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.)
Solution 1:
Let the cost of a notebook be ₹ x
Let the cost of a pen be ₹ y
Given: Cost of Notebook is twice the cost of Pen
Therefore, we can write the required linear equation in the form of,
Cost of notebook = 2 × Cost of pen x = 2y
x − 2y = 0
Question 2:
Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case:
(i) 2x + 3y=9.35
Solution
2x+3y=9.35 --------------- Equation (1)
2x+3y-9.35=0
Comparing this equation with standard form of the linear equation,ax + by + c = 0 we have,
a = 2,
b = 3,
c =-9.35
2x+3y-9.35=0
Comparing this equation with standard form of the linear equation,ax + by + c = 0 we have,
a = 2,
b = 3,
c =-9.35
(ii)\(x- {y\over 5} -10=0\)
Solution
(iii)−2x + 3y = 6
\(x- {y\over 5} -10=0\) --------------- Equation (1)
Comparing Equation (1) with standard form of the linear equation, ax + by + c = 0 we have,
a = 1,
\(b ={ -1\over5}\),
c = −10
Comparing Equation (1) with standard form of the linear equation, ax + by + c = 0 we have,
a = 1,
\(b ={ -1\over5}\),
c = −10
(iii)−2x + 3y = 6
Solution
(iv)x = 3y
−2x + 3y = 6 --------------- Equation (1)
−2x + 3y − 6 = 0
Comparing this equation with standard form of the linear equation, ax + by + c = 0 we have,
a = −2,
b = 3,
c = −6
−2x + 3y − 6 = 0
Comparing this equation with standard form of the linear equation, ax + by + c = 0 we have,
a = −2,
b = 3,
c = −6
(iv)x = 3y
Solution
(v)2x = − 5y
x = 3y --------------- Equation (1)
1x − 3y + 0 = 0
Comparing this equation with standard form of the linear equation, ax + by + c = 0 we have,
a = 1,
b = −3,
c = 0
1x − 3y + 0 = 0
Comparing this equation with standard form of the linear equation, ax + by + c = 0 we have,
a = 1,
b = −3,
c = 0
(v)2x = − 5y
Solution
(vi)3x + 2 = 0
2x = −5y --------------- Equation (1)
2x + 5y + 0 = 0
Comparing this equation with standard form of the linear equation, ax + by + c = 0 we have,
a = 2,
b = 5,
c = 0
2x + 5y + 0 = 0
Comparing this equation with standard form of the linear equation, ax + by + c = 0 we have,
a = 2,
b = 5,
c = 0
(vi)3x + 2 = 0
Solution
(vii)y − 2 = 0
3x + 2 = 0 --------------- Equation (1)
We can write Equation (1) as below,
3x + 0y + 2 = 0
Comparing this equation with ax + by + c = 0,
a = 3,
b = 0,
c= 2
We can write Equation (1) as below,
3x + 0y + 2 = 0
Comparing this equation with ax + by + c = 0,
a = 3,
b = 0,
c= 2
(vii)y − 2 = 0
Solution
(viii)5 = 2x
y − 2 = 0 --------------- Equation (1)
We can write Equation (1) as below,
0x + 1y − 2 = 0
Comparing this equation with standard form of the linear equation, ax + by + c = 0 we have,
a = 0,
b = 1,
c = −2
We can write Equation (1) as below,
0x + 1y − 2 = 0
Comparing this equation with standard form of the linear equation, ax + by + c = 0 we have,
a = 0,
b = 1,
c = −2
(viii)5 = 2x
Solution
5 = 2x --------------- Equation (1)
−2x + 0y + 5 = 0
Comparing this equation with standard form of the linear equation, ax + by + c = 0 we have,
a = −2,
b = 0,
c = 5
−2x + 0y + 5 = 0
Comparing this equation with standard form of the linear equation, ax + by + c = 0 we have,
a = −2,
b = 0,
c = 5
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