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How do you solve x plus y plus z equals 7 2x plus 3y-z equals -2 5x-6x plus 3z equals 36?
Let me see:1) x + y + z = 7 y = 7 - x- z2) 2x + 3y- z = -23) -x + 3z = 36 x = 3z - 36Subst. 3) into 2):2(3z-36)+3y -z = -2Subst 1) into 3):2(3z-36) +3(7-x-z) - z = -2Subst 3) into 2) again:6z-64 + 3(7-(3z-36) - z) -z = -2z= 67/7Subst answer into 3) -x + 3(67/7) = 36x = -51/7Subst answers into 1):-51/7 + y + 67/7 = 7y = 33/7therefore x = -51/7, y = 33/7, z = 67/7
What is the answer to 2 x plus 7 equals x plus 3?
2x + 7 = x + 3 x + 7 = 3 x = -4
7x6 plus 9?
The answer to 7 x 6 plus 9 is 51. You have to multiply 7 by 6 which equals 42 and add 9 onto it which equals 51.
How can you use the property of addition to find the value of x in x plus 3 equals 3 plus 7?
If: x+3 = 3+7 Then: x = 7
Factor 210 into a product of primes?
The prime factors are: 2 x 5 = 10
How many times can 7 go in to 360?
51 with 3 remaining 360 - 3 = 357 = 7 x 51
If the factors of a polynomial are x plus 3 and x plus 7 what values of x make that polynomial 0?
If (x + 3)(x +7) = 0 then either: x + 3 = 0 or x + 7 = 7 Hence x = -3 or -7.
X plus 7 equals -3?
4
What can you multiply to get 51 and add to get 14?
51 > 0. Since the sum 14 is positive, the factors of 51 are both positive. The problem is that 51 = 3*17 and 3 + 17 = 20 ≠ 14. But we are lucky that 14 = 7 + 7. So wee can use the difference of the squares such as (7 - A)(7 + A) = 49 - A2 = 51, where -A2 = 2 = -(i√2)2 . Thus, the numbers are 7 - i√2 and 7 + i√2. or algebraically, x + y = 14; y = 14 - x xy = 51 x(14 - x) = 51 -x2 +14x = 51 x2 -14x = -51 x2 -14x + 72 = -51 + 49 (x - 7)2 = -2 x - 7 = ± √-2 x = 7 ± i√2
10 plus x -3?
7
3 plus x plus -7 x 6?
The answer to 3 plus X plus -7 times 6 equals-38. This is considered to be a math problem.
What is the LCM of 3 6 21 and 51?
3 = 3 x 1 6 = 3 x 2 21 = 3 x 7 51 = 3 x 17 '3' is a common factor throughout. Hence 3 x 1 x 2 x 7 x 17 = 714 Is the LCM
