The document discusses equations and how to solve them. It defines an equation as a statement where two algebraic expressions are equal. It also defines a linear equation as one involving only one variable.
It then lists the four properties of equations: adding/subtracting the same quantity to both sides, multiplying/dividing both sides by the same quantity.
Next, it provides examples of how to solve different types of equations (those involving addition, subtraction, multiplication, or division of the variable) using both the traditional method of operations and the shortcut method of transposing terms.
Finally, it gives examples of solving equations with variables on both sides, word problems involving equations, and equations from applied contexts like age.
A statement whichstates that two algebraic expressions are equal is called
an equation.
96
823
553 2
xx
yx
xx
The equation involving only one variable in first order is called a linear
equation in one variable.
aa
y
x
3157
28
053
3.
PROPERTIES OF ANEQUATION
•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
4.
TO SOLVE ANEQUATION
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
5.
3.To solve anequation 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
3
9
3
3
x
2
x
2
x
2
x
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
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)
x
(3 is transposed)
7.
TO SOLVE EQUATIONSUSING MORE THAN ONE
PROPERTY
Solve: (1) 3x+8=14
Solution: 3x=14-8 (transposing 8)
=> 3x=6
=> x=6/3 (transposing 3)
=>x=2
SOLVINGAN EQUATIONWITH VARIABLEON
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
13.
• A numberis 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
3
126
14.
3) A manis 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