Subject: Compulsory Maths
The mathematical statement, which cannot be predicted as true or false statements (until the variable is replaced by any number) are known as an open statement.
x>5, x+3<4, 2y-1≥9 etc. are the example of an open statement.
Linear equations in one variable
The open statement containing 'equal to' (=) sign and can be true only for a fixed value of variable is called equation.
Further more in x+2 = 5, the equation has only one variable which is x. So, it is the equation of one variable. Also the variable x has power 1.So it is called a linear equation. Thus, x+2 = 5 is a linear equation in one variable.
Solution to equations
Suppose x+2 = 7
This equation can be true only for a fixed value of x which is 5. So, 5 is called the solution (or root) of the equation.The process of getting a solution to an equation. The process of getting a solution to an equation is called solving equation.
Applications of equations
Generally, we use equations to find the unknown value of any quantity. For this, we should consider the unknown value of the given verbal problems as the variable like x, y, a, b etc. Then the verbal problems should be translated into mathematical sentences in the form of equations. And by solving the equations we obtain the required values.
The process of getting a solution to an equation is called solving equation.
The verbal problems should be translated into mathematical sentences in the form of equations.
x + 3 = 5
Solution:
or, x + 3 = 5
or, x = 5 - 3
∴ x = 2
x - 2 = 10
Solution:
or, x - 2 = 10
or, x = 10 + 2
∴ x = 12
5x = 15
Solution:
or, 5x = 15
or, x = \(\frac{15}{5}\)
∴ x = 3
\(\frac{x}{2}\) = 7
Solution:
or, \(\frac{x}{2}\) = 7
or, x = 7 × 2
∴ x = 14
4(3 - x) = 2x - 15
Solution:
or, 4(3 - x) = 2x - 15
or, 12 - 4x = 2x - 15
or, -4x - 2x = -15 - 12
or, -6x = -27
or, x = \(\frac{27}{6}\)
∴ x = \(\frac{9}{2}\)
\(\frac{3x - 5}{3}\) + x = \(\frac{5}{6}\) - \(\frac{2x + 3}{2}\)
Solution:
or, \(\frac{3x - 5}{3}\) + x = \(\frac{5}{6}\) - \(\frac{2x + 3}{2}\)
or, \(\frac{3x - 5 + 3x}{3}\) = \(\frac{5 - 3(2x + 3)}{6}\)
or, \(\frac{6x - 5}{3}\) = \(\frac{5 - 6x - 9}{6}\)
or, \(\frac{6x - 5}{3}\) = \(\frac{-4 - 6x}{6}\)
or, 6(6x - 5) = 3 (-4 - 6x)
or, 36x - 30 = -12 - 18x
or, 36x + 18x = -12 + 30
or, 54x = 18
or, x = \(\frac{18}{54}\)
∴ x = \(\frac{1}{3}\)
Solve
x + 20% of x = Rs 180
Solution:
or, x + 20% of x = Rs 180
or, x + \(\frac{20}{100}\) × x = Rs 180
or, x + \(\frac{x}{5}\) = Rs 180
or, \(\frac{5x + x}{5}\) = Rs 180
or, \(\frac{6x}{5}\) = rs Rs 180
or, 6x = 5 × Rs 180
or, x = Rs \(\frac{900}{6}\)
or, x = Rs 150
∴ x = Rs 150
The sum of two numbers is 35. if one of the numbers is 21, find the other number.
Solution:
Let the other number be x.
Now,
or, x + 21 = 35
or, x = 35 - 21
or, x = 14
So, the required number is 14.
If the sum of three consecutive even numbers is 30, find them.
Solution:
Let the smaller even number be x.
Then, the second consecutive even number = x + 2
And, the third consecutive even number = x + 4
Now,
or, x + (x + 2) + (x + 4) = 30
or, 3x + 6 = 30
or, 3x = 30 - 6
or, x = \(\frac{24}{3}\)
or, x = 8
∴ The first even number = x = 8
The secomd even number = x + 2 = 8 + 2 = 10
Th e third even number = x + 4 = 8 + 4 = 12
So, the reuired consecutive even numbers are 8, 10 and 12.
A sum of Rs 50 is dived into two parts. If the greater part exceeds the smaller by Rs 10, find the parts of the sum.
Solution:
Let the smaller part of the sum be Rs x.
Then,
the greater part of the sum = Rs (x + 10)
Now,
or, x + (x + 10) = Rs 50
or, 2x = Rs (50 - 10)
or, x = Rs \(\frac{40}{2}\)
or, x = Rs 20
∴ The smaller part of the sum = x = Rs 20
The greater part of the sum = Rs (x + 10) = Rs (20 + 10) = Rs 30
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