-x<-x+7(x-2)
-x<-x+7x-14 Distribute the 7
-x<6x-14 Combine like terms
-7x<-14 Move all variables to one side
x>2 Divide by -7 to isolate the variable
X, x+1, x+2, x+3 are 4 consecutive integers
2(x + x+2) = 5(x+3) - 26
2(2x+2) = 5(x+3) - 26
4x + 4 = 5x + 15 - 26
4x + 4 = 5x - 11
15 = x
15, 16, 17, 18
x, x+2, x+4 consecutive even integers
2(x+4)+11 = x + 9
2x + 8 + 11 = x + 9
2x + 19 = x + 9
x = -10
-10, -8, -6
J+M=T/7
Total crackers= T
J = Jaylens crackers
M = Marvin’s crackers
Answer:
8 [ (x-3) (2x+1) ]
Step-by-step explanation:
Factoring 2x2 - 5x - 3
The first term is, 2x2 its coefficient is 2 .
The middle term is, -5x its coefficient is -5 .
The last term, "the constant", is -3
Step-1 : Multiply the coefficient of the first term by the constant 2 • -3 = -6
Step-2 : Find two factors of -6 whose sum equals the coefficient of the middle term, which is -5 .
-6 + 1 = -5 That's it
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -6 and 1
2x2 - 6x + 1x - 3
Step-4 : Add up the first 2 terms, pulling out like factors :
2x • (x-3)
Add up the last 2 terms, pulling out common factors :
1 • (x-3)
Step-5 : Add up the four terms of step 4 :
(2x+1) • (x-3)
Which is the desired factorization
