We want two consecutive even integers such that twice the smaller is 26 less than 3 times the larger.
I know how to calculate this but I can't figure out what the x is. What is the answer and how did u figure it out?? That is 2x = 3(x+2) - 26.
2007-05-02 11:46:13 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Thanks for all the great answers. Can any of you answer my other math question, too?
http://answers.yahoo.com/question/index;_ylt=AnknMQuEK8PWUHMeuR9Wcfrsy6IX?qid=20070502154746AAHWJYl
2007-05-02 12:23:07 · update #1
Let x be the smaller and (x+2) be the larger. Solve:
2x = 3(x+2) - 26
2x = 3x + 6 - 26
2x - 3x = 6 - 26
-x = -20
x = 20
The two numbers are 20 and 22.
2007-05-02 11:49:38 · answer #1 · answered by Anonymous · 1⤊ 0⤋
Beautiful eye Yommygoo...
2x = 3(x+2) - 26
2x = 3x + 6 - 26
- x = - 20
So x = 20, and integers are 20, 22
2007-05-02 18:57:42 · answer #2 · answered by Anonymous · 1⤊ 0⤋
2x = 3(x+2)-26
distribute the 3:
2x = 3x+6-26
combine like terms:
2X = 3x-20
subtract 3x:
-x = -20
multiply( or divide doesnt matter) by -1.
x=20; so the integers are 20 and 22.
2007-05-02 18:54:33 · answer #3 · answered by Ari 6 · 1⤊ 0⤋
integers are x & x+2
2x=3(x+2)-26 distribute 3
2x=3x+6-26 combine like terms
2x=3x-20 subtract 3x from each side
-x=-20 divide both sides by -1
x=20
x+2=22
the numbers are 20 & 22
2007-05-02 18:51:17 · answer #4 · answered by yupchagee 7 · 1⤊ 0⤋
2x=3(x+2)-26
2x=3x+6-26
2x=3x-20
-x=-20
x=20
and x=22
2007-05-02 19:25:49 · answer #5 · answered by Dave aka Spider Monkey 7 · 1⤊ 0⤋