Contents
- 1 Step-wise procedure for performing transposition method on linear equations in one variable:
- 1.1 Solving Linear Equations by Transposition method Example1:
- 1.2 Solving Linear Equations by Transposition method Example2:
- 1.3 Solving Linear Equations by Transposition method Example3:
- 1.4 Solving Linear Equations by Transposition method Example4:
- 1.5 Solving Linear Equations by Transposition method Example5:
Sometimes the two sides of an equation contain both variable (unknown quantity) and constants (numerals). In such cases, we first simplify two sides in their simplest forms and then transpose (shift) terms containing variable on R.H.S. to L.H.S and constant terms on L.H.S to R.H.S. By transposing (shifting) a term from one side to the other side, we mean changing its sign and carrying it to the other side.
In transposition the plus sign (+) of the term changes into minus sign (-) on the other side and vice-versa.
Step-wise procedure for performing transposition method on linear equations in one variable:
Step 1) Obtain the linear equation.
Step 2) Identify the variable(unknown quantity) and constants(numerals).
Step 3) Simplify the L.H.S. and R.H.S. to their simplest forms by removing brackets.
Step 4) Transpose all terms containing variable on L.H.S. and constant terms on R.H.S.
Step 5) Simplify L.H.S. and R.H.S. in the simplest form so that each side contains just one term.
Step 6) Solve the equation obtained in step (5) by dividing both sides by the coefficient of the variable on LHS.
Solving Linear Equations by Transposition method Example1:
Solve \(\frac{x}{2} – \frac{1}{5} = \frac{x}{3} + \frac{1}{4}\).
Solution: We have,
\(\frac{x}{2} – \frac{1}{5} = \frac{x}{3} + \frac{1}{4}\)
The denominators on two sides are 2,5,3, and 4. Their LCM is 60.
Multiplying both sides of the given equation by 60, we get
=> \(60(\frac{x}{2})\) – \(60(\frac{1}{5})\) = \(60(\frac{x}{3}) + 60(\frac{1}{4})\)
=> 30x – 12 = 20x + 15
=> 30x – 20x = 15 + 12 { On transposing 20x to LHS and -12 to RHS }
=> 10x = 27 => \(x = \frac{27}{10}\) { On dividing both sides by 10 }
Hence, \(x = \frac{27}{10}\) is the solution of the given equation.
Verification:
Substituting \(x = \frac{27}{10}\) in the given equation, we get
L.H.S. = \(\frac{x}{2} – \frac{1}{5} = (\frac{27}{10})(\frac{1}{2}) – \frac{1}{5} = \frac{27}{20} – \frac{1}{5}\)
= \(\frac{(27 – 1 x 4)}{20} = \frac{(27 – 4)}{20} = \frac{23}{20}\)
Thus, for \(x = \frac{27}{10}\), we have L.H.S. = R.H.S.
Solving Linear Equations by Transposition method Example2:
Solve \(\frac{x – 3}{5} + \frac{x – 4}{7} = 6 – \frac{2x – 1}{35}\).
Solution:
LCM of 5, 7 and 35 = 35. Therefore, multiplying both sides of the equation by 35, we get
=> 7(x – 3) + 5(x – 4) = 210 – (2x – 1)
=> 7x – 21 + 5x – 20 = 210 – 2x + 1
=> 7x + 5x – 21 – 20 = 210 + 1 – 2x
=> 12x – 41 = 211 – 2x => 12x + 2x = 211 + 41
=> 14x = 252 => x = \(\frac{252}{14}\) = 18.
Solving Linear Equations by Transposition method Example3:
Solve 0.16(5x – 2) = 0.4x + 7.
Solution: We have,
0.16(5x – 2) = 0.4x + 7
=> 0.8x – 0.32 = 0.4x + 7 { By expanding the bracket on LHS }
=> 0.8x – 0.4x = 0.32 + 7 { By transposing 0.4x to LHS and -0.32 to RHS }
=> 0.4x = 7.32
=> \(\frac{0.4x}{4} = \frac{7.32}{4}\) { By dividing both sides by 0.4 }
=> x = \(\frac{732}{40}\) = \(\frac{183}{10}\) = 18.3
Hence, x = 18.3 is the solution of the given equation.
Solving Linear Equations by Transposition method Example4:
Solve \(\frac{17 – 3x}{5} – \frac{4x + 2}{3} = 5 – 6x + \frac{7x + 14}{3}\).
Solution:
Multiplying both sides of the equation by 15 i.e., the LCM of 5 and 3, we get
3(17 – 3x) – 5(4x + 2) = 15(5 – 6x) + 5(7x + 14)
=> 51 – 9x – 20x – 10 = 75 – 90x + 35x + 70
=> 41 – 29x = 145 – 55x
=> -29x + 55x = 145 – 41
=> 26x = 104 => \(\frac{26x}{26} = \frac{104}{26}\)
=> x = 4
Thus, x = 4 is the solution of the given equation.
Solving Linear Equations by Transposition method Example5:
Solve \({(2x + 3)}^2 + {(2x – 3)}^2 = (8x + 6)(x – 1) + 22\).
Solution: We have,
\({(2x + 3)}^2 + {(2x – 3)}^2 = (8x + 6)(x – 1) + 22\)
=> \(2({(2x)^2 + 3^2}) = x(8x + 6) – (8x + 6) + 22\)
{ By using \({(a + b)}^2 + {(a – b)}^2 = 2(a^2 + b^2)\) }
=> \(2(4x^2 + 9) = 8x^2 + 6x – 8x – 6 + 22\)
=> \(8x^2 + 18 = 8x^2 – 2x + 16\)
=> \(8x^2 – 8x^2 + 2x = 16 – 18\)
=> 2x = -2 => x = -1
Hence, x = -1 is the solution of the given equation.