Answer:
Divide the second equation [tex]2x+4y=32[/tex] by 2
Step-by-step explanation:
Given a linear equation system, there are three basic operations that we can make in order to obtain a second equivalent system :
1. Making a new equivalent system changing the equations order.
2.Making a new equivalent system multiplying one of the equations by a constant k [tex](k\neq 0)[/tex] and k ∈ IR
3.Making a new equivalent system adding equations that are a linear combination of the remaining equations.
Examples : The system
[tex]\left \{ {{-x+y=2} \atop {2x+4y=32}} \right.[/tex] is equivalent to
[tex]\left \{ {{2x+4y=32} \atop {-x+y=2}} \right.[/tex] Â (We used 1.)
The system
[tex]\left \{ {{-x+y=2} \atop {2x+4y=32}} \right.[/tex] is equivalent to [tex]\left \{ {{-10x+10y=20} \atop {2x+4y=32}} \right.[/tex] (We used 2. given that we multiplied the first equation by 10)
The system
- x + y = 2
2 x + 4 y = 32
is equivalent to
- x + y = 2
2 x + 4 y = 32
x + 5 y = 34
(We used 3. given that the third new equation is a linear combination of the first and second equation. The third equation is the sum of the first and the second one)
In the exercise,
[tex]\left \{ {{-x+y=2} \atop {2x+4y=32}} \right.[/tex] is equivalent to
[tex]\left \{ {{-x+y=2} \atop {x+2y=16}} \right.[/tex]
Because we divided the second equation by 2 (Notice that this is an example of 2. because we multiply the second equation by [tex]\frac{1}{2}[/tex])
