问题1：以ax + by + c = 0的形式表示以下线性方程，并分别表示a，b和c的值：
(i)-2x + 3y = 12
(ii)x – y / 2 – 5 = 0
(iii)2x + 3y = 9.35
(iv)3x = -7y
(v)2x + 3 = 0
(vi)y – 5 = 0
(vii)4 = 3倍
(viii)y = x / 2
解决方案：
(i) -2x + 3y = 12
Rearranging,
– 2x + 3y – 12 = 0
On comparing with the given form of linear equation, ax + by + c = 0,
We get,
a = – 2
b = 3
c = -12
(ii) x – y/2 – 5 = 0
On comparing with the given form of linear equation, ax + by + c = 0,
We get,
a = 1
b = -1/2
c = -5
(iii) 2x + 3y = 9.35
Rearranging, 2x + 3y – 9.35 = 0
On comparing with the given form of linear equation, ax + by + c = 0,
We get,
a = 2
b = 3
c = -9.35
(iv) 3x = -7y
Rearranging, 3x + 7y + 0 = 0
On comparing with the given form of linear equation, ax + by + c = 0,
We get,
a = 3
b = 7
c = 0
(v) 2x + 3 = 0
Rearranging, 2x + 0y + 3 = 0
On comparing with the given form of linear equation, ax + by + c = 0,
We get,
a = 2
b = 0
c = 3
(vi) y – 5 = 0
Rearranging, 0x + y – 5 = 0
On comparing with the given form of linear equation, ax + by + c = 0,
We get,
a = 0
b = 1
c = -5
(vii) 4 = 3x
Rearranging, 3x + 0y – 4 = 0
On comparing with the given form of linear equation, ax + by + c = 0,
We get,
a = 3
b = 0
c = -4
(viii) y = x/2
Rearranging, x – 2y +0 = 0
On comparing with the given form of linear equation, ax + by + c = 0,
We get,
a = 1
b = -2
c = 0
问题2：将以下各项作为方程式写在两个变量中：
(i)2x = -3
(ii)y = 3
(iii)5倍= 7/2
(iv)y = 3 / 2x
解决方案：
(i) 2x = -3
Rearranging,
2x + 3 = 0
Now adding ‘y’ term,
2x + 0.y + 3 = 0
Required equation is,
2x + 0.y + 3 = 0
(ii) y = 3
Rearranging,
y – 3 = 0
Now adding ‘x’ term,
0.x + y – 3 = 0
Required equation is,
0.x + y – 3 = 0
(iii) 5x = 7/2
Rearranging,
10x = 7,
or 10x – 7 – 0;
Now adding ‘y’ term,
10x +0.y – 7 = 0
Required equation is,
10x + 0.y – 7 = 0
(iv) y = 3/2 x
Rearranging,
2y = 3x
or 3x – 2y = 0
Now adding the constant term,
3x – 2y + 0 = 0
Required equation is,
3x – 2y + 0 = 0
问题3：圆珠笔的成本比钢笔成本的一半低5卢比。将此语句写为两个变量的线性方程式。
解决方案：
Let the cost of a ball pen and fountain pen be x and y respectively.
According to the question the following equation can be formed,
x = y/2 − 5
or x = (y – 10)/2
or 2x = y – 10
or 2x – y + 10 = 0
The required linear equation will be 2x – y + 10 = 0.