
Solve the following equation
\[5\left( {2x - 3} \right) - 3\left( {3x - 7} \right) = 5\]
Answer
433.8k+ views
Hint: Here we need to find the value of ‘x’ which satisfies the given equation. Now we need to expand the brackets on the left-hand side of the equation. Then we need to add or subtract the like terms and constant terms on the left-hand side of the equation. Then to find the value of ‘x’ we use the transposition method.
Complete step-by-step solution:
Given, \[5\left( {2x - 3} \right) - 3\left( {3x - 7} \right) = 5\].
Now expand the brackets in LHS of the equation.
\[10x - 15 - 9x + 21 = 5\]
Now subtracting the like terms and constant on the left hand side of the equation we have,
\[x + 6 = 5\]
Thus we have a linear equation with one variable.
We transpose \[6\] which is present in the left-hand side of the equation to the right-hand side of the equation by Subtracting \[6\] on the right-hand side of the equation.
\[x = 5 - 6\]
\[ \Rightarrow x = - 1\].
This is the required answer.
Note: By simplifying we have obtained the answer for ‘x’. We can check whether the obtained value of ‘x’ is correct or not. To check we simply substitute the obtained value of ‘x’ in the given problem. If L.H.S is equal to R.H.S. then our answer is correct.
\[5\left( {2x - 3} \right) - 3\left( {3x - 7} \right) = 5\]
\[\Rightarrow 5\left( {2( - 1) - 3} \right) - 3\left( {3( - 1) - 7} \right) = 5\]
\[\Rightarrow 5\left( { - 2 - 3} \right) - 3\left( { - 3 - 7} \right) = 5\]
\[\Rightarrow 5\left( { - 5} \right) - 3\left( { - 10} \right) = 5\]
\[\Rightarrow - 25 + 30 = 5\]
\[ \Rightarrow 5 = 5\]
Hence the obtained answer is correct.
In the above, we did the transpose of addition and subtraction. Similarly, if we have multiplication, we use division to transpose. If we have division, we use multiplication to transpose. Follow the same procedure for these kinds of problems.
Complete step-by-step solution:
Given, \[5\left( {2x - 3} \right) - 3\left( {3x - 7} \right) = 5\].
Now expand the brackets in LHS of the equation.
\[10x - 15 - 9x + 21 = 5\]
Now subtracting the like terms and constant on the left hand side of the equation we have,
\[x + 6 = 5\]
Thus we have a linear equation with one variable.
We transpose \[6\] which is present in the left-hand side of the equation to the right-hand side of the equation by Subtracting \[6\] on the right-hand side of the equation.
\[x = 5 - 6\]
\[ \Rightarrow x = - 1\].
This is the required answer.
Note: By simplifying we have obtained the answer for ‘x’. We can check whether the obtained value of ‘x’ is correct or not. To check we simply substitute the obtained value of ‘x’ in the given problem. If L.H.S is equal to R.H.S. then our answer is correct.
\[5\left( {2x - 3} \right) - 3\left( {3x - 7} \right) = 5\]
\[\Rightarrow 5\left( {2( - 1) - 3} \right) - 3\left( {3( - 1) - 7} \right) = 5\]
\[\Rightarrow 5\left( { - 2 - 3} \right) - 3\left( { - 3 - 7} \right) = 5\]
\[\Rightarrow 5\left( { - 5} \right) - 3\left( { - 10} \right) = 5\]
\[\Rightarrow - 25 + 30 = 5\]
\[ \Rightarrow 5 = 5\]
Hence the obtained answer is correct.
In the above, we did the transpose of addition and subtraction. Similarly, if we have multiplication, we use division to transpose. If we have division, we use multiplication to transpose. Follow the same procedure for these kinds of problems.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Trending doubts
List some examples of Rabi and Kharif crops class 8 biology CBSE

How many ounces are in 500 mL class 8 maths CBSE

Name the states through which the Tropic of Cancer class 8 social science CBSE

a Define matter Give four examples of matter b What class 8 chemistry CBSE

How many ten lakhs are in one crore-class-8-maths-CBSE

In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
