
How do you solve \[-\dfrac{x}{3}-5=2x\]?
Answer
558k+ views
Hint: In this problem, we have to solve and find the value of x. To solve this problem, we should know cross multiplication. In a cross-multiplication method, we should multiply the numerator of one fraction to the denominator of the other and the denominator of the first term to the numerator of the other. By using cross multiplication in the left-hand side, we can solve and find the value of x.
Complete step by step answer:
We know that the given fraction to be solved is,
\[-\dfrac{x}{3}-5=2x\] ……. (1)
Now we can use the cross-multiplication method on the left-hand side. We should multiply the numerator of one fraction to the denominator of the other and the denominator of first term to the numerator of other, we get
\[\Rightarrow \dfrac{-x-15}{3}=2x\]
Now we can multiply 3 on both sides, we get
\[\begin{align}
& \Rightarrow 3\times \dfrac{-x-15}{3}=2x\times 3 \\
& \Rightarrow -x-15=6x \\
\end{align}\]
Now we can add x on both the sides, we get
\[\begin{align}
& \Rightarrow x-x-15=6x+x \\
& \Rightarrow 7x=-15 \\
\end{align}\]
Now we can divide by 7 on both sides, we get
\[\begin{align}
& \Rightarrow \dfrac{7x}{7}=-\dfrac{15}{7} \\
& \Rightarrow x=-\dfrac{15}{7} \\
\end{align}\]
Therefore, on solving \[-\dfrac{x}{3}-5=2x\], \[x=-\dfrac{15}{7}\].
Note: Students make mistakes while cross multiplying which should be concentrated. To cancel similar terms we can add, subtract, multiply or divide to simplify to solve the given problem. We can also take the numbers to the other side by changing signs instead of add or subtract. In a cross-multiplication method, we should multiply the numerator of one fraction to the denominator of the other and the denominator of the first term to the denominator of the other.
Complete step by step answer:
We know that the given fraction to be solved is,
\[-\dfrac{x}{3}-5=2x\] ……. (1)
Now we can use the cross-multiplication method on the left-hand side. We should multiply the numerator of one fraction to the denominator of the other and the denominator of first term to the numerator of other, we get
\[\Rightarrow \dfrac{-x-15}{3}=2x\]
Now we can multiply 3 on both sides, we get
\[\begin{align}
& \Rightarrow 3\times \dfrac{-x-15}{3}=2x\times 3 \\
& \Rightarrow -x-15=6x \\
\end{align}\]
Now we can add x on both the sides, we get
\[\begin{align}
& \Rightarrow x-x-15=6x+x \\
& \Rightarrow 7x=-15 \\
\end{align}\]
Now we can divide by 7 on both sides, we get
\[\begin{align}
& \Rightarrow \dfrac{7x}{7}=-\dfrac{15}{7} \\
& \Rightarrow x=-\dfrac{15}{7} \\
\end{align}\]
Therefore, on solving \[-\dfrac{x}{3}-5=2x\], \[x=-\dfrac{15}{7}\].
Note: Students make mistakes while cross multiplying which should be concentrated. To cancel similar terms we can add, subtract, multiply or divide to simplify to solve the given problem. We can also take the numbers to the other side by changing signs instead of add or subtract. In a cross-multiplication method, we should multiply the numerator of one fraction to the denominator of the other and the denominator of the first term to the denominator of the other.
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