Question

Solve for x: \[\dfrac{16}{x}-1=\dfrac{15}{x+1};x\ne 0,-1\]

Answer 1

Hint: In the above question, we have to find the value of variable x by solving the expression. At first, we will convert the fractional part into normal form by taking the LCM wherever needed and then doing the cross multiplication. Taking the like terms on one side, we will obtain two values of x for which the above equation holds.

Complete step-by-step solution:
This is a question related to algebra in mathematics. Algebra is the branch of mathematics that represents the problems in the form of mathematical expression and it involves variables, and also arithmetic operations like addition, subtraction, multiplication, and division. Algebra will be used in every branch of mathematics like trigonometry, calculus, coordinate geometry, etc. An algebraic expression is the collection of operators, variables, constants, and arithmetic operations together to form an equation.
In this question, we have to find the value of x in the expression which is shown below.
\[\dfrac{16}{x}-1=\dfrac{15}{x+1}\].
First, we will take the LCM of the denominator on the left-hand side of the equation.
\[\begin{align}
  & \dfrac{16}{x}-1=\dfrac{15}{x+1}, \\
 & \Rightarrow \dfrac{16-x}{x}=\dfrac{15}{x+1} \\
\end{align}\]
Now after taking the LCM, we will do the cross multiplication in the equation. The denominator on the left-hand side will be multiplied with the numerator on the right-hand side and the denominator on the right-hand side will be multiplied by the numerator on the left-hand side, which is as shown below.
\[(16-x)(x+1)=15x\]
Now in the above expression, first, \[16\] will be multiplied with \[x+1\] and then \[x\] will be multiplied with \[x+1\] which is as shown below.
\[\begin{align}
  & 16(x+1)-x(x+1)=15x, \\
 & \Rightarrow 16x+16-{{x}^{2}}-x=15x \\
\end{align}\]
Now we will take the term with variable x on one side and we get.
\[\begin{align}
  & {{x}^{2}}+15x-16x+x-16=0, \\
 & \Rightarrow {{x}^{2}}-x+x-16=0 \\
\end{align}\]
x variables will cancel out each other as they are equal and have opposite signs. After doing this we get.
\[{{x}^{2}}-16=0\]
As we know that,
\[\begin{align}
  & (a-b)(a+b)={{a}^{2}}-{{b}^{2}} \\
\end{align}\]
So the expression \[{{x}^{2}}-16\] can be written as \[(x-4)(x+4)\].
Hence, \[(x-4)(x+4)=0\]
First, we will put \[(x-4)\] equals to zero, and then we will put \[(x+4)\] equals to zero.
\[\begin{align}
  & x-4=0, \\
 & \Rightarrow x=4 \\
\end{align}\]
\[\begin{align}
  & x+4=0, \\
 & \Rightarrow x=-4 \\
\end{align}\]
So here we have obtained two values for x which are \[4\] and \[-4\].

Note: The terms that do not have a fixed value are known as the variable and the terms which have the fixed value are known as the constants. If two algebraic expressions have an equal sign between them then those two expressions will be equal to each other. A polynomial is also a kind of algebraic expression