An equation that can be written in the form ax + b =0, where a and b are constants, is called a linear equation in one variable.
For example, x - 4 =10 is a linear equation in x.
A value, such that, when you replace the variable with it, makes the equation true is called the solution of the equation.
If a = b, then a + c = b + c
For example:
Solve: x - 10 = -2
Solution:
x - 10 = -2
or, x - 10 + 10 = -2 + 10, addition axiom
\(\therefore\) x = 8
If a = b, then a-c = b-c
For example:
Solve for x: x+2 = 6
Solution:
x+2 =8
or, x = 8 - 2
\(\therefore\) x =6
If a = b, then ac = bc
For example:
Solve for x: \(\frac{x}{2}\) = 5
Solution:
\(\frac{x}{2}\) = 5
or, x = 5 \(\times\) 2
\(\therefore\) x = 10.
To solve word problems on linear equations apply the following steps:
Examples
An inequality is a relationship between two quantities that are not equal. In equations, one side is equal to the other side. In linear inequalities, one side is bigger than or smaller than or equal to the other side.
A linear equation in one variable has only one solution. An equality in one variable has a set of possible solutions.
The symbols used in inequality are:
Example:
Given that x is an integer. State the possible integer values of x in the following inequalities.
Solution
Solving linear inequalities in one variable
Solving linear inequalities are the same as solving linear equations with one important exception.
When you multiply or divide an inequality by a negative value, it changes the direction of the inequality.
Look at this statement: 5> 2
Suppose we multiply both sides by -1.
( -1 ) (5 )> (- 1 ) ( 2 )
- 5 <-2
-5 less than -2 because it is further to the left on the number line.
Thus, -5 < -2
The two equations having two unknowns are called Simultaneous equations. They are called simultaneous because they must be solved at the same time. For example, x+y = 10 and x-y = 2 are simultaneous equations. To solve a simultaneous equation we must use graph. For this, we have to draw the lines of the equations, The point where the lines cross is the solution.
Example:
Solve graphically: x + y = 6, 2x - y = 8
Solution:
Given equations are x + y = 6 and 2x - y = 8
x + y = 6
y = 6 - x ...........(i)
Table for y = 6 - x:
x | 3 | 0 | 2 |
y | 3 | 6 | 4 |
Again, 2x -y = 8
2x - 8 = y
y = 2x - 8 ..........(ii)
Table for y = 2x - 8
x | 4 | 6 | 3 |
y | 0 | 4 | -2 |
Draw the lines of the equations on a graph. Since, both lines intersect at P(6, 2), solution of the given equations is x = 6 and y = 2
A general quadratic equation can be written in the form ax^{2}+ bx + c = 0, where x represents a variable or an unknown and a, b and c are constants with a≠ 0.(If a = 0, the equation is the linear equation.). For example: 2x^{2} = 5x +2 = 0, is a quadratic equation where a = 2, b = 5 and c=2
Solution of quadratic equations by factoring
To make the equation true, the solution of a quadratic equation consists of all numbers (roots). All quadratic equations have 2 solutions (i.e 2 roots).
Sometimes it is easy to find solutions or roots of a quadratic equation by factoring. Indeed, the basic principle to be used is: If a and b are real numbers such that ab = 0, then a = 0 or b = 0.
Examples
Solution:
Here given, 7x=21
Dividing by 7 in both side
\(\frac{7x}{7}\)=\(\frac{21}{7}\)
∴x=3
Solution:
Here given, x-8=9
Adding 8 on both side
x-8 +8=9+8
∴x=17
Solution:
Here given,
3x+4=13
Subtracting 4 from both sides
3x+4-4=13-4
or, 3x=9
Dividing 3 on both side
\(\frac{3x}{3}\)=\(\frac{9}{3}\)
∴x=3
Solution:
5x-14=8
or, 5x-14+14=8+14 [Adding 14 on both side]
or, 5x=22
Dividing by 5 on both side
\(\frac{5x}{5}\)=\(\frac{22}{5}\)
∴ x = \(\frac{22}{5}\)
Solution:
8x+9=10
or, 8x+9-9=10-9 [Subtracting 9 from both side]
or, 8x=1
or, \(\frac{8x}{8}\)=\(\frac{1}{8}\) [Dividing by 8 on both side]
∴ x = \(\frac{1}{8}\)
Solution:
13x-14 = 12
or, 13x-14+14 = 12+14 [Adding 14 on both side]
or, 13x = 26 [Dividing by 13 on both side]
or, \(\frac{13x}{13}\) = \(\frac{26}{13}\)
∴ x = 2
Solution:
2\(\frac{1}{2}\)% of 500 = x
or, 500×\(\frac{5}{2}\) = 100x( By cross multiplication)
or, \(\frac{500×5}{2×100}\) = x
or, \(\frac{25}{2}\) = x
∴ x = 12.5
Solution:
13% of x =6.5
or, x×\(\frac{13}{100}\) = 6.5
or, x×13 6.5×100
or, x = \(\frac{6.5×100}{13}\)
or, x =\(\frac{65×100}{130}\)
or, x = 5×10
∴ x = 50
Solution:
33% of x+x = 266
or, x+x×\(\frac{33}{100}\) = 266
or, \(\frac{100x+33x}{100}\) = 266
or, \(\frac{133x}{100}\) = 266
or, x = \(\frac{266×100}{133}\)
or, x = 2×100
∴ x = 200
(frac{x + 1}{6})=7
Solve:
13x - 14 = 12
Solve:
(frac{10x + 8}{6}) = 17
Solve:
9 + 14x = 27 - 11x
Solve:
4x + 7 = 3x + 10
Solve:
4(x+4) = 3(x-1)
Solve:
(frac{x-2}{x+2}) = (frac{4}{3})
Solve:
10% of x.
Solve:
10% of x = 35
Solve:
50% of x+x = 381
If the sum of 2 number is 20 and one number is 4 times greater than other.Find the value of numbers?
Solve:
(frac{3 - 4x}{5 - 4x}) = (frac{7}{2})
The sum of three consecutive odd numbers is 57. Find the numbers.
Two-third of a number is 20 less than the original number. Find the number.
Solve:
17 - 8y = 5 - 20y
You must login to reply
Algebra
If x 1\x=6,Find the value of x³ 1\x³
Mar 26, 2017
0 Replies
Successfully Posted ...
Please Wait...
puja
solve graphically2x-y=5 and x-y=1
Mar 19, 2017
0 Replies
Successfully Posted ...
Please Wait...
Samriddha
A number is such that it is as much greater than 112 as it is less than it. Find the number
Feb 08, 2017
0 Replies
Successfully Posted ...
Please Wait...
L.K. Mahato
A is 25 years older than B,and A's age is as much above 20 as B's age is below 85. Find their ages.
Jan 30, 2017
0 Replies
Successfully Posted ...
Please Wait...