Explaination needed .
Q.1: Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:
(i) x – y/5 – 10 = 0
(ii) -2x+3y = 6
(iii) y – 2 = 0
Q.2. Write four solutions for each of the following equations:
(i) 2x + y = 7
(ii) πx + y = 9
Q.3: Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.
Q.4: Draw the graph of each of the following linear equations in two variables:
(i)y = 3x
Q.5: If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.
Q.6: Show that the points A (1, 2), B ( – 1, – 16) and C (0, – 7) lie on the graph of the linear equation y = 9x – 7.
Class 9 Mathematics
Chapter 4 .
Note : Explaination needed .
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Answers
Step-by-step explanation:
Answers:-
Q-1:-
i) Given equation is x – y/5 – 10 = 0
=>( 5x -y-50)/5 = 0
=> 5x-y-50 = 0×5
=> 5x-y-50=0
On Comparing this with ax+by+c = 0 then
a = 5 , b = -1 and c = -50
(ii) Given equation is -2x+3y = 6
=> -2x +3y-6 = 0
On Comparing this with ax+by+c = 0 then
a = -2 , b = 3 and c = -6
iii) Given equation is y – 2 = 0
=> 0x + y -2 = 0
On Comparing this with ax+by+c = 0 then
a = 0 , b= 1 and c = -2
Q-2:-
i)Given equation is 2x+y = 7
Put x = 0 then
=> 2(0)+y = 7
=> 0+y = 7
=> y = 7
1st solution= (0,7)
Put y = 0 then
=>2x +0 = 7
=> x = 7/2
2nd solution = (7/2,0)
Put x = 1 then
2(1)+y = 7
=> 2+y = 7
=>y = 7-2
=> y = 5
3rd solution = (1,5)
Put y = 1 then
2x+1 = 7
=> 2x = 7-1
=>2x = 6
=>x = 6/2
=>x = 3
4th solution=(3,1)
ii)Given equation is πx+y=9
Put x = 0 then
=> π(0)+y = 9
=> 0+y = 9
=> y = 9
1st solution= (0,9)
Put y = 0 then
=>πx +0 = 9
=> x = 7/π
2nd solution = (9/π,0)
Put x = 1 then
π(1)+y = 9
=> π+y = 9
=>y = 9-π
=> y = 5
3rd solution = (1,9-π)
Put y = 1 then
πx+1 = 9
=> πx = 9-1
=>πx = 8
=>x = 8/π
=>x = 3
4th solution=(8/π,1)
Q-3:-
Given equation is 2x + 3y = k.
if x = 2, y = 1 is a solution then it satisfies the given equation
=> 2(2)+3(1)=k
=> 4+3=k
k=7
The value of k = 7
Q-4:-
See the attachment
Q-5:-
Given equation is 3y = ax + 7
If the point (3, 4) lies on the graph of the equation then it is a solution
Put x = 3 and y = 4 then
=> 3(4)=a(3)+7
=> 12 = 3a+7
=> 12-7 = 3a
=>3a = 5
=> a = 5/3
The value of a = 5/3
Q-6:-
Given equation is y = 9x – 7.
If A (1, 2), B ( – 1, – 16) and C (0, – 7) lie on the graph of the linear equation y = 9x – 7 then they are solutions.
Put x = 1 and y = 2 then
2=9(1)-7
= 2=2
and Put x = -1 and y=-16
=> -16 = 9(-1)-7
=>-16=-16
and Put x =0 and y = -7
=> -7 = 9(0)-7
=>-7=-7
A ,B,C are the solutions of y = 9x-7
A (1, 2), B ( – 1, – 16) and C (0, – 7) lie on the graph of the linear equation y = 9x – 7 then they are solutions.
