Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

How do you solve \[6 - 5\left( {2x - 3} \right) = 4x + 7\] ?

seo-qna
Last updated date: 23rd Jul 2024
Total views: 383.7k
Views today: 7.83k
Answer
VerifiedVerified
383.7k+ views
Hint: Here in this question, the given equation is the linear equation with the equation with variable x on both sides. To solve this equation first keep x to the one side of the equation by using a distributive property and further simplify by using basic mathematics operations to get the required solution.

Complete step by step solution:
To solve this linear equation with variables on both sides. The most important thing to remember when solving such equations is that whatever you do to one side of the equation, you must do to the other side. Using this rule, it is easy to move variables around so that you can isolate them and use basic operations to find their value.
Now consider the given linear equation
\[ \Rightarrow 6 - 5\left( {2x - 3} \right) = 4x + 7\]
Firstly, remove bracket in the equation using distributive property
\[ \Rightarrow 6 - 10x + 15 = 4x + 7\]
On simplification we get
\[ \Rightarrow 21 - 10x = 4x + 7\]
To isolate coefficients of x variables on one side of the equations by adding 10x on both sides.
\[ \Rightarrow 21 - 10x + 10x = 4x + 7 + 10x\]
On further simplification we get
\[ \Rightarrow 21 = 14x + 7\]
Subtract 7 on both side of equations, then
\[ \Rightarrow 21 - 7 = 14x + 7 - 7\]
\[ \Rightarrow 14 = 14x\]
To solve the equation for x. Divide both side of equation by 14, then
\[ \Rightarrow 1 = x\]
\[\therefore x = 1\]
Hence, the value of x in the given linear equation \[6 - 5\left( {2x - 3} \right) = 4x + 7\] is \[x = 1\] .
So, the correct answer is “ x = 1”.

Note: While solving the equation we shift or transform the terms either from LHS to RHS or from RHS to LHS we should take care of the sign. Because while shifting or transforming the terms the sign of the term will change. If we miss out the sign we may go wrong while finding the variable or solving.