Question

How do you solve $3x-y=3,$ $2x+y=2$ by graphing and classify the system?

Answer 1

Hint: For each question you need to solve for two points as here you have two equations therefore number of points for pointing the graph will be $4.$
For equation $\left( 1 \right)$ firstly put $x=0$ and then $y=0$ for graphing the points. For classifying the system, focus on the number of solutions i.e. one solution, an infinite number of solutions, no solution. System of equations can be classified by the number of solutions. No solution, a system of equations can be classified by number of solutions. If a consistent system has exactly one solution therefore this type of system is called an independent consistent system.

Complete step by step solution:As we know that, given two equations are $3x-y=3\ \text{and}\ \text{2x+y=2}$
Here, you have to solve for two points for each equation,
Equation no:-$1$ i.e. $3x-y=3...\left( 1 \right)$
Solve for the first point.
Let $x=0$ in equation no:- $1$
$3\left( 0 \right)-y=3$
$0-y=3$
$-y=3-0$
$-y=3$
$y=-3$
Or $\left( 0,-3 \right)$
For second point,
For $y=0$ in equation no:-$1,$
$3x-0=3$
$3x=3$
$x=\dfrac{3}{3}$
$x=1$
Or $\left( 1,0 \right)$
For
Graph, $\left\{ \left( 3x-y-3 \right) \right\}\ \left\{ {{x}^{2}}+{{\left( y+3 \right)}^{2}} \right\}-\left\{ {{\left( x-1 \right)}^{2}}+y-0.075 \right\}=0$
$\left[ -20,20,-10,10 \right]$
$2x+y=2\ .....\left( 2 \right)$
Here some for the first point in ${{2}^{\text{nd}}}$ equation
Let $x=$$0$, in equation no:- $2$
$2\times 0+y=2$
$0+y=2$
$\therefore y=2$
Or it can also be written as co$.21$
Let $y=0$ in equation no:-$2$
$2x+0=2$
$2x=2$
$x=2$
$\therefore x=1\ \text{or}\ \left( 1,0 \right)$
Graph $\left\{ \left( 2x+y-2 \right)\left( 3x-y-3 \right)\left( {{x}^{2}}+{{\left( y-2 \right)}^{2}} \right)-0.075.{{\left( x-1 \right)}^{2}}+{{y}^{2}}-0.075 \right\}$
$=0\left[ -20,20,-10,10 \right]$
You can see the points across at $\left( 1,0 \right)$
Graph $\left\{ \left( 2x+y-2 \right)\left( 3x-y-3 \right){{\left( x-1 \right)}^{2}}+{{y}^{2}}-0.05 \right\}$
$=0\left[ -10,10,-5,5 \right]$
$\Rightarrow $ classify the system.
A system of two linear equations can have one solution, an infinite number of solutions, or no solution. System of equations can be classified by the number of solutions. If a system has at least one solution. It is said to be consistent. The system has exactly one solution, it is independent.

Additional Information:
It is possible for a system of equations to have no solution because a point on a coordinate graph because a point on a coordinate graph to solve the equation may not exist. This concept may seem hard to understand at first, but if you think about it, it comes easily.
The only way this can happen if there is no intersection point, as this point would answer. And of course, if there is no intersection, then the lines are parallel.
For example:-
$y=4x+19$
$y=4x+7$
Here, the slopes are the same, so the lines are parallel and never merest. That being said. There is no solution to this problem.

Note:
As there are two equations given in the problem you have to solve for plotting the graph.
First take equation No.$1.$ And take $x=0$ for finding first point and take $y=0$ or finding second point as you take $x=0$ you will get y points and for $y=0$ you will get $x=0$ points.
Remember that if a system has at least one solution, it is said to be consistent.
If a consistent system has exactly one solution. It is called an independent consistent system.