I'm back . . . no tears please. I've shed enough over this for all of us.
Can someone help me with this.
Find four consecutive integers such that, twice the sum of the two larger integers, exceeds three times the first by ninety-one.
. . .x = 2(3x + 4x) - 91
. . .x = 6x + 8x - 91
. . .x = 14x - 91
. . .-13x/13 = -91/13
. . .x = 7
Checking:
. . .2(3(7)+ 4(7)) - 91
. . .98 - 91 wrong
What am I doing wrong in the setup?
by the way, the answers are 81, 82, 83, and 84. Thank you!
Can someone help me with this.
Find four consecutive integers such that, twice the sum of the two larger integers, exceeds three times the first by ninety-one.
. . .x = 2(3x + 4x) - 91
. . .x = 6x + 8x - 91
. . .x = 14x - 91
. . .-13x/13 = -91/13
. . .x = 7
Checking:
. . .2(3(7)+ 4(7)) - 91
. . .98 - 91 wrong
What am I doing wrong in the setup?
by the way, the answers are 81, 82, 83, and 84. Thank you!