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I am trying to solve
|2y+7|=|-3y-5|

Anyone up to the math challenge?

2006-12-07 03:56:49 · 10 answers · asked by Ken H 1 in Science & Mathematics Mathematics

10 answers

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

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