Today our topic of discussion is Solving Linear Equations.
Solving Linear Equations
Solving Linear Equations
In case of solving equations, some rules are to be applied. If the rules are known, solution of equations becomes easier. The rules are as follows:
- If the same number or quantity is added to both sides of an equation, two sides remain equal.
- If the same number or quantity is subtracted from both sides of an equation, two sides remain equal.
- If both sides of an equation are multiplied by the same number or quantity, the two sides remain equal.
- If both sides of an equation are divided by same non- zero mumber or quantity, the two sides remain equal.
The rules stated above may be expressed in terms of algebraic expressions as follows:
If x = a and c ≠ 0 ,
(i) x + c = a + c (ii) x – c = a – c (iii) xc = ac (i)
x/c = a/c Besides,if a,b and c are three quantities, if a =b + c ,a – b = c and if a + c = b ,a =b – c
This law is known as transposition law and different equations can be solved by applying this law. If the terms of an equation are in fractional form and if the degree of the variables in each numerator is I and the denominator in each term is constant, such equations are linear equations.
Example 1. Solve: (5x)/7 – 4/5 = x/5 – 2/7
Solution: (5x)/7 – 4/5 = x/5 – 2/7 or, (5x)/7 – x/5 = 4/5 – 2/7 o*r_{r} * (25x – 7x)/35 = (28 – 10)/35 * o*r_{r} * (18x)/35 = 18/35 [by transposition]
or, 18x = 18or, x = 1
∴ Solution is 1
Now, we shall solve such equations which are in quadratic form. These equations are transformed into their equivalent equations by simplifications and lastly the equations is transformed into linear equation of the form ax = b Again, even if there are variables in the denominator, they are also transformed into linear equation by simplification.
Example 2. Solve: (y – 1)(y + 2) = (y + 4)(y – 2)
Solution: (y – 1)(y + 2) = (y + 4)(y – 2)
or, y² – y + 2y – 2 = y ^ 2 + 4y – 2y – 8
or, y – 2 = 2y – 8
or, y – 2y = – 8 + 2 [by transposition]
0r ,-y=-6
or, y = 6
Solution is y = 6
Example 3. Solve and write the solution set: (6x + 1)/15 – (2x – 4)/(7x – 1) = (2x – 1)/5
Solution: (6x + 1)/15 – (2x – 4)/(7x – 1) = (2x – 1)/5
or, (6x + 1)/15 – (2x – 1)/5 = (2x – 4)/(7x – 1) [by transposition]
or,(6x + 1 – 6x + 3)/15 = (2x – 4)/(7x – 1)
or, 4/15 = (2x – 4)/(7x – 1)
or, 15(2x – 4) = 4(7x – 1) [by cross-multiplication]
or, 0.3x – 60 = 28x – 4 x * 0.3x – 28x = 60 – 4 [by transposition]
or, 2x = 56
Solution is x = 28
or, 28 and solution set is S = {28}
Example 4. Solve:1/(x – 3) + 1/(x – 4) = 1/(x – 2) + 1/(x – 5)
Solution: 1/(x – 3) + 1/(x – 4) = 1/(x – 2) + 1/(x – 5)
(x – 4 + x – 3)/((x – 3)(x – 4))
= (x – 5 + x – 2)/((x – 2)(x – 5)) (2x – 7)/(x ² – 7x + 12)
= (2x – 7)/(x² – 7x + 10)
Values of the fractions of two sides are equal. Again, numerators of two sides are equal, but denominators are unequal. In this case, if only the value of the numerators is zero, two sides will be equal.
2x – 7 = 0
or , 2x=7
∴ Solution is x = 7/2
∴ x = 7/2
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