
How do you solve \[x + 5x + 4 = 3(2x - 1)\]?
Answer
534k+ views
Hint: In the given problem we need to solve this for ‘x’. We can solve this using the transposition method. The common transposition method is to do the same thing (mathematically) to both sides of the equation, with the aim of bringing like terms together and isolating the variable (or the unknown quantity). That is we group the ‘x’ terms one side and constants on the other side of the equation.
Complete step-by-step solution:
Given, \[x + 5x + 4 = 3(2x - 1)\].
Firstly we expand the brackets in right hand side of the equation,
\[ \Rightarrow x + 5x + 4 = 6x - 3\]
Adding like terms we have,
\[ \Rightarrow 6x + 4 = 6x - 3\]
We transpose ‘6x’ which is present in the right hand side of the equation to the left hand side of the equation by subtracting ‘6x’ on the left hand side of the equation.
\[ \Rightarrow 6x - 6x + 4 = - 3\]
That is we have,
\[ \Rightarrow 4 = - 3\]
Which is impossible, hence the given equation is not valid and it is inconsistent.
Hence, no solution.
Note: In 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. A consistent system of equations has at least one solution and an inconsistent system has no solutions.
Complete step-by-step solution:
Given, \[x + 5x + 4 = 3(2x - 1)\].
Firstly we expand the brackets in right hand side of the equation,
\[ \Rightarrow x + 5x + 4 = 6x - 3\]
Adding like terms we have,
\[ \Rightarrow 6x + 4 = 6x - 3\]
We transpose ‘6x’ which is present in the right hand side of the equation to the left hand side of the equation by subtracting ‘6x’ on the left hand side of the equation.
\[ \Rightarrow 6x - 6x + 4 = - 3\]
That is we have,
\[ \Rightarrow 4 = - 3\]
Which is impossible, hence the given equation is not valid and it is inconsistent.
Hence, no solution.
Note: In 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. A consistent system of equations has at least one solution and an inconsistent system has no solutions.
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