The lines whose equations are 2x + y = 3z and x + y = 6z intersect at which point? (3z, -3z) (-3z, 9z) (9z, -15z)

The lines whose equations are 2x + y = 3z and x + y = 6z intersect at the point (-3z,9z)

Step-by-step explanation:

When two straight lines intersect, they coincide only at one point. This point is described through a set of "x" and "y" coordinates, usually expressed as a point (x, y). Because both lines pass through that point, you know that the coordinates (x, y) must satisfy both equations. Therefore it is possible to establish a system of two equations with two unknowns or variables :

This system of equations can be solved by several methods. In this case it is decided to solve it by the substitution method. This method consists of isolating one of the variables and replacing its expression in the other equation. In this case, the value of "y" is isolated from the second equation of the system of equations, obtaining:

y=6*z-x

Substituting in the first equation of the system of equations, the following first degree equation is obtained:

2*x+6*z-x=3*z

This equation is solved:

2*x-x=3*z-6*z

x=-3*z =-3z

In this way the value of "x" is obtained, whose value is -3*z (or -3z). To obtain the value of "y", the value of "x" obtained in the equation where the value of "y" was previously isolated is substituted:

y=6*z-x

y=6*z-(-3*z)

y=6*z+3*z

y=9*z=9z

In this way the value of "y" is obtained, whose value is 9*z (or 9z).

The lines whose equations are 2x + y = 3z and x + y = 6z intersect at the point (-3z,9z)

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v