Question

Write four solutions for each of the following equations:
i) 2x + y = 7
ii) \[\pi x+y+=9\]
iii) x = 4y

Answer 1

Hint: Put the value of ‘x’ or ‘y’ (any one) as a real number (rational or irrational) and hence get the other variable by solving the given equation. Put 4 different values of ‘x’ or ‘y’ (only one of them) and hence get another variable, to get the four solutions of each of them.

Complete Step-by-Step solution:
We know that the given equations in the problem are of two variables ‘x’ and ‘y’. so, we cannot get exact values of variables as we have only one relation in two variables, it means there will always be a ‘x’ for any ‘y’ and vice-versa will also true. So, there will be infinite solutions of the each equation given in the problem.
Now, coming to the question:
i) 2x + y = 7
So, here we can put any real value of ‘x’ or ‘y’ and hence get other accordance. So, as we need to find four solutions, let us put four different values of ‘x’ and hence try to get four values of ‘y’ corresponding to them.
So, let
a) x = 0
So, put x = 0 in the given equation i.e.
2x + y = 7
So, on putting x = 0, we get
2(0) + y = 7
0 + y = 7
y = 7
hence, one solution of the given equation is
x = 0, y = 7
b) Put x = 1
On putting x = 1 in the equation 2x + y = 7, we get
2(1) + y = 7
2 + y = 7
Subtract ‘2’ from each side of the equation, we get
2 + y – 2 = 7 – 2
 y + 2 – 2 = 5
y = 5.
One more solution of the given equation is
x = 1, y = 5.
c) Put x = 2
On putting x = 2 in the equation 2x + y = 7, we get
2(2) + y = 7
4 + y = 7
Subtract 4 from both the sides, we get
4 + y – 4 = 7 – 4
y = 7 – 4 = 3
y = 3
Hence, one more solution would be x = 2, y = 3.
d) Put $x=\dfrac{3}{7}$
On putting $x=\dfrac{3}{7}$to the equation 2x + y = 7, we get
$\begin{align}
  & 2\left( \dfrac{3}{7} \right)+y=7 \\
 & \dfrac{6}{7}+y=7 \\
\end{align}$
Subtract ‘$\dfrac{6}{7}$’ from both the sides, we get
$\begin{align}
  & \dfrac{6}{7}-\dfrac{6}{7}+y=\dfrac{7}{1}-\dfrac{6}{7} \\
 & y=\dfrac{49-6}{7}=\dfrac{43}{7} \\
\end{align}$
Hence, one more solution of the given equation is
$\left( \dfrac{3}{7},\dfrac{43}{7} \right)$.

ii) $\pi x+y=9$
a) Put x = 0
On putting x = 0 to the given expression, we get
$\pi \left( 0 \right)+y=9$
y = 9
Hence, one solution of the given expression is
x = 0, y = 9.
b) Put x = 1
On putting x = 1, to the given equation, we get
$\begin{align}
  & \pi \left( 1 \right)+y=9 \\
 & \pi +y=9 \\
\end{align}$
Subtract $\pi $ from both the sides, we get
So, one more solution of the equation is x = 1, $y=9-\pi $.
c) Put x = -2
On putting x = -2 to the given expression, we get
$\begin{align}
  & \pi \left( -2 \right)+y=9 \\
 & -2\pi +y=9 \\
\end{align}$
On adding $'2\pi '$ both the sides of the equation, we get
$\begin{align}
  & -2\pi +y+2\pi =9+2\pi \\
 & y=9+2\pi \\
\end{align}$
Hence, one more solution of the equation is
x = -2 and $y=9+2\pi $
d) Put $x=\dfrac{1}{\pi }$
On putting $x=\dfrac{1}{\pi }$ to the given equation, we get
$\pi \left( \dfrac{1}{\pi } \right)+y=9$
1 + y = 9
Subtract ‘1’ from both the sides, we get
1 + y – 1 = 9 – 1
y = 8
Hence, one more solution would be $\left( \dfrac{1}{\pi },8 \right)$.

iii) x = 4y
a) Put x = 0
On putting x = 0 to the given equation, we get
0 = 4y
$y=\dfrac{0}{4}=0$ .
Hence, one solution of the equation is x = 0, y = 0
b) Put x = 1
On putting x = 1 to the given equation, we get
1 = 4y,
4y = 1
Divide, the whole equation by ‘4’ we get
$\dfrac{4y}{4}=\dfrac{1}{4} \Rightarrow y=\dfrac{1}{4}$
Hence, one solution would be x = 1, $y=\dfrac{1}{4}$
c) Put x = 4
On putting x = 4 to the given equation, we get
4 = 4y
Divide the whole equation by ‘4’, we get
$\dfrac{4}{4}=\dfrac{4y}{4},$ y = 1
Hence, one solution of the given equation would be x = 4 and y = 1.
d) Put $x=-\sqrt{3}$
On putting $x=-\sqrt{3}$to the given expression, we get
$-\sqrt{3}=4y$
Divide the whole equation by 4, we get
$\dfrac{-\sqrt{3}}{4}=\dfrac{4y}{4},y=\dfrac{-\sqrt{3}}{4}$
Hence, one solution of the given equation would be $x=-\sqrt{3},y=\dfrac{-\sqrt{3}}{4}$

Note: One may put any value of 'x’ in the solution as there will exist infinite pairs of x and y which will satisfy each given equation. So, don’t confuse the values of ‘x’ in the solution. One may put their own values of x and hence get other solutions.
Take care while solving the equation for getting other variables after putting the value of one variable. Students may make mistakes with those parts as well.