Question

How do you determine the number of solutions in a linear system without solving $3x - 6y = 9$ and $y = \dfrac{1}{2}x - \dfrac{3}{2}$ ?

Answer 1

Hint: we can identify the number of solutions in a linear system of equations by:
$1.$ Rewriting the given system of equations into the slope intercept form.
$2.$ Then the slope and y intercept of the equation will determine the number of solutions in a linear system of equations.

Complete step by step answer:
Let us start with the very first step to rewrite the given equation in the slope- intercept form.
But before proceeding to that we should be clear with the slope- intercept form of equation.
See, any equation which is given in the form $y = mx + c$ is the slope intercept form of the equation.
Now, we will start by writing the main equations that are given to us in the question:
$ \Rightarrow 3x - 6y = 9$……………………. let it be eq. $(1)$
$ \Rightarrow y = \dfrac{1}{2}x - \dfrac{3}{2}$…………………… Let it be eq. $(2)$
Firstly, rewriting the equation (1) in slope- intercept form of equation:
$ \Rightarrow 3x - 6y = 9$
It can be done by relocating all the terms that are not containing y to the RHS of the equation:
$ \Rightarrow - 6y = - 3x + 9$
Simplify by dividing both side of the equation with -6:
$ \Rightarrow \dfrac{{ - 6y}}{{ - 6}} = \dfrac{{ - 3}}{{ - 6}}x + \dfrac{9}{{ - 6}}$
Cancel the common factor of -6:
$ \Rightarrow \dfrac{{ - \not{6}y}}{{ - \not{6}}} = \dfrac{{ - 3}}{{ - 6}}x + \dfrac{9}{{ - 6}}$
Simplify and rewrite the expression:
$ \Rightarrow y = \dfrac{{ - 1}}{{ - 2}}x + \dfrac{9}{{ - 6}}$
$ \Rightarrow y = \dfrac{{\not{ - }1}}{{\not{ - }2}}x + \dfrac{9}{{ - 6}}$
Cancelling the same sign from the numerator and denominator
$ \Rightarrow y = \dfrac{1}{2}x - \dfrac{3}{2}$………………….. eq. $(3)$
Similarly, rewriting the equation (2) in slope- intercept form of equation:
$ \Rightarrow y = \dfrac{1}{2}x - \dfrac{3}{2}$
Here, we can see the equation (2) is already in a slope- intercept form so we will not evaluate it further.
$ \Rightarrow y = \dfrac{1}{2}x - \dfrac{3}{2}$………………… eq. $(4)$
As from the equation (3) and (4) we will write the slope and y intercept:
From eq. $(3)$
$ \Rightarrow y = \dfrac{1}{2}x - \dfrac{3}{2}$
Where, slope ${m_1} = \dfrac{1}{2}$and $y$- intercept ${c_1} = - \dfrac{3}{2}$
from eq. $(4)$
$ \Rightarrow y = \dfrac{1}{2}x - \dfrac{3}{2}$
Where, slope \[{m_2} = \dfrac{1}{2}\]and $y$- intercept ${c_2} = - \dfrac{3}{2}$
As ${m_1} = {m_2}$ and ${c_1} = {c_2}$ we can say the linear system of equations has infinitely many solutions.

Note: To identify the number of solutions for a system of equations we should be familiar with the 3 properties:
$1.$ If ${m_1} = {m_2}$and ${c_1} \ne {c_2}$, then there is no solution.
$2.$ If ${m_1} \ne {m_2}$and ${c_1} \ne {c_2}$, then there is only one solution.
$3.$ If ${m_1} = {m_2}$and ${c_1} = {c_2}$, then there are infinitely many solutions.