Write four solutions for each of the following equations:
(i) 2x + y = 7
(ii) πx + y = 9
(iii) x= 4y
Answers
For finding the solutions of linear equation in two variables we use the following steps:
1.Put an put an arbitrary value of x or y in given equation and find the corresponding value of other variable.
Thus, we get one pair of solution of given equation.
2.Repeat repeat step 1 for another arbitrary value of x or y and get other pair of solution of given equation.
For convenience , we put x= 0 & get corresponding value of y to find one pair of solution.
Also we put y = 0 and get corresponding value of x to find 2nd pair solution.
---------------------------------------------------------------------------------------------------
Solution:
(a)2x + y = 7......(1)
On putting x=0 in equation 1
2x+y=7
2×0+y=7
0+y=7
Y=7
So( 0, 7) is a solution of given equation.
let y=0
2x+y=7
2x+0=7
2x=7
X= 7/2
So(7/2,0) is a solution of given equation.
let x = 1
2x+y=7
2×1+y=7
2+y=7
Y= 7-2
Y=5
(1,5) is the solution.
let x = 2
2x+y=7
2×2+y=7
4+y=7
Y= 7-4
Y= 3
The solution is (2,3)
Hence, 4 out of the infinitely many solutions of the given equation are(0,7), (7/2,0),(1,5),(2,3)
(b) πx + y = 9
let x = 0
πx + y = 9
π×0+y =9
0+y= 9
Y=9
(0,9) is the solution.
Let y = 0
πx + y = 9
πx+0=9
πx=9
x= 9/π
(9/π,0) is the solution.
Let x = 1
πx + y = 9
π×1+y=9
π+y=9
Y= 9-π
(1,9-π) is the solution.
Let x = -1
πx + y = 9
π×-1+y=9
-π+y=9
Y= 9+π
(-1,9+π) is the solution.
Hence, 4 out of the infinitely many solutions of the given equation are(0,9), (9/π,0),(1,9-π),(-1,9+π)
(c) x= 4y
Let x = 0
X=4y
0=4y
Y=0/4=0
Y=0
(0,0) is the solution.
Let x= 1
X=4y
1=4y
Y= 1/4
(1,1/4) is the solution.
Let y = 1
X=4y
X=4×1
X=4
(4,1) is the solution.
Let x = 2
X=4y
2=4y
Y=2/4
Y=1/2
(2,1/2) is the solution.
Hence, 4 out of the infinitely many solutions of the given equation are(0,0), (1,1/4),(4,1),(2,1/2)
==========================================================
Hope this will help you....
Step-by-step explanation:
Linear Equations
Exercise 4.2
Question 1: Which one of the following statements is true and why?
y
=
3
x
+
5
has
(i) a unique solution
(ii) only two solutions
(iii) infinitely many solutions.
Solution: Since;
y
=
3
x
+
5
is a linear equation in two variables. We know that a linear equation in two variables has infinite solutions.
Hence option (iii) is true.
Question 2: Write four solutions for each of the following equations:
Solution: (i)
2
x
+
y
=
7
Or,
y
=
7
–
2
x
…………equation (1)
When x = 0, Putting the value in equation (1)
y
=
7
–
2
×
0
Or,
y
=
7
–
0
Or,
y
=
7
Putting the value
x
=
1
in equation (1)
y
=
7
–
2
×
1
Or,
y
=
7
–
2
Or,
y
=
5
Putting the value
x
=
2
in equation (1)
y
=
7
–
2
×
2
Or,
y
=
7
−
4
Or,
y
=
3
Putting the value
x
=
3
in equation (1)
y
=
7
−
2
×
3
Or,
y
=
7
−
6
Or,
y
=
1
Hence, four solutions for equation
2
x
+
y
=
7
are (0,7), (1,5), (2,3), (3,1).
(ii)
π
x
+
y
=
9
Or,
y
=
9
−
π
x
……………… (1)
When x = 0, putting the value in equation (1)
y
=
9
−
π
0
Or,
y
=
9
–
0
=
9
Putting the value
x
=
1
in equation (1)
Y
=
9
−
π
Putting the value
x
=
−
1
in equation (1)
y
=
9
+
π
Putting the value
x
=
2
in equation (1)
y
=
9
−
2
π
Hence, four solutions for the given equation are; (0, 9), (1, 9 - π), (- 1, 9 + π), (2, 9 - 2π)
(iii)
x
=
4
y
…………equation (1)
When
y
=
0
, Putting the value in equation (1)
x
=
4
×
0
Or,
x
=
0
Putting the value
y
=
1
in equation (1)
x
=
4
×
1
Or,
x
=
4
Putting the value
y
=
−
1
in equation (1)
x
=
4
×
−
1
Or,
x
=
−
4
Putting the value
y
=
2
in equation (1)
x
=
4
×
2
Or,
x
=
8
Hence, four solutions for equation
x
=
4
y
are (0, 0), (1, 4), (-4,-1), (8, 2).
Question 3: Check which of the following are solutions of the equation
x
–
2
y
=
4
and which are not:
Solution: (i) (0, 2)
The given equation is
x
–
2
y
=
4
L.H.S
=
x
–
2
y
Putting the value
x
=
0
and
y
=
2
to verify the solution (0, 2)
=
0
–
2
×
2
=
0
–
4
=
−
4
≠
R.H.S.
So, (0, 2) is not a solution of the equation
x
–
2
y
=
4
(ii) (2, 0)
Solution: The given equation is;
x
–
2
y
=
4
L.H.S
=
x
–
2
y
Putting the value
x
=
2
and
y
=
0
to verify the solution (2, 0)
=
2
–
2
×
0
=
2
–
0
=
2
≠
R.H.S.
So, (2, 0) is not a solution of the equation
x
–
2
y
=
4
(iii) (4, 0)
Solution: The given equation is;
x
–
2
y
=
4
L.H.S
=
x
–
2
y
Putting the value
x
=
4
and
y
=
0
to verify the solution (4, 0)
=
4
–
2
×
0
=
4
–
0
= 4 = R.H.S.
So, (4, 0) is a solution of the equation x – 2y = 4.
(iv)
(
√
2
,
4
√
2
)
Putting the value
x
=
√
2
and
y
=
4
√
2
to verify the solution:
=
√
2
−
2
×
4
√
2
=
√
2
−
8
√
2
=
7
√
2
≠
RHS
Or,
(
√
2
,
4
√
2
)
is not a solution for the equation
x
–
2
y
=
4
(v) (1, 1)
Solution: The given equation is;
x
–
2
y
=
4
L.H.S
=
x
–
2
y
Putting the value
x
=
1
and
y
=
1
to verify the solution (1, 1)
=
1
–
2
×
1
=
1
–
2
=
−
1
≠
R.H.S.
So, (1, 1) is not a solution of the equation
x
–
2
y
=
4
Question 4: Find the value of k if x = 2, y = 1 is a solution of the equation
2
x
+
3
y
=
k
Solution: The given equation is;
2
x
+
3
y
=
k
Putting the given value
x
=
2
and
y
=
1
in the equation
2
×
2
+
3
×
1
=
k
4
+
3
=
k
7
=
k
Hence,
k
=
7
Prev
❣️❣️ it's a part of ncert exercise
hope it helps u dear mark as brainliest please ❣️❣️❣️