Question

Solve $\dfrac{27}{x-2}+\dfrac{31}{y+3}=85$ ; $\dfrac{31}{x-2}+\dfrac{27}{y+3}=89$

Answer 1

Now here to solve this question we can use the inverse of variables to be different variables to make the equations simplified. Now that we have simplified equations with 2 variables each we can use the method of solving simultaneous equations to solve both of them. In simultaneous solving of equations we need to try to combine all terms into one simple equation with one variable by the arithmetic functions like addition, subtraction, multiplication, division, etc. By solving we will get the value of the 2 variables through which we can then solve and find the value of x and y.

Complete step-by-step answer:
The two equation given to us here are;
$\dfrac{27}{x-2}+\dfrac{31}{y+3}=85$ ; $\dfrac{31}{x-2}+\dfrac{27}{y+3}=89$
Now to simplify both equations we can put the values of variables to be
$\dfrac{1}{x-2}=m$ and $\dfrac{1}{y+3}=n$
Therefore substituting these values we get that
$27m+31n=85$ ---- equation I
$31m+27n=89$ ---- equation II

Now to solve these equations we will first start by adding both equation I and II which gives us
$(27+31)m+(31+27)n=85+89$
Solving
$58m+58n=174$
Now dividing both sides by $58$ we get
$m+n=3$---- equation III
Now to find the second equation we subtract equation II from I and we get
$(27-31)m+(31-21)n=85-89$
Now solving
$-4m+4n=-4$
Dividing both sides by $4$ we are left with
$-m+n=-1$---- equation IV
Now therefore the two equations we are left with are equation III and IV
$m+n=3$
$-m+n=-1$
Now if we add them both we are left with
$(1-1)m+(1+1)n=(3-1)$
Therefore
$2n=2$
Therefore
$n=1$
Substituting this value of n in equation III we get
$m+n=3$
$m+1=3$
$m=2$
Now since we know both values of m and n which is $m=2$,$n=1$ we can find the values of x and y
$\dfrac{1}{x-2}=m$ , $\dfrac{1}{y+3}=n$
Substituting
$\dfrac{1}{x-2}=2$ ; $\dfrac{1}{y+3}=1$
Now cross multiplying in both equations we get
$1=2(x-2)$ ; $1=1(y+3)$
Opening the brackets
$1=2x-4$ ; $1=y+3$
Solving
$2x=5$ ; $y=-2$
Dividing the first equation with 2 on both sides
$x=\dfrac{5}{2}$ ; $y=-2$

Note: An alternative method to solve this is by multiplying the equation I by $31$ and equation II by $27$ and then subtract to get the value of n and substituting the value of n in equation I to then get the value of m. But doing this will make the calculation very difficult. This is why we prefer the above mentioned method to not make any calculation error in solving.