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Question: Solve the equations: \(\dfrac{5}{x-1}+\dfrac{1}{y-2}=2\) and \(\dfrac{6}{x-1}-\dfrac{3}{y-2}=1\)....

Solve the equations: 5x−1+1y−2=2\dfrac{5}{x-1}+\dfrac{1}{y-2}=2 and 6x−1−3y−2=1\dfrac{6}{x-1}-\dfrac{3}{y-2}=1.

Explanation

Solution

For solving this question you should know about the general mathematical solutions of equations. To reduce the equations, we will put a term equal to a variable such that after doing this, we will get a pair of linear equations. Then we will solve these pairs of linear equations by method of substitution.

Complete step-by-step solution:
According to our question, we have to solve the equations 5x−1+1y−2=2\dfrac{5}{x-1}+\dfrac{1}{y-2}=2 and 6x−1−3y−2=1\dfrac{6}{x-1}-\dfrac{3}{y-2}=1. In this question there are four parts, so we will solve each part one by one. To reduce the equations into a pair of linear equations, we will first take that term which are common in both the equations and then we will equate these terms to some other variables. After doing this, we will solve these questions with the help of a substitution method. After finding the values of assumed variables equal to the term and find the values of x and y. It is given that,
5x−1+1y−2=2\dfrac{5}{x-1}+\dfrac{1}{y-2}=2 and 6x−1−3y−2=1\dfrac{6}{x-1}-\dfrac{3}{y-2}=1
Here we will assume,  1x−1=a~\dfrac{1}{x-1}=a and 1y−2=b\dfrac{1}{y-2}=b
Thus the new equations that we will get are,
5a+b=2………(i) 6a−3b=1………(ii) \begin{aligned} & 5a+b=2\ldots \ldots \ldots \left( i \right) \\\ & 6a-3b=1\ldots \ldots \ldots \left( ii \right) \\\ \end{aligned}
From (i), we can say that,
5a+b=2 ⇒b=2−5a \begin{aligned} & 5a+b=2 \\\ & \Rightarrow b=2-5a \\\ \end{aligned}
Now, we will put this value of b in equation (ii), thus we will get as,
6a−3(2−5a)=1 ⇒6a−6+15a=1 ⇒21a=7 ⇒a=721=13………(iii) \begin{aligned} & 6a-3\left( 2-5a \right)=1 \\\ & \Rightarrow 6a-6+15a=1 \\\ & \Rightarrow 21a=7 \\\ & \Rightarrow a=\dfrac{7}{21}=\dfrac{1}{3}\ldots \ldots \ldots \left( iii \right) \\\ \end{aligned}
Putting this in equation (i), we get,
5(13)+b=2 ⇒b=2−5(13) ⇒b=2−53=6−53=13………(iv) \begin{aligned} & 5\left( \dfrac{1}{3} \right)+b=2 \\\ & \Rightarrow b=2-5\left( \dfrac{1}{3} \right) \\\ & \Rightarrow b=2-\dfrac{5}{3}=\dfrac{6-5}{3}=\dfrac{1}{3}\ldots \ldots \ldots \left( iv \right) \\\ \end{aligned}
Now we will compare the values of a and b with the assumed values. So,
As  1x−1=a~\dfrac{1}{x-1}=a, so  1x−1=13~\dfrac{1}{x-1}=\dfrac{1}{3}
⇒x−1=1a⇒x=1+1a ⇒x=1+1(13)=1+3=4 \begin{aligned} & \Rightarrow x-1=\dfrac{1}{a}\Rightarrow x=1+\dfrac{1}{a} \\\ & \Rightarrow x=1+\dfrac{1}{\left( \dfrac{1}{3} \right)}=1+3=4 \\\ \end{aligned}
And,
1y−2=b⇒y−2=1b⇒y=2+1b ⇒y=2+1(13)⇒y=2+3=5 \begin{aligned} & \dfrac{1}{y-2}=b\Rightarrow y-2=\dfrac{1}{b}\Rightarrow y=2+\dfrac{1}{b} \\\ & \Rightarrow y=2+\dfrac{1}{\left( \dfrac{1}{3} \right)}\Rightarrow y=2+3=5 \\\ \end{aligned}
So, the values of x=4x=4 and y=5y=5.

Note: While solving this question you should be careful about comparing the values of the assumed values with each other. We will then find the value for that very carefully and compare that with the same assumed value, otherwise our question will be wrong if any of these mistakes are there in the solution.