LINEAR EQUATIONS : LINEAR EQUATIONS IN ONE VARIABLE
Slide 2: EQUATION A statement which states that two algebraic expressions are equal is called an equation. LINEAR EQUATIONS IN ONE VARIABLE The equation involving only one variable in first order is called a linear equation in one variable.
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
4. In an equation a term in division is taken to the other side it is multiplied.
i.e
=> y=6×4=24 (4 is transposed) (3 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
Slide 11: (2) Solve 2(x-5) + 3(x-2) = 8+7(x-4)
Solution: 2x-10+3x-6=8+7x-28 (removing the brackets)
=> 5x-16 = 7x-20
=> 5x-7x = -20+16
=> -2x = -4
=> x = -4/-2
=> x = 2
Slide 12: 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 =
=>x=3
Son’s age = 3 years
Father’s age = 3 + 26=29years