Linear Equations

Consider an equation $$ax+by=c$$. The slope of this equation is $$\frac{-a}{b}$$

The slope-intercept form $$y=mx+b$$

If the equation is in the form given above, m will be the slope of the given linear equation.

$$\dfrac{a}{x}+\dfrac{b}{y}=\dfrac{1}{z}$$ 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 $$\left(x-az\right)\left(y-bz\right)$$

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
• 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

Given a system of three linear equations in three variables:

$$a_1x + b_1y + c_1z = d_1$$
$$a_2x + b_2y + c_2z = d_2 $$
$$a_3x + b_3y + c_3z = d_3$$

Suppose two equations represent the same line, i.e., one is a multiple of the other:

$$a_2 = k a_1,\quad b_2 = k b_1,\quad c_2 = k c_1,\quad d_2 = k d_1 \quad (k \neq 0)$$

Since two of the three equations are the same, they together act as only one equation.
Hence, the system effectively has only two independent equations in three variables.

$$\boxed{\text{Therefore, the system does not have a unique solution.}}$$

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

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

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.