Question

How do you solve $\dfrac{3}{x} = \dfrac{7}{{10}}$ ?

Answer 1

Hint: A linear equation is an equation for a straight line. The term which is involved in a linear equation is either a constant or a single variable or product of a constant. The two variables can never be multiplied. All linear equations have a line graph. Linear equations are the same as linear functions. The general form of writing a linear equation is $y = mx + c$ and $m$ is not equal to zero, where $m$ is the slope and $c$ is the point on which it cuts the y-axis.$y = mx + c$ is also known as equation of the line in slope-intercept form. Here in this question, we are supposed to solve the given expression for the value of $x$.

Complete step by step answer:
Given expression is$\dfrac{3}{x} = \dfrac{7}{{10}}$,
In order to simplify and get the value of$x$, we first need to get rid of the fractions. Here in this case, we can do it by cross multiplying them.
$\dfrac{3}{x} = \dfrac{7}{{10}}$
To cross multiply, we multiply the left numerator that is $3$ by the right denominator that is $10$, then the left denominator that is $x$ by the right numerator that is $7$.
$3 \times 10 = 7 \times x \\
\Rightarrow 30 = 7x$
Now, solving it we get,
$7x = 30 \\
\therefore x = \dfrac{{30}}{7} \\ $
Hence, the value of $x$ for which $\dfrac{3}{x} = \dfrac{7}{{10}}$ is $\dfrac{{30}}{7}$.

Note: Now that we know the value of $x$ is $\dfrac{{30}}{7}$, there is a way to double check our answer. In order to double check the solution we are supposed to substitute the value of $x$ which we got as $\dfrac{{30}}{7}$ in the given equation, $\dfrac{3}{x} = \dfrac{7}{{10}}$
$\dfrac{3}{x} = \dfrac{7}{{10}} \\
\Rightarrow \dfrac{3}{{\dfrac{{30}}{7}}} = \dfrac{7}{{10}} \\ $
$\Rightarrow \dfrac{3}{{30}} \times \dfrac{7}{1} = \dfrac{7}{{10}} \\
\Rightarrow \dfrac{1}{{10}} \times \dfrac{7}{1} = \dfrac{7}{{10}} \\
\Rightarrow \dfrac{7}{{10}} = \dfrac{7}{{10}} \\ $
Now, the left-hand side is equal to the right-hand side. So, we can conclude that our solution or the value of $x$ which we calculated was correct.