
How do you solve x in \[ax + b = 0\] ?
Answer
543.9k+ views
Hint: The given equation is the linear equation with one variable x that can be solve by add or subtract the necessary term from each side of the equation to isolate the term with the variable x, then multiply or divide each side of the equation by the appropriate value solve for the variable x while keeping the equation balanced then solve the resultant balance equation for the x value.
Complete step-by-step answer:
The equation is an algebraic equation and it contains only one variable. The variable is x. The variable acts as a constant. So we solve for the variable x.
Consider given equation
\[ \Rightarrow \,\,\,ax + b = 0\]
Where x is the variable a is the coefficient of x and b is the constant term
Rearrange the equation by subtracting b from both sides of the equation, then
\[ \Rightarrow \,\,\,\,\,ax + b - b = 0 - b\]
On subtracting we get
\[ \Rightarrow \,\,\,ax + 0 = - b\]
So it is written as
\[ \Rightarrow \,\,\,ax = - b\]
To solve the above equation for x, divide both side by a, then
\[ \Rightarrow \,\,\,\dfrac{{ax}}{a} = \dfrac{{ - b}}{a}\]
On rewriting we have
\[ \Rightarrow \,\,\,\dfrac{a}{a}.\,x = \dfrac{{ - b}}{a}\]
As we know the value of \[\dfrac{a}{a}\] is 1
\[\therefore \,\,\,\,\,x = \dfrac{{ - b}}{a}\]
Hence, by solving the value of variable x in the linear equation \[ax + b = 0\] is \[\dfrac{{ - b}}{a}\] .
Hence we have solved for x and obtained the solution.
Therefore \[x = \dfrac{{ - b}}{a}\]
So, the correct answer is “ \[x = \dfrac{{ - b}}{a}\] ”.
Note: In general the algebraic equation or expression is defined as a combination of variables and constants. In some equations the variables act as a constant, in this question also the a and b are variable but we consider it as a constant. While transforming or shifting the terms from LHS to RHS or RHS to LHS the sign of the term will change and we should take care of it.
Complete step-by-step answer:
The equation is an algebraic equation and it contains only one variable. The variable is x. The variable acts as a constant. So we solve for the variable x.
Consider given equation
\[ \Rightarrow \,\,\,ax + b = 0\]
Where x is the variable a is the coefficient of x and b is the constant term
Rearrange the equation by subtracting b from both sides of the equation, then
\[ \Rightarrow \,\,\,\,\,ax + b - b = 0 - b\]
On subtracting we get
\[ \Rightarrow \,\,\,ax + 0 = - b\]
So it is written as
\[ \Rightarrow \,\,\,ax = - b\]
To solve the above equation for x, divide both side by a, then
\[ \Rightarrow \,\,\,\dfrac{{ax}}{a} = \dfrac{{ - b}}{a}\]
On rewriting we have
\[ \Rightarrow \,\,\,\dfrac{a}{a}.\,x = \dfrac{{ - b}}{a}\]
As we know the value of \[\dfrac{a}{a}\] is 1
\[\therefore \,\,\,\,\,x = \dfrac{{ - b}}{a}\]
Hence, by solving the value of variable x in the linear equation \[ax + b = 0\] is \[\dfrac{{ - b}}{a}\] .
Hence we have solved for x and obtained the solution.
Therefore \[x = \dfrac{{ - b}}{a}\]
So, the correct answer is “ \[x = \dfrac{{ - b}}{a}\] ”.
Note: In general the algebraic equation or expression is defined as a combination of variables and constants. In some equations the variables act as a constant, in this question also the a and b are variable but we consider it as a constant. While transforming or shifting the terms from LHS to RHS or RHS to LHS the sign of the term will change and we should take care of it.
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