8/3x-2 + 45/4y+3=5 a and 12/3x-2 - 30/4y+3 = 1

Question

It is a question from the topic linear equations in two variables.

Answer 1

Let A=1/(3x-2) and B=1/(4y+3), so the equations become: 

8A+45B=5 and 12A-30B=1.

Solve these first.

Multiply first by 2 and the second by 3: 16A+90B=10 and 36A-90B=3. Now add them together:

52A=13 so A=1/4.

Multiply first by 3 and the second by 2: 24A+135B=15 and 24A-60B=2 and subtract:

195B=13, so B=1/15.

From these we have 1/4=1/(3x-2) so 3x-2=4 and 1/15=1/(4y+3), so 4y+3=15. Therefore 3x=6, x=2 and 4y=12, y=3.

Solution: x=2, y=3

CHECK

Substitute values into the original equations:

8/4+45/15=2+3=5 OK.

12/4-30/15=3-2=1 OK. 

Answer 2

8/3x-2 + 45/4y+3=5 a and 12/3x-2 - 30/4y+3 = 1

The solution is easier if we can spot that a simple substitution can be made.

Let u = 3x - 2 ams let v = 4y + 3, then our equations become

8/u + 45/v = 5

12/u - 30/v = 1

Multiplying both equations by uv gives us,

8v + 45u = 5uv

12v - 30u = uv

multiply 2nd equation by 5

8v + 45u = 5uv

60v - 150u = 5uv

subtract 1st eqn from the 2nd,

52v - 195u = 0

u = (52/195)v = (4/15)v

Substituting for u = (4/15)v into the very first of the two eqns above, viz. 8v + 45u = 5uv, then

 

8v + 45(4/15)v = 5(4/15)v^2

8v + 12v = (4/3)v^2

60v = 4v^2

v^2 - 15v = 0

v(v - 15) = 0

v = 0, v = 15

Since v is used as a denominator, then we must ignore the solution v = 0, leaving us with but one solution, v = 15

Hence, u = 4, and v = 15

From which, x = 2, y = 3