h = 6\].
Note: The above method is very easy and the question is solved very quickly. But we have another method also. We can make the variable \[h\] alone by shifting \[6\]t o the other side. On that side \[6\] gets subtracted from \[30\] and the result is \[24\], and then shifts \[4\] also to the other side. The number \[4\] gets divided with a result that is \[24\] and the final answer is \[6\]. This method is also easy to solve.
Recently Updated Pages
Master Class 7 English: Engaging Questions & Answers for Success

Master Class 7 Maths: Engaging Questions & Answers for Success

Master Class 7 Science: Engaging Questions & Answers for Success

Class 7 Question and Answer - Your Ultimate Solutions Guide

Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Master Class 9 Social Science: Engaging Questions & Answers for Success

Trending doubts
What are the factors of 100 class 7 maths CBSE

The value of 6 more than 7 is A 1 B 1 C 13 D 13 class 7 maths CBSE

Convert 200 Million dollars in rupees class 7 maths CBSE

What were the major teachings of Baba Guru Nanak class 7 social science CBSE

Find the largest number which divides 615 and 963 leaving class 7 maths CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE


