Question

How do you solve the system $ 3\dfrac{x}{8} - 3\dfrac{y}{5} = \dfrac{{33}}{{80}} $ and $ 4\dfrac{x}{7} + 4\dfrac{y}{5} = \dfrac{{37}}{{35}} $ ?

Answer 1

Hint: We are given the system of two linear equations with two variables. But we can see that the terms are in the form of mixed fraction. Therefore, first we have to convert all the terms in the simple fraction. After that, we will convert the equation in the simple form by taking LCM and then solve them to get our final answer.

Complete step by step solution:
We will first simplify both the equations.
Let us consider the first equation $ 3\dfrac{x}{8} - 3\dfrac{y}{5} = \dfrac{{33}}{{80}} $ .
We will convert both the terms into simple fractions.
 $ 3\dfrac{x}{8} = \dfrac{{\left( {8 \times 3} \right) + x}}{8} = \dfrac{{24 + x}}{8} $ and $ 3\dfrac{y}{5} = \dfrac{{\left( {5 \times 3} \right) + y}}{5} = \dfrac{{15 + y}}{5} $
Therefore, the equation becomes:
 $ \dfrac{{24 + x}}{8} - \dfrac{{15 + y}}{5} = \dfrac{{33}}{{80}} $
We know that the LCM of 8, 5 and 80 is 80.
 \[
   \Rightarrow \dfrac{{10\left( {24 + x} \right)}}{{80}} - \dfrac{{16\left( {15 + y} \right)}}{{80}} = \dfrac{{33}}{{80}} \\
   \Rightarrow 10\left( {24 + x} \right) - 16\left( {15 + y} \right) = 33 \\
   \Rightarrow 240 + 10x - 240 - 16y = 33 \\
   \Rightarrow 10x - 16y = 33 \;
 \]
Thus, our first equation is \[10x - 16y = 33\] .
Now, we will consider the second equation $ 4\dfrac{x}{7} + 4\dfrac{y}{5} = \dfrac{{37}}{{35}} $ .
We will convert both the terms into simple fractions.
 $ 4\dfrac{x}{7} = \dfrac{{\left( {7 \times 4} \right) + x}}{7} = \dfrac{{28 + x}}{7} $ and $ 4\dfrac{y}{5} = \dfrac{{\left( {5 \times 4} \right) + y}}{5} = \dfrac{{20 + y}}{5} $
Therefore, the equation becomes:
 $ \dfrac{{28 + x}}{7} + \dfrac{{20 + y}}{5} = \dfrac{{37}}{{35}} $
We know that the LCM of 7, 5 and 35 is 35.
 \[
   \Rightarrow \dfrac{{5\left( {28 + x} \right)}}{{35}} + \dfrac{{7\left( {20 + y} \right)}}{{35}} = \dfrac{{37}}{{35}} \\
   \Rightarrow 5\left( {28 + x} \right) + 7\left( {20 + y} \right) = 37 \\
   \Rightarrow 140 + 5x + 140 + 7y = 33 \\
   \Rightarrow 5x + 7y = 33 - 280 \\
   \Rightarrow 5x + 7y = - 243 \;
 \]
Thus, our second equation is \[5x + 7y = - 243\] .
Now, we will multiply the second equation with 2 and then subtract it from the first equation.
 $
   \Rightarrow 10x - 16y - \left( {10x + 14y} \right) = 33 - \left( { - 243 \times 2} \right) \\
   \Rightarrow 10x - 16y - 10x - 14y = 33 + 486 \\
   \Rightarrow - 30y = 519 \\
   \Rightarrow y = \dfrac{{519}}{{ - 30}} \\
   \Rightarrow y = - 17.3 \;
  $ $ $ $ $
Now we will put this value in the first equation.
 \[
  10x - 16y = 33 \\
   \Rightarrow 10x - 16( - 17.3) = 33 \\
   \Rightarrow 10x + 276.8 = 33 \\
   \Rightarrow 10x = 33 - 276.8 \\
   \Rightarrow x = \dfrac{{ - 243.8}}{{10}} \\
   \Rightarrow x = - 24.38 \;
 \]
Thus, our final answer is \[x = - 24.38\] and $ y = - 17.3 $.
So, the correct answer is “ \[x = - 24.38\] and $ y = - 17.3 $”.

Note: Here, we have used the elimination method for solving the system of two linear equations with two variables. In this method, we have to make the coefficient of one of the variables same and then add or subtract to eliminate that variable depending on its sign. After elimination, we will get the value of another variable. Finally, by putting the obtained value in one of the equations, we can determine the value of the remaining variable.