Question

The cost of a notebook is 5 less than twice the cost of a pen. Taking cost of notebook as x
and cost of pen as y, write a linear equation for the given statement and draw the graph for the
same.
​

Answer 1

1.   The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement.       (Take the cost of a notebook to be Rs. x and that of a pen to be Rs. y).

Sol: Let the cost of a notebook = Rs x            The cost of a pen = y            According to the condition, we have                            [Cost of a notebook] = 2 × [Cost of a pen]            i.e.            [x] = 2 × [Y]            or              x = 2y            or              x – 2y = 0       Thus, the required linear equation is × – 2y = 0.

2.   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)        (ii)        (iii) –2x + 3y = 6       (iv) x = 3y      (v) 2x = –5y               (vi) 3x + 2 = 0          (vii) y – 2 = 0           (viii) 5 = 2x

Sol: (i) We have                    Comparing it with ax + bx + c = 0, we have a = 2, b = 3 and  

     (ii) We have              

     

     Comparing with ax + bx + c = 0, we get      

Note: Above equation can also be compared by:Multiplying throughout by 5,          or 5x – y – 50 = 0or 5(x) + (–1)y + (–50) = 0Comparing with ax + by + c = 0, we get a = 5, b = –1 and c = –50.

(iii) We have             –2x + 3y = 6      ⇒                      –2x + 3y – 6 = 0      ⇒                      (–2)x + (3)y + (–6) = 0      Comparing with ax + bx + c = 0, we get a = –2, b = 3 and c = –6.

(iv) We have             x = 3y      x – 3y = 0      (1)x + (–3)y + 0 = 0      Comparing with ax + bx + c = 0, we get a = 1, b = –3 and c = 0.

(v) We have 2x = –5y      ⇒                      2x + 5y =0      ⇒                      (2)x + (5)y + 0 = 0      Comparing with ax + by + c = 0, we get a = 2, b = 5 and c = 0.

(vi) We have 3x + 2 = 0      ⇒                      3x + 2 + 0y = 0      ⇒                      (3)x + (10)y + (2) = 0      Comparing with ax + by + c = 0, we get a = 3, b = 0 and c = 2.

(vii) We have y – 2 = 0      ⇒                      (0)x + (1)y + (–2) = 0      Comparing with ax + by + c = 0, we have a = 0, b = 1 and c = –2.

(viii) We have 5 = 2x      ⇒                      5 – 2x = 0      ⇒                      –2x + 0y + 5 = 0      ⇒                      (–2)x + (0)y + (5) = 0      Comparing with ax + by + c = 0, we get a = –2, b = 0 and c = 5.

EXERCISE: 4.2

1.   Which one of the following options is true, and why?            y = 3x + 5 has

      (i) a unique solution, (ii) only two solutions, (iii) infinitely many solutionsSol: Option (iii) is true because a linear equation has an infinitely many solutions.

2.   Write four solutions for each of the following equations:

     (i) 2x + y = 7      (ii) πx + y = 9      (iii) x = 4ySol: (i) 2x + y = 7      When x = 0, 2(0) + y = 7      ⇒                      0 + y = 7      ⇒                      y =7      ∴ Solution is (0, 7).      When x = 1, 2(1) + y = 7      ⇒                      y = 7 – 2      ⇒                      y = 5      ∴ Solution is (1, 5).      When x = 2, 2(2) + y = 7      ⇒                      y = 7 – 4      ⇒                      y = 3      ∴ Solution is (2, 3).      When x = 3, 2(3) + y = 7      ⇒                      y = 7 – 6      ⇒                      y = 1      ∴ Solution is (3, 1).(ii) πx + y = 9      When x = 0 π(0) + y = 9      ⇒                      y = 9 – 0      ⇒                      y = 9      ∴ Solution is (0, 9).      When × = 1, π(1) + y = 9      ⇒                      y = 9 – π      ∴ Solution is {1, (9 – π)}      When x = 2, π(2) + y = 9      ⇒                      y = 9 – 2π      ∴ Solution is {2, (9 – 2π)}      When × = –1, π(–1) + y = 9      ⇒                      – π + y = 9      ⇒                      y = 9 + π      ∴ Solution is {–1, (9 + π)}(iii) x = 4y      When x = 0, 4y = 0      ⇒                      y = 0      ∴ Solution is (0, 0).      When x = 1, 4y = 1      ⇒                      y = 0      ∴ Solution is (0, 0)      When x = 4, 4y = 4      ⇒                            

Answer 2

Answer:

$x-2y+5=0$ is the required linear equation.

Step-by-step explanation:

Let the cost of $a$ notebook be Rs.$x$

Let the cost of $a$ pen be Rs.$y$

According to the question,

The cost of a notebook is $5$ less than twice the cost of a pen,

⇒ cost of a notebook = 2 × cost of a pen - 5

⇒ $x=2y-5$

⇒ $x-2y+5-0$

Therefore, the linear equation for the given statement is $x-2y+5=0$.

Graph of the linear equation.

Consider the linear equation as follows:

$x-2y+5=0$

(i) For $x=0$,

⇒ $0-2y+5=0$

⇒ $-2y=-5$

⇒ $y=5/2$

⇒ $y=2.5$

(ii) For $x=1$,

⇒ $1-2y+5=0$

⇒ $-2y+6=0$

⇒ $y=6/2$

⇒ $y=3$

(iii) For $y=0$,

⇒ $x-2(0)+5=0$

⇒ $x+5=0$

⇒ $x=-5$

The points $(x,y)$ for a linear equation is calculated in a table as follows:

$x$         0            1           -5

$y$        2.5          3           0

Plot all the points on the graph.

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