To answer by systems of elimination, you add equations together to eliminate variables if they have the same amount of negative variables in one as the positive variables in another. For example, in number one.
x - y = 11 2x + y = 19
Both equations contain a variable expressed in the same numbers of negative and positive in each of them, one positive y and negative y. This lets us add the equations together. x + 2x = 3x 11 + 19 = 30 3x = 30 x = 10
Now that we know that x = 10, we can plug it into the first equation, x - y = 11 to find y.
x - y = 11 10 - y = 11 y = -1
For the second one,
4x + 8y = 20 -4x + 2y = 30
Negate the x's since there is the same and opposite amount, while adding the rest of the equations together.
8y + 2y = 10y 20 + 30 = 50 10y = 50 y = 5
Plug it into the initial equation to find x
4x + 8(5) = 20 4x + 40 = 20
Subtract 40 from both sides
4x = -20 x = -5
For the third equation,
-6x + 5y = 1 6x +4y = -10
Negate the x's, add the rest together
5y + 4y = 9y 1 - 10 = -9 9y = -9 y = -1
Plug it into the first equation to find x
-6x + 5y = 1 -6x + 5(-1) = 1 -6x - 5 = 1
Add five to both sides
-6x = 6 Divide both sides by negative 6 x = -1
When the problems do not have the same value of opposite versions of current numbers, for example:
3x + 4y = 6 -6x +g = 6
You can multiply an entire equation to get similar numbers of a variable. For example, you can multiply the first equation entirely by 2 to get the same number of positive x's to the negative x's in the second one, changing it to 6x x 8y = 12.
( y - 4 ) ( y² + 4 y + 16 ) = = y³ + 4 y² + 16 y - 4 y² - 16 y - 64 = y³ - 64 If the result is the polynomial of the form: y³ + 4 y² + a y - 4 y² - a y - 64 a = 16
