Basically you could divide it by 2 and then halving 34 would leave you with 17 so , you could say 17x2=34
Hope this was helpful
I believe the correct answer is true. <span>When solving a system of linear equations, try to algebraically form one equation that has only one variable. In this way, you can solve the value of that variable and eventually solve the other variables. Hope this answers the question. Have a nice day.</span>
2y^3 – 2y – 10y + 10 + y^2 – 1 < 0 [the terms are simply reorganized again]
factor 2y from the first two terms, -10 from the second two terms
2y (y^2 -1) – 10 (y-1) + y^2 – 1 < 0
2y (y+1)(y–1) – 10 (y-1) + (y+1)(y–1) < 0 [ because y^2 – 1 = (y+1)(y–1) ]
factor out (y-1) from all the terms
(y-1) [2y(y+1)-10+ y+1] < 0
(y-1) [(y+1) (2y+1) - 10] < 0
Let us simplify (y+1) (2y+1) - 10 < 0 now
(y-1) (2y^2+y+2y+1-10) < 0
(y-1) (2y^2 +3y -9 < 0
(y-1) (2y^2 +6y -3y - 9) < 0 [ because 3y = 6y -3y] j
Answer:
Step-by-step explanation:
We can use the process of elimination as
5x + 4y = 24
5(x + 7y = 11) —> 5x + 35y = 55
5x + 4y = 24
-(5x + 35y = 55)
—————————
-31y = -31
-31y/-31 = -31/-31
y=1
Answer:
26
Step-by-step explanation:
