Total students in the class=y
Leth the boys in the class= x, then equation between the students and the boys
\(y=\frac{4}{3}x\)
Now

Now, draw a graph between these points.

From the graph it is clear that there are 30 boys in a class of 40 students.

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.

Equation of p is x=-1
Equation of r is y=-2
Yes, the equation of q is 
x+y=1
Area \(=\frac{1}{2}\times4\times 4=8\) sq.units

Given equation is 3x+4y=6
(i) When it cuts x-axis then
put y=0, i.e 3x=6
x=2
Hence point on x-axis is (2,0)
(ii) when it cuts y-axis then
put x=0 i.e., 4y=6
y=3/2
hence point on y-axis is \((0,{3\over2})\)
3x+4y=6
\(\Rightarrow y=\frac{6-3x}{4}\)

\({2\over3}x-y=2\)
\(\Rightarrow 2x-3y=6\)
\(\Rightarrow 2x=3y+6\)
\(\Rightarrow x=\frac{3y+6}{2}\)
(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)

According to the question, the linear
equation for the given problem is
=> y = 500 + 20x ... (1)
Table of solution

We plot the points (0, 500) and (5, 600) on a graph paper and join the same by a ruler to get the line which is the graph of the equation (1).

We have
y=4x+3
Table of solution

We plot the points (0,3) and (1, 7) on a graph paper and join the same by a ruler to get the line, which is the graph of the equation y = 4x + 3.
This gives the work wages graph of the given equation.
Now, on x-axis, take a point P(4, 0). From P draw a line parallel to y-axis intersecting the work wage graph at Q. From Q, draw a line parallel to x-axis to intersect the y-axis at R. We see that R is (0, 19).
Hence, the wages of a laborer who puts in 4 hours of work is Rs.19.

x = 3 represents a line parallel to y-axis at a distance of 3 units to the right of the origin.
4x= 3y
\(\Rightarrow\ \ \ y={4x\over 3}\)
Table of solution

We plot the points (0,0) and (3, 4) on a graph paper and join the same by a ruler to get the line which is the graph of the equation 4x = 3y.

Area of the triangle GAB formed by the given two lines and the x-axis \(={3\times 4\over2}=6\) square units

4x + 3y = 36     .....(1)
⇒ 3y=36-4x
⇒ \(y={36-4x\over 3}\)
Table o solution

We plot points (0, 12) and (9, 0) on a graph paper and join the same by a ruler to get the line which is the graph of the equation (1).
Mark a point P(3, 0) on x-axis. From P draw PQ II y-axis to intersect the graph at Q. From Q, draw QR II x-axis to intersect the y-axis at R(O, 8).

Hence, v= 8 when x = 3.
Again, mark a point A(O, 6) on the y-axis. From A draw AB II x-axis to intersect the graph at B. From B, draw BC II y-axis to intersect the x-axis at \(C\left({9\over 2},0\right)\).
Hence \(x={9\over 2}\) when y=6