$$\dfrac{a}{x}+\dfrac{b}{y}=\dfrac{1}{z}$$ Such equations can be simplied as $$\left(x-az\right)\left(y-bz\right)=a.b.z^2$$

Further, we  can factorise $$a.b.z^2$$ which are the possible values for $$\left(x-az\right)\left(y-bz\right)$$

Equations with three variables: Let the equations be $$ a_{1}x+b_{1}y+c_{1}z$$=$$d_{1}$$, $$ a_{2}x+b_{2}y+c_{2}z$$=$$d_{2}$$ and $$ a_{3}x+b_{3}y+c_{3}z$$=$$d_{3}$$. Here we define the following matrices: $$$D=\begin{bmatrix} a_{1} & b_{1} & c_{1}\\ a_{2} & b_{2} & c_{2}\\ a_{3} & b_{3} & c_{3}\end{bmatrix}$$$, $$$D_{x}=\begin{bmatrix} d_{1} & b_{1} & c_{1}\\ d_{2} & b_{2} & c_{2}\\ d_{3} & b_{3} & c_{3}\end{bmatrix}$$$, $$$D_{y}=\begin{bmatrix} a_{1} & d_{1} & c_{1}\\ a_{2} & d_{2} & c_{2}\\ a_{3} & d_{3} & c_{3}\end{bmatrix}$$$, $$$D_{z}=\begin{bmatrix} a_{1} & b_{1} & d_{1}\\ a_{2} & b_{2} & d_{2}\\ a_{3} & b_{3} & d_{3}\end{bmatrix}$$$

• Find the determinant of $$D$$ = $$a_{1}(b_{2}c_{3}-c_{2}b_{3})$$ + $$b_{1}(c_{2}a_{3}-a_{2}c_{3})$$ + $$c_{1}(a_{2}b_{3}$$-$$b_{2}a_{3})$$
• Similarly find the determinant of Dx, Dy and Dz
• If $$ \textrm{determinant of D}\neq 0$$, there exists a unique solution
• If $$ \textrm{determinant of D} = 0$$ and at least one of the determinants of Dx, Dy or Dz are non-zero then the system of equations has no solution
• If $$ \textrm{determinant of D} = 0$$ and all three determinants Dx, Dy and Dz are zero then there are infinitely many solutions

In questions, sometimes we are given information that can be written as axe + by + cz =d, ex + fy + gz = h, and we will be asked to find the value of lx + my + nz.

Here, the trick is to write lx + my + nz in terms of ax + by + cz and ex + fy + gz, for example: lx + my + nz = p( ax + by + cz) + q(ex + fy + gz) => lx + my + nz = p*d + q*h

• If the system of equations has n variables with n-1 equations then the solution is indeterminate
• If the system of equations has n variables with n-1 equations with some additional conditions like the variables are integers then the solution may be determinate
• If the system of equations has n variables with n-1 equations then some combination of variables may be determinable.
  
• For Example, if ax+by+cz=d and mx+ny+pz=q if a,b, and c are in Arithmetic progression and m,n and p are in AP then the sum x+y+z is determinable

Equations with 2 variables: Consider two equations ax+by=c and mx+ny=p. Each of these equations represent two lines on the x-y coordinate plane. The solution of these equations is the point of intersection.

• If $$ \frac{a}{m}=\frac{b}{n}\neq\frac{c}{p}$$: This means that both the equations have the same slope but different intersect and hence are parallel to each. Hence, there is no point of intersection and no solution.
• If $$ \frac{a}{m}\neq\frac{b}{n}$$: They have different slopes and hence must intersect at some point. This results in a Unique solution.
• $$ \frac{a}{m}=\frac{b}{n}=\frac{c}{p}$$: The two lines have the same slope and intercept. Hence they are the same lines. As they have infinite points common between them, there are infinite many solutions possible.

Solving linear equations:

Follow these basic steps to solve linear equations:

• Aggregate the constant terms and variable terms
• For equations with more than one variable, eliminate variables by substituting equations in their place.
• Hence, for two equations with two variables x and y, express y in terms of x and substitute this in the other equation.
• For example, x+y=14 and x+4y=26 => x=14-y. Substituting this in equation 2, we get 14-y+4y=26. Hence y=4 and x=10.

For equations of the form ax+by=c and mx+ny=p, find the LCM of b and n. Multiply each equation with a constant to make the y term coefficient equal to the LCM. Then subtract equation 2 from equation 1.

Example:

Let 2x+3y=13 and 3x+4y=18 are the given equations (1) and (2).

• LCM of 3 and 4 is 12.
  
• Multiplying (1) by 4 and (2) by 3 we get 8x+12y=52 and 9x+12y=54.
  
• (2)-(1) gives x=2, y=3