问题1.笔记本的成本是笔的成本的两倍。在两个变量中编写一个线性方程式以表示该语句。
解决方案：
Let’s take the cost of a notebook as Rs. x and the cost of a pen as Rs. y.
Given that cost of a notebook (x) is twice the cost of a pen(y).
So, x = 2y.
x – 2y = 0
x – 2y = 0 is the linear equation in two variables that represent the statement.
问题2.将以下线性方程式以ax + x + c = 0的形式表示，并分别表示a，b和c的值：
(i)2x + 3y = 9.35
解决方案：
2x + 3y – 9.35 = 0 (Transposing 9.35 to LHS)
(2)x + (3)y + (-9.35) = 0 (converting it in the form ax + by + c = 0)
On comparing equation with ax + by + c = 0,
We get a = 2, b = 3, c = -9.35
(ii)x – y / 5 -10 = 0
解决方案：
(5x – y – 50)/5 = 0 (Multiply and divide the whole equation by 5)
5x – y – 50 = 0
(5)x + (-1)y + (-50) = 0 (converting it in the form ax + by + c = 0)
On comparing equation with ax + by +c =0,
We get a = 5, b = -1, c = -50
(iii)-2x + 3y = 6
解决方案：
-2x + 3y – 6 = 0 (Transposing 6 to LHS)
(-2)x + (3)y+ (-6) = 0 (converting it in the form ax + by + c = 0)
On comparing equation with ax + by +c =0,
We get a = -2, b = 3, c = -6
(iv)x = 3y
解决方案：
x – 3y = 0 (Transposing 3y to LHS)
(1)x + (-3)y + 0 = 0 (converting it in the form ax + by + c = 0)
On comparing equation with ax + by + c = 0,
We get a = 1, b = -3, c = 0
(v)2x = -5y
解决方案：
2x + 5y = 0 (Transposing 5y to LHS)
(2)x + (5)y + 0 = 0 (converting it in the form ax + by + c = 0)
On comparing equation with ax + by +c =0,
We get a = 2, b = 5, c = 0
(vi)x + 2 = 0
解决方案：
(1)x + (0)y + 2 = 0 (converting it in the form ax + by + c = 0)
On comparing equation with ax + by +c =0,
We get a = 1, b = 0, c = 2
(vii)y – 2 = 0
解决方案：
(0)x + (1)y + (-2) = 0 (converting it in the form ax + by + c = 0)
On comparing equation with ax + by +c =0,
We get a = 0, b = 1, c = -2
(viii)5 = 2倍
解决方案：
5 – 2x = 0 (Transposing 2x to LHS)
(-2)x + (0)y + 5 = 0 (converting it in the form ax + by + c = 0)
On comparing equation with ax + by + c = 0,
We get a = -2, b = 0, c = 5