Question

How do you solve $\left| \dfrac{x}{7} \right|=3$?

Answer 1

Hint: In this question, we are given an algebraic equation in terms of x and we need to solve it to find the value of x. For this we will first try to remove the modulus from the equation which will result in giving two equation because if $\left| a \right|=b$ then this means a = b or a = -b. So we will form two equations and solve them to find the two values of x which satisfies this equation. We will use operations such as multiplication of numbers on both sides and a constant term on the right side. The constant term will give us the value of x which satisfies the equation.

Complete step by step solution:
Here we are given the equation as $\left| \dfrac{x}{7} \right|=3$. We need to solve it to find the value of x which satisfies this equation. As we can see, the equation consists of modulus so let us first remove it. We know that, if $\left| a \right|=b$ then we have two solutions i.e. a = b or a = -b. So let us use it on the given equation to form two equations which will give us two values of x. Here we have $\left| \dfrac{x}{7} \right|=3$. So it can be written in two equations as $\dfrac{x}{7}=3\text{ and }\dfrac{x}{7}=-3$.
Now let us solve this equation one by one to get the two values of x which will satisfy the original equation.
Taking the first equation we have $\dfrac{x}{7}=3$.
To have just variable x on one side of the equation and a constant term on the right side, let us remove the constant 7 from the left side. For this, let us multiply both sides of the equation by 7. We get $7\times \dfrac{x}{7}=7\times 3$.
7 can be cancelled from the numerator and the denominator on the left side and $7\times 3$ gives 21. So we get x = 21 which is the required form and thus one of the values of x is 21.
Taking the second equation we have $\dfrac{x}{7}=-3$.
Operating this equation the same way i.e. multiplying both sides of the equation by 7 we get $7\times \dfrac{x}{7}=7\times -3$.
Cancelling 7 from the numerator and the denominator on the left side and solving $7\times 3$ as 21 on the right side we get x = -21 which is required form and thus another value of x is -21.
Hence the required two values of x are 21 and -21.

Note: Students should not forget about negative signs while removing the modulus. Note that putting a negative sign inside modulus gives us the positive value as answer only, so we need to consider both possibilities. Putting x = 21 in $\left| \dfrac{x}{7} \right|=3$ we see that $\left| \dfrac{21}{7} \right|=3$. As $\dfrac{21}{7}$ is already positive, so removing modulus we have $\dfrac{21}{7}=3$ 21 divided by 7 gives 3 so 3 = 3. Thus 21 is the correct answer. Similarly if we put x = -21 in $\left| \dfrac{x}{7} \right|=3$ we get $\left| \dfrac{-21}{7} \right|=3$. Modulus will absorb negative signs and we are left with $\dfrac{21}{7}=3\Rightarrow 3=3$. Thus -21 is the correct value of x.