Question

Solve for x and y:

$\frac{2}{(3x+2y)}+\frac{3}{(3x-2y)}=\frac{17}{5},\frac{5}{(3x+2y)}+\frac{1}{(3x-2y)}=2.$

Answer 1

solve for x and y .

2/(3x+2y)+3/(3x-2y)=17/5

5/(3x+2y)+1/(3x-2y)=2

Let1/(3x+2y)=p
and
1/(3x−2y)=q
then the given equations become:

2p+3q=17/5.....(1)

5p+q=2............(2)

Multiplying equation (1) by 5 and equation (2) by 2,

weget,
10p+15q=17............(3)

10p+2q=4...........(4)
Subtracting equation(4) fromequation(3),
weget,
13q=13⇒q=13/13⇒
q=1
Putting q=1 inequation(3),
weget,
10p+15×1=17⇒10p+15=17⇒10p=17−15⇒10p=2⇒p=2/10⇒p=1/5

Therefore,
1/(3x+2y)=1/5
and
1/(3x−2y)=1

⇒3x+2y=5.........(5) and
3x−2y=1..........(6) Adding equation (5) and equation (6), weget, 6x=6⇒x=6/6
⇒x=1
Putting x=1 inequation(5), weget, 3×1+2y=5⇒3+2y=5⇒2y=5−3⇒2y=2⇒y=2/2=1
Hence,
thesolutionis
x=1
and
y=1