liear equations in one variable, its properties. solutions of simple equations, solving word problems based on simple equations.

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LINEAR EQUATIONS :

LINEAR EQUATIONS IN ONE VARIABLE

Slide 2:

SIMPLE EQUATIONS
If the expressions x-3 &5-x are equal it is written as x-3=5-x; which is an equation.
Seven subtracted from a number(y), equals eight. => y-7=8

Slide 3:

PROPERTIES OF AN EQUATION
If same quantity is added to both sides of the equation, the sums are equal.
Thus: x=7 => x+a=7+a
If same quantity is subtracted from both sides of an equation, the differences are equal
Thus: x=7 => x-a=7-a
If both the sides of an equation are multiplied by the same quantity, the products are equal.
Thus: x=7 => ax=7a
If both the sides of an equation are divided by the same quantity, the quotients are equal.
Thus: x=7 => x÷a=7÷a

Slide 4:

TO SOLVE AN EQUATION
1.To solve an equation of the form x+a=b
E.g.: Solve x+4=10
Solution: x+4=10 => x+4-4=10-4 (subtracting 4 from both the sides)
=> x=6
2.To solve an equation of the form x-a=b
E.g.: Solve y-6=5 equal.
Solution: y-6=5 => y-6+6=5+6 (adding 6 to both sides)
=> y=11
1.To

Slide 5:

3.To solve an equation of the form ax=b
E.g.: Solve 3x=9
Solution: 3x=9 =>
=> x = 3
4. To solve an equation of the form x/a=b
E.g.: Solve = 6
Solution: =6 => ×2=6×2
=> x=12

Slide 6:

SHORT- CUT METHOD (SOLVING AN EQUATION BY TRANSPOSING TERMS)
1. In an equation, an added term is transposed (taken) from one side to the other, it is subtracted.
i.e., x+4=10
=> x=10-4=6 (4 is transposed)
2. In an equation, a subtracted term is transposed to the other side, it is added.
i.e., y-6=5
=>y=5+6=11 (6 is transposed)
3. In an equation, a term in multiplication is transposed to the other side, it is divided.
i.e., 3x=12
=> x=12/3=4 (3 is transposed)
4. In an equation a term in division is taken to the other side it is multiplied.
i.e., y/4=6
=> y=6×4=24 (4 is transposed)

Slide 7:

TO SOLVE EQUATIONS USING MORE THAN ONE PROPERTY
Solve: (1) 3x+8=14 Solution: 3x=14-8 (transposing 8)
=> 3x=6
=> x=6/3 (transposing 3)
=>x=2

Slide 8:

2a-3=5
Solution: 2a=5+3 (transposing 3)
=> 2a=8
=> a = 8/2 (transposing2)
=> a = 4

Slide 9:

(3) 5n/8 =20
Solution: 5n=20×8
=> n =204×8/51
=> n=4×8=32

Slide 10:

SOLVING AN EQUATION WITH VARIABLE ON BOTH THE SIDES
Transpose the terms containing the variable, to one side and the constants to the other side
.
E.g.: (1) Solve 10y-3=7y+9
Solution: 10y-7y = 9+3 (transposing 7y to the left & 3 to the right)
=> 3y = 12
=> y = 12/3
=> y = 4

SOLVING WORD PROBLEMS
A number increased by 8 equal 15. Find the number?
Solution: Let the number be ‘x’
Given, the number increased by 8 equal 15.
=> x+8 = 15
=> x = 15-8
=> x = 7

Slide 13:

A number is decreased by 15 and the new number so obtained is multiplied by 3; the result is 81.Find the number?
Solution: Let the number be ‘x’
The number decreased by 15 = x-15
The new number (x-15) multiplied by 3 = 3(x-15)
Given 3(x-15) = 81
=> 3x-45 = 81
=> 3x = 81 + 45
=> 3x = 126
=> x =
=> x = 42

Slide 14:

3) A man is 26 years older than his son. After 10 years, he will be three times as old as his son. Find their present ages
.
Solution: let son’s present age= x years
Then father’s age = x+26
After ten years,
Son’s age = x+10
Father’s age = x+26+10 =x+36
Given, x+36 = 3(x+10)
=> x+36 =3x+30
=> x-3x =30-36
=> -2x =-6
=> x = -6/-2
=>x=3
Son’s age = 3 years
Father’s age = 3 + 26=29years

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