You have two cases:
2y + 7 = (-3y - 5)
or
2y + 7 = -(-3y - 5)
Solve each independently to get the values of y.
In the first case:
2y + 7 = -3y - 5
2y + 3y = -5 - 7
5y = -12
y = -12/5
In the second case:
2y + 7 = -(-3y - 5)
2y + 7 = 3y + 5
7 - 5 = 3y - 2y
2 = y
y = 2
Double checking:
|2(2) + 7| = 11
|-3(2) - 5| = 11
|2(-12/5) + 7| = 2 1/5
|-3(-12/5) - 5| = 2 1/5
So the answers are:
y = -12/5
or
y = 2
(Note: technically there are four combinations (++, --, +-, -+), but ++ and -- are equivalent and +- and -+ are equivalent. Just multiply both sides by -1 to see that ++ transforms into -- and +- transforms into -+.)
2006-12-07 03:58:42
·
answer #1
·
answered by Puzzling 7
·
0⤊
1⤋
Case I : 2y+7>0 & -3y-5>0
i.e y>-7/2 & y<-5/3
equation reduces to
2y+7=-3y-5
=> 5y=-12
=> y=-12/5 lies in above range => is a solution
similarly do for other three cases
Case II 2y+7>0 & -3y-5<0
2y+7 = -(-3y-5)
...
Case III 2y+7<0 & -3y-5>0
...
Case IV 2y+7<0 & -3y-5<0
...
2006-12-07 04:06:11
·
answer #2
·
answered by chinky 1
·
0⤊
0⤋
There are two possibilities:
2y + 7 = -3y - 5
and
2y + 7 = -(-3y - 5)
You must solve both.
2y + 7 = -3y - 5
5y = -12
y = -12/5
2y + 7 = -(-3y - 5)
2y + 7 = 3y + 5
2 = y
Plug them both back in the original equation to verify that they both work.
2006-12-07 04:00:09
·
answer #3
·
answered by Jim Burnell 6
·
0⤊
0⤋
LHS= |2y+7| = 2y+7
RHS= |-(3y+5)| = 3y+5
since LSH=RHS
2y+7=3y+5
ie y=2
2006-12-07 04:10:44
·
answer #4
·
answered by Anonymous
·
0⤊
0⤋
=> 2y + 7 = -3y-5 or 2y + 7 = -(-3y-5)
2y + 7 = -3y -5 => 5y = -12 y = - 12/5
2y + 7 = -(-3y-5) => y = 2
so, y = -12/5 or 2
2006-12-07 04:04:06
·
answer #5
·
answered by Anonymous
·
0⤊
0⤋
2y+7 = -3y -5
5y=-12
y= -12/5 = - 2.4
Also 2y+7 = 3y +5
y =2
Check |2*2 +7| = | -3*2 -5| so |11| = | -11|
|2(-2.4) +7| = |-3(-2.4) -5| so |2.2| = |2.2|
2006-12-07 04:08:42
·
answer #6
·
answered by ironduke8159 7
·
0⤊
0⤋
I think you can move the constant out of the abs value so
|2y| + 7 = |-3y| + 5
2|y| + 7 = 3|-y| + 5
and |y| = |-y| so
2|y| + 7 = 3|y| + 5
and you can solve that for |y|.
2006-12-07 04:00:59
·
answer #7
·
answered by Anonymous
·
0⤊
1⤋
|2y+7|^2=|-3y-5|^2
4y^2 + 28y + 49 = 9y^2 + 25 + 30y
5y^2 + 2y - 24=0
Hence: y1=( -2 + sqr(2^2+ (4*5*24)))/10
y2=(-2 - sqr(2^2 + (4*5*24)))/10
2006-12-07 04:11:20
·
answer #8
·
answered by SaturnReLnArimani 2
·
0⤊
0⤋
2y+7=-3y-5
5y=-12
y=-12/5
-2y-7=-3y-5
y=2
2y+7=3y+5
-y=-2
y=2
-2y-7=3y+5
-5y=12
y=-12/5
2006-12-07 04:02:09
·
answer #9
·
answered by raj 7
·
0⤊
0⤋
- use excel 2 generate this function, draw the graphs.
- were the lines asr getting crossed is an answer
2006-12-07 04:01:24
·
answer #10
·
answered by Me _AA4Ro 2
·
0⤊
0⤋