Question

Solve for \[x:\dfrac{7}{2}x - \dfrac{5}{2}x = \dfrac{{20}}{3}x + 10\]

Answer 1

Hint:
Here we will first find the L.C.M of fractions on both sides and solve the values. Then we will simplify the terms by cross multiplication and basic mathematical operation to get the required answer.

Complete step by step solution:
We have to solve \[\dfrac{7}{2}x - \dfrac{5}{2}x = \dfrac{{20}}{3}x + 10\].
First we will take L.C.M on both sides as,
\[ \Rightarrow \dfrac{{7x - 5x}}{2} = \dfrac{{20x + 3 \times 10}}{3}\]
Simplifying the equation, we get
\[ \Rightarrow \dfrac{{2x}}{2} = \dfrac{{20x + 30}}{3}\]
Now, we will cross multiply the denominator from both sides. Therefore, we get,
\[\begin{array}{l} \Rightarrow 3\left( {2x} \right) = 2 \times \left( {20x + 30} \right)\\ \Rightarrow 6x = 40x + 60\end{array}\]
Now, taking all the \[x\] term one side of equal to sign, we get,
\[ \Rightarrow 6x - 40x = 60\]
Subtracting the terms, we get
\[ \Rightarrow - 34x = 60\]
Dividing both sides by \[ - 34\], we get
\[ \Rightarrow x = \dfrac{{60}}{{ - 34}}\]
Dividing denominator and numerator by 2, we get
\[ \Rightarrow x = - \dfrac{{30}}{{17}} \approx - 1.765\]

So value of \[x \approx - 1.765\]

Note:
Here, the given equation is a linear equation. A linear equation is defined as an equation with the highest degrees of variable as 1. We can also solve the equation by another method.
As we can see that the fraction has 2 on one side and 3 on the other side, so we will multiply 6 on both sides as 6 is the product of 3 and 2. Therefore, we get
 \[\left( {\dfrac{7}{2}x - \dfrac{5}{2}x} \right) \times 6 = \left( {\dfrac{{20}}{3}x + 10} \right) \times 6\]
Multiplying the terms using the distributive property, we get
\[\begin{array}{l}\dfrac{{2x}}{2} \times 6 = \dfrac{{20x}}{3} \times 6 + 10 \times 6\\ \Rightarrow 6x = 40x + 60\end{array}\]
Subtracting \[40x\] from both sides, we get
\[ \Rightarrow 6x - 40x = 60\]
\[ \Rightarrow - 34x = 60\]
Dividing both sides by \[ - 34\], we get
\[ \Rightarrow x = - \dfrac{{60}}{{34}}\]
Dividing denominator and numerator by 2, we get
\[ \Rightarrow x = - \dfrac{{30}}{{17}} \approx - 1.765\]
So value of \[x \approx - 1.765\]
We are getting the same answer. So as we can see no matter what operation we apply we will always get the same result just the condition is the operation should be applied on both sides on an “equal” sign.