A presentation on Linear equation for class 10

2. Standard form of Linear Equation in
two variable
ax + by +c= 0
● The degree of each variable is 1.
● The variables are added or subtracted.
● a or b are not equal to zero.
● Besides x and y, other commonly used variables are m and
n, a and b, and r and s.
● There are no radicals(√) in the equation.
● Every linear equation graphs as a line.

3. Examples of linear equations
2x + 4y =8
6y = 3 – x
x = 1
-2a + b = 5
4
7
3
x y
 


4. Examples of Nonlinear Equations
4x2 + y = 5
xy + x = 5
s/r + r = 3
4x 
The exponent is 2
There is a radical in the equation
Variables are multiplied
Variables are divided
The following equations are NOT in the
standard form of ax + by +c=0:

5. x and y -intercepts
● The x-intercept is the point where a line crosses the x-
axis.
The general form of the x-intercept is (x, 0). The y-
coordinate will always be zero.
● The y-intercept is the point where a line crosses the y-
axis.
The general form of the y-intercept is (0, y). The x-
coordinate will always be zero.

6. Finding the x-intercept
● For the equation 2x + y = 6, we know that y must
equal 0. What must x equal?
● Plug in 0 for y and simplify.
2x + 0 = 6
2x = 6
x = 3
● So (3, 0) is the x-intercept of the line.

7. Finding the y-intercept
● For the equation 2x + y = 6, we know that x must
equal 0. What must y equal?
● Plug in 0 for x and simplify.
2(0) + y = 6
0 + y = 6
y = 6
● So (0, 6) is the y-intercept of the line.

8. Find the x and y- intercepts
of x = 4y – 5
● x-intercept:
● Plug in y = 0
x = 4y - 5
x = 4(0) - 5
x = 0 - 5
x = -5
● (-5, 0) is the
x-intercept
● y-intercept:
● Plug in x = 0
x = 4y - 5
0 = 4y - 5
5 = 4y
= y
● (0, )is the
y-intercept
5
4
5
4

9. Solution: Substitution Method
• Given:
y +2x= 4…….(i)
7x – 2y = 3………(ii)
From (i) y= -2x+4
Putting y in equation(ii)
7x – 2(-2x + 4 = 3
7x + 4x – 8 = 3
11 x = 11
x = 1
Now again putting the value of x in equation (i)
y = -2(1) + 4 = 2
• Solution
(1, 2)

10. Solution: Eliminating One Variable
• Given:
3x + 4y = -10 ……..(i)
5x – 2y = 18……….(ii)
(i) x 1= 3x + 4y = -10
(ii) x 2 =(10x – 4y) = 36
Adding these two equation
13x = 26
x = 2
Putting value of x in equation (i)
3(2) + 4y = -10
4y = -16
y = -4
Therefore the solution is (2, -4)

11. Compare the Ratio Graphic
representation
Algebraic
interpretation
Linear equations
Intersection lines at
one point
Exactly one solution
or
unique solution
Consistent(Unique
solution)
Coincident line Infinity solution or
many solutions
Dependent and
consistent
Parallel lines No solution Inconsistent

13. A chemist has two solutions which is 50% acid and another which is 25% acid.
How much of each should be mixed to make 10 liters of 40% acid solution.
Ans: Let 50 % acids in the solution be x Let 25 % of
other solution be y
Total Volume in the mixture = x + y
A.P.Q:
x + y = 10 --------(1)
A.P.Q: 50%of x + 25%of y = 40% ×10
2x + y = 16 --------(2)
So x = 6 & y = 4

14. Q.In an election contested between A and B, A obtained votes
equal to twice the no. of persons on the electoral roll who did not
cast their votes & the number of person who did not caste vote
was equal to twice the majority of A over B. If there were 18000
persons on the electoral roll. How many voted for B.Ans: Let x and y be the no. of votes for A & B respectively.
The no. of persons who did not vote = (18000 – x – y)
APQ:
x = 2(18000 – x – y)
=> 3x + 2y = 36000 ---------------(1)
Again APQ
(18000 – x – y) = 2 (x – y)
3x – y = 18000 ----------------(2)
On solving we get, y = 6000 and x = 8000
Vote for B = 6000

15. Q. The population of the village is 5000. If in a year, the number of males were to increase by
5% and that of a female by 3% annually, the population would grow to 5202 at the end of the
year. Find the number of males and females in the village.
APQ,
X + Y = 5000………………..(i)
APQ
X+(5% of x) + y +( 3% of y)=5202
X +x/20 + y +3y/100=5202……………………….(ii)
On solving 1 & 2 we get x= 2600 y=2400
No. of males = 2600
No. of females = 2400

16. Q. The income of a and b are in the ratio of 3:2 and their expenditures are in the ratio
of 5:3 . If each saves rs 1000 the income of B is?
let income of A is 3x
and income of B is 2x
same way expenditure ratio is 5:3
let expenditure of A is 5y
and expenditure of B is 3y
saving of A will be 3x-5y =1000 (i) multiplied by 3
saving of B will be 2x-3y =1000 (ii) multiplied by 5
we will get
9x-15y =3000
10x-15y =5000
by elimination method
-x =-2000
x =2000
income of A will be 3x=6000
income of A = 6000
income of B will be 2x =2x2000 =4000

17. Q. When the son will be as old as the father today their ages will add up to 126 years.
When the father was old as the son is today, their ages add upto 38 years. Find their
present ages
let the son’s present age be x
Father’s age be y
Difference in age (y – x)
Of this difference is added to the present age of son, then son will be as old as the
father now and at that time, the father’s age will be [ y + (y – x)]
APQ:
[x + (y – x)] + [y (y – x)] = 126
[y + (x – y)] + [x + (x – y)] = 38
Solving we get the value of x and y
Q.A says to B "my present age is five times your that age when I was as old as you are now." If
the sum of their present ages is 48 years. Find their present ages
Let Age of A=x years and age of B= y years
APQ-1
x + y =48………………….(i)
APQ-2
X = 5{ y –(x –y)}………………….(ii)
On solving
We get A=30 years and B= 18 years

18. Q.A train covers a certain distance at a uniform speed. if the train had been 10 km/hr
faster, it would have taken 2 hour less than the scheduled time. if the train was
slower by 10 km/hr, it would have taken 3 hours more than the scheduled time.
Write the two equation
Let the speed=x and the time taken=y therefore Distance=speed . Time=xy
First statement if speed increased by 10 km per hour then distance covered in 2
hours. Distance= speed ×time
(X+10)(y-2)=xy
Xy -2x +10y -20=xy
-2x +10y=20
x -5y=-10………………(i)
And second statement if speed decreased by 10 km per hour them distance covered
in 03 hours
(x-10)(y+3)=xy
3x -10y=30………………….(ii)
On Solving we get
Distance= 50×12
Distance= 600 kms.