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Question: How do you solve the system of equations \(\dfrac{1}{2}x+3y=\dfrac{5}{6}\) and \(\dfrac{1}{3}x-5y=\d...

How do you solve the system of equations 12x+3y=56\dfrac{1}{2}x+3y=\dfrac{5}{6} and 13x−5y=169\dfrac{1}{3}x-5y=\dfrac{16}{9} ?

Explanation

Solution

We need to multiply each linear equation given in the question, with a number in such a way that one of the coefficients of either x or y become the same. This will allow us to remove one variable to get the value of the other. Then, we will use the value that we found in one of the linear equations to get the value of the remaining variable.

Complete step by step answer:
First, let’s solve the both equations for y. We will multiply 2 and 3 to the first and second equation respectively to get the coefficient of x to be 1, i.e. to make them similar.
First equation ⇒12x+3y=56\Rightarrow \dfrac{1}{2}x+3y=\dfrac{5}{6}
⇒(12x+3y=56)×2\Rightarrow \left( \dfrac{1}{2}x+3y=\dfrac{5}{6} \right)\times 2
⇒x+6y=53.......(i)\Rightarrow x+6y=\dfrac{5}{3}.......(i)
Second equation ⇒13x−5y=169\Rightarrow \dfrac{1}{3}x-5y=\dfrac{16}{9}
⇒(13x−5y=169)×3\Rightarrow \left( \dfrac{1}{3}x-5y=\dfrac{16}{9} \right)\times 3
⇒x−15y=163......(ii)\Rightarrow x-15y=\dfrac{16}{3}......(ii)
Now, let us subtract equation (i) from equation (ii) to first get the value of “y”. By doing so, x will get cancelled and we will get an equation in terms of y, thus, helping us to get the value of it.
⇒(x+6y=53) −(x−15y=163)‾  0+21y=−113 \begin{aligned} & \Rightarrow \left( x+6y=\dfrac{5}{3} \right) \\\ & \underline{-\left( x-15y=\dfrac{16}{3} \right)} \\\ & \text{ }0+21y=\dfrac{-11}{3} \\\ \end{aligned}
⇒21y=−113 ∴y=−1163....(iii) \begin{aligned} & \Rightarrow 21y=\dfrac{-11}{3} \\\ & \therefore y=\dfrac{-11}{63}....(iii) \\\ \end{aligned}
Now, let us put the value of y from equation (iii) in equation (i), we will get,
⇒x+6(−1163)=53 ⇒x−2(1121)=53 ⇒x−2221=53 ⇒x=53+2221 ⇒x=5721 \begin{aligned} & \Rightarrow x+6\left( \dfrac{-11}{63} \right)=\dfrac{5}{3} \\\ & \Rightarrow x-2\left( \dfrac{11}{21} \right)=\dfrac{5}{3} \\\ & \Rightarrow x-\dfrac{22}{21}=\dfrac{5}{3} \\\ & \Rightarrow x=\dfrac{5}{3}+\dfrac{22}{21} \\\ & \Rightarrow x=\dfrac{57}{21} \\\ \end{aligned}
or
∴x=197\therefore x=\dfrac{19}{7}
So, the solutions of the given two linear equations is x=197x=\dfrac{19}{7} and y=−1121y=\dfrac{-11}{21} .

Note:
Alternatively, we can first find the value of x by removing y from both the equations in the similar manner as mentioned above. We will still get the same answer.
See that, the values of x and y that we calculated are known as the solutions of the given linear equations, i.e. if we put these values in the L.H.S. of both the equations, they will give us the same result which is 0.