The given equations that make up the systems are:
-3a+7b=-16...................(Equation 1)
-9a+5b=16...................(Equation 2)
To find the values of a and b we will have to solve the two equations. To solve the equations let us employ the method of elimination.
Let us multiply (Equation 1) by -3. This will give us:
[tex] -3\times (-3a+7b=-16)=9a-21b=48 [/tex]...........(Equation 3)
Let us now add (Equation 2) to (Equation 3) to eliminate a. This will give us:
-16b=64
[tex] \therefore b=-4 [/tex]
Plugging this value of b in any of the equations we will get the value of a. Let us plug it in (Equation 1). This gives:
[tex] -3(-4)+7b=-16 [/tex]
[tex] 12+7b=-16 [/tex]
[tex] \therefore b=\frac{-28}{7}=-4 [/tex]
Thus, out of the given options, the first option is the correct one. The values of a and b are both equal to –4.
