There we discussed about Solving Simple Equations.
That knowledge is a prerequisite here.
Transposition in Linear Equations
If you look at the solution of Example 1 of Linear Equations above, the solution of x - 5 = 10 is x = 10 + 5. i.e. -5 present on the L.H.S. is taken to the R.H.S. as +5.
Similarly, in Example 2 of Linear Equations above, the solution of x + 9 = 4 is x = 4 - 9. i.e. +9 present on the L.H.S. is taken to the R.H.S. as -9.
In Example 3 of Linear Equations above, the solution of p⁄3 = 5 is p = 5 x 3. i.e. 3 dividing the L.H.S. is taken to the R.H.S. as 3 multiplying.
Similarly, In Example 4 above, the solution of 5y = -20 is y = -20⁄5. i.e. 5 multiplying the L.H.S. is taken to the R.H.S. as 5 dividing.
These ideas can be generalised.
Any term of an equation may be moved (transposed) to the other sidewith the sign changed.
Any non zero quantity multiplying a side may be moved (transposed) to the other side with the quantity dividing the other side. Any non zero quantity dividing a side may be moved (transposed) to the other side with the quantity multiplying the other side.
Changing a term or quantity from one side of an equation to the other side is called transposition.
Example 1 of Easy Algebra
Solve the Equation x + 9 = 9.
Solution: The given equation is x + 9 = 9. Transposing +9 from the L.H.S. to the R.H.S., we get x = 9 - 9 = 0. From this it is clear that
If the same term (same both in magnitude and sign) is present on both sides of an equation, then it can be cancelled.
Example 2 of Easy Algebra
Solve the Equation 3x + 12 = 9.
Solution: The given equation is 3x + 12 = 9. Transposing +12 from the L.H.S. to the R.H.S., we get 3x = 9 - 12 = -3. Transposing 3 which is multiplying the L.H.S. to the R.H.S., we get x = -3⁄3 = -1. Ans.
Example 3 of Easy Algebra
Solve the Equation 4p + 12 = 36 - 2p.Solution to Example 3 of Easy Algebra : The given equation is 4p + 12 = 36 - 2p. Transposing -2p from the R.H.S. to the L.H.S., and 12 from the L.H.S. to the R.H.S., we get 4p + 2p = 36 - 12 ⇒ 6p = 24. Transposing 6 which is multiplying the L.H.S. to the R.H.S., we get p = 24⁄6 = 4. Ans.
The method we adopted here in solving may be summarised as follows.
To solve a simple equation, the terms containing unknown quantity (variable) should be collected on the L.H.S. of the equation and constants on the other side. Then the coefficientof the variable should be reduced to unity.
Solution to Example 4 of Easy Algebra : The given equation is 9y + 5 = 15y - 1 Transposing 15y from the R.H.S. to the L.H.S., and 5 from the L.H.S. to the R.H.S., we get 9y - 15y = -1 - 5 ⇒ -6y = -6. Transposing -6 which is multiplying the L.H.S. to the R.H.S., we get p = -6⁄-6 = 1. Ans.
Exercise : Easy Algebra
Solve the Equation 3(x + 2) - (x - 8) = 3(x + 8)
Solve the Equation 4m - 10 = 34
Solve the Equation 2y + 5 = 8
Solve the Equation 9 - 7x = 5 - 3x
Progressive Learning of Math : Easy Algebra
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