3|3x+2|=51
I think my teacher graded this wrong on my "test"
can anyone help me out by telling me that I was right in figuring it out?
2007-10-03 12:37:55 · 8 answers · asked by ? 2 in Education & Reference ➔ Homework Help
x= 5 OR -19/3
you would be wrong if you didn't say both answers because....
1. if x=5
3|3(5)+2|=51
3|15+2|=51
3|17|=51
3*17=51
51=51
TRUE
AND
2. if x= -19/3
3|3(-19/3)+2|=51
3|-19+2|=51
3|-17|=51
3*17=51
51=51
TRUE
2007-10-03 12:56:39 · answer #1 · answered by malmal 2 · 0⤊ 0⤋
Distribute the 3 into everything that's in the brackets. You'll get: 9x+6=51, then just solve for x by taking away 6 from each side and dividing both by 9.
2007-10-03 12:43:58 · answer #2 · answered by Anonymous · 0⤊ 1⤋
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2016-12-17 16:27:25 · answer #3 · answered by ? 4 · 0⤊ 0⤋
-51=9x+6=51
-57=9x=45
-57/9=x=5
Next, plug in the answers in original equation to check; the ones that work are right. Make sure to check though because sometimes there are cases of no solution.
Remember, that because of the absolute, you need to put -51 also!
2007-10-03 12:45:55 · answer #4 · answered by Mechiko 2 · 0⤊ 1⤋
3|3x+2|=51
÷3 ÷3
|3x+2|=17
3x+2=17
-2 -2
3x=15
x=3
lol and im only in geometry but this stuff was in algebra one
2007-10-03 12:44:13 · answer #5 · answered by penguinfreek 2 · 0⤊ 1⤋
I assume that | and | are absolute value brackets.
|3x+2|=17
therefore
3x+2=17
or 3x+2 = -17
You have to solve for both.
2007-10-03 12:44:10 · answer #6 · answered by Meredith 4 · 0⤊ 0⤋
3|3x+2|=51
3x+2=17
3x=15
x=5
2007-10-03 12:41:52 · answer #7 · answered by wwwtoha 3 · 0⤊ 1⤋
Did you say that x=5?
2007-10-03 12:42:38 · answer #8 · answered by hottotrot1_usa 7 · 0⤊ 0⤋