Question

11) How many solution(s) of the equation 2x + 1 = x – 3 are there on the:
(i) Number line (ii) Cartesian plane

12) Find the solution of the linear equation x + 2y = 8 which represents a point on
(i) x-axis (ii) y-axis

13) For what value of c, the linear equation 2x + cy = 8 has equal values of x and y for its
solution?

Can you please answer this question

Answer 1

11)

Only One solution possible for the Number line

       $\tt 2x+1 = x-3 \\\\ \Rightarrow x = -4$

And for Cartesian Plane

the equation of the line is $\tt x+4=0$

So, there's only one solution possible for this equation is $\tt (x=-4\ \ and \ \ y= 0)$

12)

We have Equation as $\tt x+2y = 8$

Now

i) Here we have said that the point is on x axis

So, y = 0

Now putting value of y in the equation

$\tt x + 2(0) = 8 \\\\ \tt \Rightarrow x = 8$

So, the point is (8,0)

ii) Here we have said that point is on y axis

So, x = o

Now putting value of x in the equation

$\tt x+2y=8 \\\\ \Rightarrow 0 + 2y =8 \\\\ \Rightarrow 2y =8 \\\\ \Rightarrow y =4$

So, the point is (0,4)

13)

We have Given that x = y

So, replacing y at the place of x

So,

$\tt 2x+cx = 8 \\\\ \tt \Rightarrow x(2+c) = 8 \\\\ \tt \Rightarrow x = \dfrac{8}{(2+c)}$

Now observing c, we get that x is only positive integer, when c = 0 or c= 2 or c= 6

So, Taking all the cases

In Case 1

c = 0

2x + 0y = 8

=> 2x = 8

=> x = 4

So, here x = 4 and y = any number possible

Si, $\tt x \neq y$

In Case 2

c = 2

2x + 2y = 8

=> x + y = 4

x = y = 2 (x and y are equal here)

So, c = 2

In Case 3

c= 6

2x + 6y = 8

=> x + 3y = 4

x = y = 1 (x and y are equal here)

So, c = 6

So, possible values for c = 2, 6