x+√2=0 has how many solutions..
One solution
No solution
Infinity
Two solution
Answers
Answered by
5
Question 1: Which one of the following statements is true and why?
y
=
3
x
+
5
y
=
3
x
+
5
has
(i) a unique solution
(ii) only two solutions
(iii) infinitely many solutions.
Solution: Since;
y
=
3
x
+
5
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
2
x
+
y
=
7
Or,
y
=
7
–
2
x
y
=
7
–
2
x
…………equation (1)
When x = 0, Putting the value in equation (1)
y
=
7
–
2
×
0
y
=
7
–
2
×
0
Or,
y
=
7
–
0
y
=
7
–
0
Or,
y
=
7
y
=
7
Putting the value
x
=
1
x
=
1
in equation (1)
y
=
7
–
2
×
1
y
=
7
–
2
×
1
Or,
y
=
7
–
2
y
=
7
–
2
Or,
y
=
5
y
=
5
Putting the value
x
=
2
x
=
2
in equation (1)
y
=
7
–
2
×
2
y
=
7
–
2
×
2
Or,
y
=
7
−
4
y
=
7
-
4
Or,
y
=
3
y
=
3
Putting the value
x
=
3
x
=
3
in equation (1)
y
=
7
−
2
×
3
y
=
7
-
2
×
3
Or,
y
=
7
−
6
y
=
7
-
6
Or,
y
=
1
y
=
1
Hence, four solutions for equation
2
x
+
y
=
7
2
x
+
y
=
7
are (0,7), (1,5), (2,3), (3,1).
(ii)
π
x
+
y
=
9
π
x
+
y
=
9
Or,
y
=
9
−
π
x
y
=
9
-
π
x
……………… (1)
When x = 0, putting the value in equation (1)
y
=
9
−
π
0
y
=
9
-
π
0
Or,
y
=
9
–
0
=
9
y
=
9
–
0
=
9
Putting the value
x
=
1
x
=
1
in equation (1)
Y
=
9
−
π
Y
=
9
-
π
Putting the value
x
=
−
1
x
=
-
1
in equation (1)
y
=
9
+
π
y
=
9
+
π
Putting the value
x
=
2
x
=
2
in equation (1)
y
=
9
−
2
π
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
x
=
4
y
…………equation (1)
When
y
=
0
y
=
0
, Putting the value in equation (1)
x
=
4
×
0
x
=
4
×
0
Or,
x
=
0
x
=
0
Putting the value
y
=
1
y
=
1
in equation (1)
x
=
4
×
1
x
=
4
×
1
Or,
x
=
4
x
=
4
Putting the value
y
=
−
1
y
=
-
1
in equation (1)
x
=
4
×
−
1
x
=
4
×
-
1
Or,
x
=
−
4
x
=
-
4
Putting the value
y
=
2
y
=
2
in equation (1)
x
=
4
×
2
x
=
4
×
2
Or,
x
=
8
x
=
8
Hence, four solutions for equation
x
=
4
y
x
=
4
y
are (0, 0), (1, 4), (-4,-1), (8, 2).
y
=
3
x
+
5
y
=
3
x
+
5
has
(i) a unique solution
(ii) only two solutions
(iii) infinitely many solutions.
Solution: Since;
y
=
3
x
+
5
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
2
x
+
y
=
7
Or,
y
=
7
–
2
x
y
=
7
–
2
x
…………equation (1)
When x = 0, Putting the value in equation (1)
y
=
7
–
2
×
0
y
=
7
–
2
×
0
Or,
y
=
7
–
0
y
=
7
–
0
Or,
y
=
7
y
=
7
Putting the value
x
=
1
x
=
1
in equation (1)
y
=
7
–
2
×
1
y
=
7
–
2
×
1
Or,
y
=
7
–
2
y
=
7
–
2
Or,
y
=
5
y
=
5
Putting the value
x
=
2
x
=
2
in equation (1)
y
=
7
–
2
×
2
y
=
7
–
2
×
2
Or,
y
=
7
−
4
y
=
7
-
4
Or,
y
=
3
y
=
3
Putting the value
x
=
3
x
=
3
in equation (1)
y
=
7
−
2
×
3
y
=
7
-
2
×
3
Or,
y
=
7
−
6
y
=
7
-
6
Or,
y
=
1
y
=
1
Hence, four solutions for equation
2
x
+
y
=
7
2
x
+
y
=
7
are (0,7), (1,5), (2,3), (3,1).
(ii)
π
x
+
y
=
9
π
x
+
y
=
9
Or,
y
=
9
−
π
x
y
=
9
-
π
x
……………… (1)
When x = 0, putting the value in equation (1)
y
=
9
−
π
0
y
=
9
-
π
0
Or,
y
=
9
–
0
=
9
y
=
9
–
0
=
9
Putting the value
x
=
1
x
=
1
in equation (1)
Y
=
9
−
π
Y
=
9
-
π
Putting the value
x
=
−
1
x
=
-
1
in equation (1)
y
=
9
+
π
y
=
9
+
π
Putting the value
x
=
2
x
=
2
in equation (1)
y
=
9
−
2
π
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
x
=
4
y
…………equation (1)
When
y
=
0
y
=
0
, Putting the value in equation (1)
x
=
4
×
0
x
=
4
×
0
Or,
x
=
0
x
=
0
Putting the value
y
=
1
y
=
1
in equation (1)
x
=
4
×
1
x
=
4
×
1
Or,
x
=
4
x
=
4
Putting the value
y
=
−
1
y
=
-
1
in equation (1)
x
=
4
×
−
1
x
=
4
×
-
1
Or,
x
=
−
4
x
=
-
4
Putting the value
y
=
2
y
=
2
in equation (1)
x
=
4
×
2
x
=
4
×
2
Or,
x
=
8
x
=
8
Hence, four solutions for equation
x
=
4
y
x
=
4
y
are (0, 0), (1, 4), (-4,-1), (8, 2).
Answered by
0
Answer:
y = 3 x + 5 y = 3 x + 5 is a linear equation in two variables.
Step-by-step explanation:
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