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Method
In this method we eliminate either x or y, get the value of other variable and then substitute that value in either of the original equations to get the value of the other variable. Let us look at it.
We are given two equations 3x + 4y = 10 and 4x + y = 9. We write them as follows.
3x + 4y = 10 ........(1)
4x + y = 9 ........(2)
In this example, let us eliminate y. Multiply equ. (2) by 4. We have
4(4x + y = 9)
which is 16x + 4y = 36 ........(3)
We observe that the coefficients of y in equations (1) and (3) are one and the same, and therefore, we subtract equ.(1) from equ.(3).
16x + 4y = 36
- (3x + 4y = 10)
13x + 0 = 26
At this point one should remember that the signs of equation (1) should be changed before subtracting.
13x = 26
x = 26/13 = 2
We now substitute the value of x = 2 in either equ. (1) or (2). Let us substitute in equ. (2).
4x + y = 9
4(2) + y = 9
8 + y = 9
y = 9 - 8 = 1
That is, the values of x and y for which both the equations are satisfied are x = 2 and y = 1.
We substitute these values in equ. (1) and check.
3x + 4y = 10
3(2) + 4(1) = 10
6 + 4 = 10
10 = 10
That is, LHS = RHS.
We're here going to take a brief detour and notice solutions to non-constant coefficient, second order differential equations of the form. p (t) y′′ + q (t ) y′ + r (t ) y = 0
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