Any want to explain how solving 3 var systems of equation using elimination works again?

solve a system of equations when multiplication is necessary to eliminate a variable. recognize systems that have no solution or an infinite number of solutions. the elimination method for solving systems of linear equations uses the addition property of equality. you can add the same value to each side of an equation.

1. is 4 and 2. is 3

step-by-step explanation:

so if you look at the first one you see that 1 is to 4 and 2 is to 8 and so on, so you can see for every 1 there is a times 4 so if it were 3 then it would be 3*4 and it is 12

for number 2 it is 3 because 1 is to 3 and 2 is to 6 so for every 1 there is 3 so if it were 3 then there would be 3*3 = 9

t = 3

step-by-step explanation:

put t = 3 in this expression :

(3²-1)(3²-2)(3² - 3)(3² - 4) = 8×7×6×5 = 56×30 = 1680