
How do you solve $ 6x + 30 + 4x = 10\left( {x + 3} \right) $ ?
Answer
534k+ views
Hint: In order to determine the value of variable $ x $ in the above equation, first apply the distributive property of multiplication $ A\left( {B + C} \right) = AB + AC $ on the right-hand side. Combining like terms by Using the rules of transposing terms to transpose terms having $ x $ on the left-hand side and constant value terms on the right-Hand side of the equation. While combining terms, you will see no term having variable $ x $ is left, so it is not a value linear equation concluding that no solution exists for the given equation.
Complete step by step solution:
We are given a linear equation in one variable $ 6x + 30 + 4x = 10\left( {x + 3} \right) $ .and we have to solve this equation for variable ( $ x $ ).
$ \Rightarrow 6x + 30 + 4x = 10\left( {x + 3} \right) $
Applying distributive law of multiplication on right-side of the equation to resolve the parenthesis and expand the terms as $ A\left( {B + C} \right) = AB + AC $
$ \Rightarrow 6x + 30 + 4x = 10\left( x \right) + 10\left( 3 \right) $
Simplifying the equation, we get
$ \Rightarrow 6x + 30 + 4x = 10x + 30 $
Now combining like terms on both of the sides. Terms having $ x $ will on the Left-Hand side of the equation and constant terms on the right-hand side.
Let’s recall one basic property of transposing terms that on transposing any term from one side to another the sign of that term gets reversed .
After transposing terms our equation becomes
$
\Rightarrow 6x + 4x - 10x = 30 - 30 \\
\Rightarrow 10x - 10x = 30 - 30 \;
$
As you can see all the terms having $ x $ are getting cancelled. The above equation is not a valid linear equation. Therefore, no solution exists for the given equation
Note: Linear Equation: A linear equation is a equation which can be represented in the form of $ ax + c $ where $ x $ is the unknown variable and a,c are the numbers known where $ a \ne 0 $ .If $ a = 0 $ then the equation will become constant value and will no more be a linear equation .
The degree of the variable in the linear equation is of the order 1.
Every Linear equation has 1 root.
Complete step by step solution:
We are given a linear equation in one variable $ 6x + 30 + 4x = 10\left( {x + 3} \right) $ .and we have to solve this equation for variable ( $ x $ ).
$ \Rightarrow 6x + 30 + 4x = 10\left( {x + 3} \right) $
Applying distributive law of multiplication on right-side of the equation to resolve the parenthesis and expand the terms as $ A\left( {B + C} \right) = AB + AC $
$ \Rightarrow 6x + 30 + 4x = 10\left( x \right) + 10\left( 3 \right) $
Simplifying the equation, we get
$ \Rightarrow 6x + 30 + 4x = 10x + 30 $
Now combining like terms on both of the sides. Terms having $ x $ will on the Left-Hand side of the equation and constant terms on the right-hand side.
Let’s recall one basic property of transposing terms that on transposing any term from one side to another the sign of that term gets reversed .
After transposing terms our equation becomes
$
\Rightarrow 6x + 4x - 10x = 30 - 30 \\
\Rightarrow 10x - 10x = 30 - 30 \;
$
As you can see all the terms having $ x $ are getting cancelled. The above equation is not a valid linear equation. Therefore, no solution exists for the given equation
Note: Linear Equation: A linear equation is a equation which can be represented in the form of $ ax + c $ where $ x $ is the unknown variable and a,c are the numbers known where $ a \ne 0 $ .If $ a = 0 $ then the equation will become constant value and will no more be a linear equation .
The degree of the variable in the linear equation is of the order 1.
Every Linear equation has 1 root.
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