Question

Solve the equation \[\dfrac{{(2 - 7x)}}{{(1 - 5x)}} = \dfrac{{(3 + 7x)}}{{(4 + 5x)}}\] and find the value of x.

Answer 1

Hint: In this question, we will simply put the formula of cross multiplication and will make this equation simple so that the calculations become easy. Cross multiplication is a method in which we will multiply the numerator of LHS with the denominator of RHS and vice versa. After simplifying the equation, we will use the splitting the middle term method so that the equations can be divided into two parts to get the required solution of x.

Complete step-by-step solution:
Let’s have a look at how the cross-multiplication method works. Suppose we are being given that
\[\dfrac{A}{B} = \dfrac{C}{D}\]
So, according to this, we will first multiply the numerator of LHS i.e. A with the denominator of RHS i.e., D, and vice versa
\[ \Rightarrow A \times D = C \times B\]
Similarly, in this question, we suppose that \[A = (2 - 7x)\], \[B = (1 - 5x)\], \[C = (3 + 7x)\] and finally \[D = (4 + 5x)\]. So, we have been given,
\[ \Rightarrow \dfrac{{(2 - 7x)}}{{(1 - 5x)}} = \dfrac{{(3 + 7x)}}{{(4 + 5x)}}\]
Now putting these values in the above equation, we get
\[ \Rightarrow \dfrac{{(2 - 7x)(4 + 5x)}}{{(1 - 5x)}} = (3 + 7x)\]
\[ \Rightarrow (2 - 7x)(4 + 5x) = (3 + 7x)(1 - 5x)\]
By opening the brackets and multiplying each term, we get
\[ \Rightarrow 2(4) + 2(5x) - 4(7x) - (7x)(5x) = 3(1) - 3(5x) + 1(7x) - (7x)(5x)\]
By simplifying the equation
\[ \Rightarrow 8 + 10x - 28x - 35{x^2} = 3 - 15x + 7x - 35{x^2}\]
Adding and subtracting the similar terms and arranging them in ascending order, we get
\[ \Rightarrow - 35{x^2} - 18x + 8 = - 35{x^2} - 8x + 3\]
Now, by canceling the terms,
\[ \Rightarrow - 18x + 8 = - 8x + 3\]
Taking similar terms on one side,
\[ \Rightarrow - 18x + 8x = 3 - 8\]
\[ \Rightarrow - 10x = - 5\]
Taking -10 on the other side, we get
\[ \Rightarrow x = \dfrac{{ - 5}}{{ - 10}}\]
\[ \Rightarrow x = \dfrac{1}{2}\]
Hence, the required value of x is \[\dfrac{1}{2}\].

Note: We usually obtain a quadratic equation while solving a cross-multiplication problem since we have to multiply two linear equations. In this situation, however, the terms with squares are canceled out, leaving only a linear equation. If a similar question arises, then splitting the middle term method will be used.