Question

If $\dfrac{7}{2}x+\dfrac{5}{2}y=5;4x+2y=7$ then what is the value of x-y?

Answer 1

Hint: In this question, we are given two equations in terms of two variables x and y. We have to find the value of x-y. For this, we will use a substitution method to solve this equation to find the value of x and y. After that, we will take the difference between x and y to find the required answer. In the substitution method, we first find the value of y in terms of x from one equation and then put it in another equation to get equation in terms of x only. We then solve this equation and find the value of x, using the value of x in any of the equations, we find the value of y.

Complete step by step answer:
Here we are given equation as,
\[\begin{align}
  & \dfrac{7}{2}x+\dfrac{5}{2}y=5\cdots \cdots \cdots \left( 1 \right) \\
 & 4x+2y=7\cdots \cdots \cdots \left( 2 \right) \\
\end{align}\]
Let us first simplify equation (1),
\[\Rightarrow \dfrac{7}{2}x+\dfrac{5}{2}y=5\]
Taking $\dfrac{1}{2}$ common from left side of the equation we get:
\[\Rightarrow \dfrac{1}{2}\left( 7x+5y \right)=5\]
On cross multiplying we get:
\[\begin{align}
  & \Rightarrow \left( 7x+5y \right)=5\times 2 \\
 & \Rightarrow \left( 7x+5y \right)=10 \\
\end{align}\]
Now let us find the value of y from this equation in terms of x. Taking 7x on opposite side, we get:
\[\Rightarrow 5y=10-7x\]
Dividing both sides by 5, we get:
\[\Rightarrow y=\dfrac{10-7x}{5}\cdots \cdots \cdots \left( 3 \right)\]
Substituting the value of y from equation (3) into the equation (2) we will have:
\[\begin{align}
  & \Rightarrow 4x+2\left( \dfrac{10-7x}{5} \right)=7 \\
 & \Rightarrow 4x+\dfrac{20-14x}{5}=7 \\
\end{align}\]
Taking LCM as 5 on left side of equation we get:
\[\Rightarrow \dfrac{20x+20-14x}{5}=7\]
Cross multiplying we get:
\[\begin{align}
  & \Rightarrow 20x+20-14x=35 \\
 & \Rightarrow 6x+20=35 \\
 & \Rightarrow 6x=35-20 \\
 & \Rightarrow 6x=15 \\
 & \Rightarrow x=\dfrac{15}{6} \\
 & \Rightarrow x=\dfrac{5}{2} \\
\end{align}\]
Hence the value of x is equal to $\dfrac{5}{2}$.
Now replacing x value in equation (2) we get:
\[\begin{align}
  & \Rightarrow 4\left( \dfrac{5}{2} \right)+2y=7 \\
 & \Rightarrow 10+2y=7 \\
 & \Rightarrow 2y=7-10 \\
 & \Rightarrow 2y=-3 \\
 & \Rightarrow y=\dfrac{-3}{2} \\
\end{align}\]
Then the value of y is equal to $\dfrac{-3}{2}$.
Now, we need to find the value of x-y. So let us take difference between x and y, we get:
\[\begin{align}
  & \Rightarrow x-y=\dfrac{5}{2}-\left( \dfrac{-3}{2} \right) \\
 & \Rightarrow x-y=\dfrac{5}{2}+\dfrac{3}{2} \\
 & \Rightarrow x-y=\dfrac{5+3}{2} \\
 & \Rightarrow x-y=\dfrac{8}{2} \\
 & \Rightarrow x-y=4 \\
\end{align}\]

Hence, the required value of x-y is equal to 4.

Note: Students should take care of signs while solving the equations. Students can also use elimination methods for solving these equations. Elimination method is given as:
Simplified equations are: \[\begin{align}
  & \dfrac{7}{2}x+\dfrac{5}{2}y=5\cdots \cdots \cdots \left( 1 \right) \\
 & 4x+2y=7\cdots \cdots \cdots \left( 2 \right) \\
\end{align}\]
Multiplying (1) by 4 and (2) by 7, we get:
$28x+20y=40\text{ and }28x+14y=49$.
Since coefficient of x in both equations are same, so subtracting these equations we get:
\[\begin{align}
  & \Rightarrow 28x+20y-28x+14y=40-49 \\
 & \Rightarrow 20y-14y=-9 \\
 & \Rightarrow 6y=-9 \\
 & \Rightarrow y=\dfrac{-9}{6} \\
 & \Rightarrow y=\dfrac{-3}{2} \\
\end{align}\]
Putting values of y in any of the equations will give us a value of x. Putting value of y in equation (2) we get:
\[\begin{align}
  & \Rightarrow 4x+2\left( \dfrac{-3}{2} \right)=7 \\
 & \Rightarrow 4x-3=7 \\
 & \Rightarrow 4x=10 \\
 & \Rightarrow x=\dfrac{10}{4} \\
 & \Rightarrow x=\dfrac{5}{2} \\
\end{align}\]
This way we have obtained the same values of x and y as obtained by substitution method. Students can check their answer by putting these values in any of the given equations and verify the left side to be equal to the right side.