Question:
Solve the systems of equations by the addition method
3x - 4y = 7
5x - 8y = 8
Lets denote the first equation by R1, that is:
R1 : 3x - 4y = 7,
Now, lets denote the second equation by R2, that is:
R2 : 5x - 8y = 8
Now. Lets multiply R1 by a multiple of five. That is R1 = 5 R1. Then we have the new system of equations:
5(3x - 4y = 7)
5x - 8y = 8
That is equivalent to say
15x - 20y = 35
5x - 8y = 8
Now, Let's subtract to R1 the second equation multiply by 3, that is: R1 = R1 - 3 R2. So we obtain:
(15 x - 20 y ) - 3(5x - 8y) = 35 - 3(8),
that is
(15 x - 20 y ) - (15x - 24y ) = 35 - 24,
that is equivalent to
(15x - 15x) - 20y + 24y = 11,
that is equivalent to
4y = 11
so y = 11/4 ( EQUATION 1)
REPLACING EQUATION 1 IN EQUATION R1 OF OUR ORIGINAL EQUATION SYSTEM WE OBTAIN
3x - 4(11/4) = 7
that is
3x - 11 = 7
that is
3x = 7 + 11 = 18
that is
x = 18 / 3
So the final answer is x = 18 / 3 and y = 11/4. THIS METHOD IS EQUIVALENT TO THE GAUSS-JORDAN ELIMINATION.