Transcript Linear Equations in One Variable

Linear Equation
in
One Variable
Linear Equation in One Variable
A linear equation in one variable is an equation that
can be written in the form
ax + b = 0
Where a  0
For example:
5x – 4 = 0, 2x+1=0, 8x + 9 = 2
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
Solving Equations
To solve an equation means
to find all values that make
the equation a true
statement. Such values
are called solutions, and the
set of all solutions is called
the solution set.
Steps to Solve a Linear Equation in One
Variable
1. Simplify both sides of the equation.
• Clear parentheses.
• Consider clearing fractions or decimals (if any are present) by
multiplying both sides of the equation by a common
denominator of all terms.
• Combine like terms.
2. Use the addition or subtraction property of equality to collect the
variable terms on one side of the equation.
3. Use the addition or subtraction property of equality to collect the
constant terms on the other side of the equation.
4. Use the multiplication or division property of equality to make the
coefficient of the variable term equal to 1.
5. Check your answer.
Solve
11x  2  5(x  2)
11x  2  5x  10
-5x
-5x
6x  2  10
-2
-2
6x  12
6 x  12

6
6
x  2
* Linear expression :
2x,2x +1,3y-7,12-5z,5/4(x-4)+10
* Not linear expression :
1 + z + z²
(a)An algebraic equation is an equality involving variables.
It has a equality sign. The expression on the left of the
equality sign is the left hand side (LHS). The expression
on the right of the equality sign is the right hand side
(RHS).
(b)In an equation on the values of the expression on the
LHS and RHS are equal. This happens to be true only for
certain values are the solution of the equation.
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
=> y=11
(adding 6 to both sides)
3.To solve an equation of the form ax=b
E.g.: Solve 3x=9
Solution: 3x=9
=>
3x 9

3 3
=> x = 3
4. To solve an equation of the form x/a=b
E.g.: Solve x = 6
2
x =6
Solution:
2
=>
x
×2=6×2
2
=> x=12
Conditional Equations
The equation x+ 4 = 6 is a
conditional equation
because it is true on the
condition that x = 2. For
other values of x, the
statement x + 4 = 6 is
false.
Contradiction
Solve:
x+1= x+2
1 = 2 is a contradiction. This equation has no
solution. And it is NOT linear.
Identities
• An equation that has all real numbers as a
solution is an Identity.
For example,
X+4=X+4
4 = 4 is a true statement.
Therefore, the solution is all
real numbers.
Important points
• An algebraic is an equality variables. It says that the
values of the expression on one side of the equality
sign is equal to the value of the expression on the
other side.
• A linear equation may have for its solution any
rational number.
• The utility of linear equation is in their diverse
applications ; different problems on number
• The utility of linear equation is their diverse
applications ; different problems on numbers ,ages
,perimeters ,combination of currency notes , and so
on can solved using linear equation.
• An equation may have linear expression on both
sides.
Work sheet
Solve: x − 2 = 7
Ans. Transposing 2 to R.H.S, we obtain
x=7+2=9
Solve : y + 3 = 10
Ans. Transposing 3 to R.H.S, we obtain
y = 10 − 3 = 7
Solve : 14y − 8 = 13
Ans. Transposing 8 to R.H.S, we obtain
14y = 13 + 8
14y = 21
Dividing both sides by 14, we obtain
Qn . If you subtract from a number and multiply the result by, you get.
What is the number?
Ans. Let the number be x. According to the question,
On multiplying both sides by 2, we obtain
On transposing to R.H.S, we obtain
Therefore, the number is .
Solve the linear equation
L.C.M. of the denominators, 3 and 5, is 15.
Multiplying both sides by 15, we obtain
5(x − 5) = 3(x − 3)
⇒ 5x − 25 = 3x − 9 (Opening the brackets)
⇒ 5x − 3x = 25 − 9
⇒ 2x = 16