Mathematics, 03.12.2019 19:31, mayamcmillan11

Solve the systems of equations 6x+3y=9 2x+3y=1

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Solve by elimination.
The goal is to cancel out one of the variables in order to easily solve for the other variable.

Do this by changing the equations so that the coefficients of either x or y add up to 0.

Notice the coefficients of y are 3 and 3, if we make one of them negative then they add up to 0. 3+ (-3) = 0

Multiply 2nd equation by -1.

6x +3y = 9
-2x -3y = -1

4x +0y = 8

Solve for x
4x = 8
x = 8/4 = 2

Plug x=2 back into one of original equations to find y.
---> 2(2) + 3y = 1
---> 4 + 3y = 1
---> 3y = -3
---> y = -1

Therefore solution is (2,-1)

answered: Guest

a complex number is a number that can be expressed in the form a + bi, where a and b are real numbers, and i is a solution of the equation x2 = −1. because no real number satisfies this equation, i is called an imaginary number. has no real solution, since the square of a real number cannot be negative.

answered: Guest

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