The line parallel to the y-axis at a distance of 4 units from the origin and on the right of the y-axis is given by x=4.
Putting x=4 in 2x+3y=6, we get
2X4+3y=9
3y=9-8
3y=1
\(y={1\over3}\)
The required point \((4,\frac{1}{3})\)

Let Age of Amit= x years
Age of Akhil= y years
(i) According to the question the linear equation for the above situation isx+y=25
y=25-x

(ii) From the graph when Amit's age=14 years, then Akhil's age =11 years.

x+y=10\(\Rightarrow\)y=10-x

2x-y=5
y=2x-5

Plot these point on the graph paper

\({2\over3}x-y=2\)
\(\Rightarrow 2x-3y=6\)
\(\Rightarrow 2x=3y+6\)
\(\Rightarrow x=\frac{3y+6}{2}\)
(i) When the line cuts x-axis then put y=0
i.e., 2x=6
x=3
Hence point is (3,0)
(ii) When the line cuts y-axis then put x=0
i.e., 3y+6=0
y=-2
Hence point is (0,-2)

5x-2=3x-8
2x=-6
x=-3
(i) Point P(-3,0) represents the solution x=-3 on the number line.

(ii) Line AB represents the solution in the cartesian plane.

\(3x-12+{3\over7}x=2(x-1)\)
\(\Rightarrow\ \ {24\over7}x-12=2x-2\)
\(\Rightarrow\ \ \ {24\over 7}x-2x=12-2\)
\(\Rightarrow\ \ \ {10x\over 7}=10\)
\(x=7\)
The graph of this equation is a line parallel to y-axis at a distance of 7 units to the right of origin O.

The given linear equation
\(\Rightarrow\ \ y={2\over3}x+{1\over3}\ \ \ \ . . . .(1)\)
Table of solution

We plot the points (1, 1) and (4, 3) on a graph paper and join the same by a ruler to get the line which is the graph of the equation \(y={2\over3}x+{1\over3}\)

From graph, we see that the point (7, 5) lies on the graph, so it is a solution of the linear equation.

2x+5y=20
\(x={5\over2}y\)
\(\therefore 2\left({5\over2},y\right)+5y=20\)
⇒ 10y=20
⇒ y=2
\(\therefore x={5\over 2}(2)=5\)
Hence the required point is (5, 2).

if(-m, 3) is a solution of the equation
4x+9y-3=0, then
4(-m)+9(3)-3=0
⇒  -4m+27-3=0
⇒ -4m+24=0
⇒  4m=24
⇒ \(m={24\over 4}=6\)

7=2x
2x-7=0
2x+0y-7=0
Comparing with ax+by+c=0, we get
a=2
b=0
​​​​​​​c=-7