Question

Someone ask me to solve this.

Answer 1

X=2h, y=3k

Substitute these values into equations.

y+2x = 4 ------> 3k+2*2h=4 ----->  3k +4h =4

2/y - 3/2x = 1-----> 2/3k -3/(2*2h) = 1 ------> 2/3k - 3/4h =1

We have a system of equations now.

 3k +4h =4          ------> 3k = 4-4h ( Substitute 3k in the 2nd equation.)
2/3k - 3/4h =1

2/(4-4h) -3/4h = 1
2/(2(2-2h)) - 3/4h = 1
1/(2-2h) -3/4h - 1=0

4h/4h(2-2h) -3(2-2h)/4h(2-2h) - 4h(2-2h)/4h(2-2h) =0

(4h- 3(2-2h) - 4h(2-2h))/4h(2-2h) = 0
Numerator should be = 0
4h- 3(2-2h) - 4h(2-2h)=0

Denominator cannot be = 0
4h(2-2h)≠0

Solve equation for numerator=0
4h- 3(2-2h) - 4h(2-2h)=0
4h - 6+6h-8h+8h² =0
8h² +2h -6=0
4h² + h-3 =0
(4h-3)(h+1)=0
4h-3=0, h+1=0
h=3/4  or h=-1

Check which
4h(2-2h)≠0
1) h= 3/4  , 4*3/4(2-2*3/4)=3*(2-6)= -12 ≠0, so we can use h= 3/4
2)h=-1, 4(-1)(2-2*(-1)) =-4*4=-16 ≠0, so we can use h= -1, also.

h=3/4, then  3k = 4-4*3/4 =4 - 3=1 , 3k =1, k=1/3
h=-1, then  3k = 4-4*(-1) =8 , 3k=8, k=8/3 

So,
if h=3/4, then  k=1/3,
and if h=-1, then  k=8/3 .