y = -4
x = -2
Step-by-step explanation:
1. Approach
To solve a system by the process of elimination, one wants to manipulate an equation such that when one adds the equations in a system, one of the variables cancels out. The easiest way to do this is to select a variable, then find the LCM (least common multiple) of the two coefficients of the selected variables. Then multiply each of the equations by the respect factor that is needed to get the coefficient of the variable to the LCM. In the event that the cofficients do not have inverse signs, multiply one of the equations by (-1). No all that is left is to add the equations, use inverse operations to solve for the other variable, then back substitute to find the value of the selected varibale.
2.Set up the process
3x - 2y = 2
5x - 5y = 10
Selected variable; x
LCM of the coefficents (3, 5) = 15
Multiply the first equation by 5, and the second by 3
(3x - 2y = 2) *5
(5x - 5y = 10)*3
Resulting system
15x - 10y = 10
15x - 15y = 30
Now, multiply the first equation by (-1) to make the coefficents additive inverses
(15x - 10y = 10) (-1)
15x - 15y = 30
-15x + 10y = -10
15x - 15y = 30
3. Solving
Add the two resulting systems,
-15x + 10y = -10
15x - 15y = 30
-5y = 20
Inverse operations
-5y = 20
/-5 /-5
y = -4
Back solve
3x - 2y = 2
(3x) - 2(-4) = 2
Simplify,
3x + 8 = 2
Inverse operations,
3x + 8 = 2
-8 -8
3x = -6
/3 /3
x = -2
