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How do you solve $\dfrac{x}{2}+\dfrac{x}{3}=10$ ?

Last updated date: 26th Feb 2024
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IVSAT 2024
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Hint: The equation given in the above question is a linear equation in one variable, that is x. The question says that we have to solve the given equation in x. In other words, we have to find the values of x which will satisfy the given equation. Try to find the value of x by performing some mathematical operations.

Complete step by step solution:
The equation says that $\dfrac{x}{2}+\dfrac{x}{3}=10$. Let us analyse the above equation and try to simplify it. The first thing that we can do is take the x as a common term from the two fractions on the left hand side of the equation.With this the above equation will simplify to
$x\left( \dfrac{1}{2}+\dfrac{1}{3} \right)=10$ …. (i)
Then we can just calculate the value of the summation inside the bracket. We know that
Substitute this value in equation (i).
Then the equation will further simplify to $x\left( \dfrac{5}{6} \right)=10$.
Now, multiply both the sides of the equation by factor of $\dfrac{6}{5}$ so that the left hand side will have x term only.
$x\left( \dfrac{5}{6} \right)\times \dfrac{6}{5}=10\times \dfrac{6}{5}$
$\Rightarrow x=2\times 6$
With this, we get that $x=12$

Therefore, the solution of the given equation is $x=12$ or we can also say that the value $x=12$ satisfies the given equation.

Note:There are many ways in which we can solve this question. We can also directionally add the two fractions on the left hand side of the equation and calculate the value of x. However, students must note that the number of solutions to a given equation in one variable is always less than or equal to the degree of the polynomial in the given equation.