We now have an algebraic expression with only one variable, which can be solved. once we have that, we can plug it back into one of the original equations (or the expression derived in part a) to solve for the other variable. when this is done with the system of two equations from parts a and b, what is the solution?

A=1

B=-2

Explanation:

Part A and B of the question wasn't given, however, I attached the relevant parts to solve this question as follows.

From part B as attached, it shows that the right option is C which is

2A+3B=-4

Substituting B with 3A-5 then we form the second equation as shown

2A+3(3A-5)=-4

By simplifying the above equation, we obtain

2A+9A-15=-4

Re-arranging, then

11A=-4+15

Finally

11A=11

A=1

To obtain B, we already know that 3A-5 so substituting the value of A into the above then we obtain

B=3(1)-5=-2

Therefore, required values are 1 and -2

-4 on

the components of a vector are the parts of the vectors that are to each other.

b perpendicular