Question

How do you solve \[\dfrac{x}{2}\, = \,\dfrac{{3x}}{4} + 5\] ?

Answer 1

Hint:In this question we have to solve for the value of $x$from the given linear equation of one variable. To solve the linear equation of one variable, we will first move all variables term to one side of the equation and all constant terms of the equation to the other side of the equation by performing necessary operations.

Complete step by step solution:
Let us try to solve this question in which we have to find the value of $x$in the given equation .This is a linear equation in one variable. We will use fraction addition to solve for the value of $x$ in equation \[\dfrac{x}{2}\, = \,\dfrac{{3x}}{4} + 5\] .

To simplify this term we will first move all terms with variables to the L.H.S of the equation and constants to R.H.S. After rearranging terms of the given equation. We get,
$\dfrac{x}{2} - \dfrac{{3x}}{4} = 5$ $eq(1)$

Since $\dfrac{{3x}}{5}$is in addition in R.H.S, so it will be subtracted in L.H.S.

Now, we will perform fraction addition in $eq(1)$ we get,
$\dfrac{x}{2} - \dfrac{{3x}}{4} = 5$

Since$L.C.M(2,4) = 4$ and performing step of fraction we have,
$
\dfrac{{2 \cdot x - 3x}}{4} = 5 \\
\dfrac{{2x - 3x}}{4} = 5 \\
\dfrac{{ - x}}{4} = 5 \\
$
Now, multiplying both sides of the equation $\dfrac{{ - x}}{4} = 5$by$ - 4$. We get the value of$x$.
$
( - 4) \cdot \left( {\dfrac{{ - x}}{4}} \right) = ( - 4) \cdot 5 \\
x = - 20 \\
$
Hence by solving the given equation \[\dfrac{x}{2}\, = \,\dfrac{{3x}}{4} + 5\] . We get the value of$x = -20$. We can also check that our answer is correct or not but putting the value of $x$in the equation.

Putting value of $x = - 20$in equation \[\dfrac{x}{2}\, = \,\dfrac{{3x}}{4} + 5\] , we get
$
\dfrac{{ - 20}}{2} = \dfrac{{3 \cdot ( - 20)}}{4} + 5 \\
- 10 = - 15 + 5 \\
- 10 = - 10 \\
$
Since L.H.S is equal to R.H.S. Our value of $x = - 20$is correct.

Note: The questions in which we are given a linear equation in one variable and we have to solve for the value of the variable very easily. While solving for a linear equation in one variable you just have to be clear about the sign of constant and variable terms where most students make mistakes.