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A system of linear equations can have either zero, one, or infinitely many solutions, depending on the ways in which the individual equations relate. In this section, the qualities that make each number of solutions possible will be explored.

When a system of linear equations has two equations and two variables, the system can have zero, one or infinitely many solutions.

Such lines are said to be coincidental, and, as they have the same slope and y-intercept, they are different versions of the same line. One example of a system that has an infinite number of solutions is

Show Solution *expand_more*

A system of equations has infinitely many solutions when the equations describe the same line. In other words, the lines must have the same slope and y-intercept. The first line has a slope of 2. Thus, the slope of the second line must also be 2. This means, m=2. The system can now be written
Notice that the y-intercepts of the lines have already been given. They are (0,5) and (0,-2). Since these values are different, and they cannot be changed, the lines are not coincidental. Therefore, the system will never have infinitely many solutions. In fact, since the lines are parallel, the system has no solution.

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The given system will have zero, one, or infinitely many solutions. To determine which, it can be helpful to identify the slopes of the lines. Since the first equation is given in slope-intercept form, we can see its slope is
To determine the slope of the second line, let's isolate y.
The system can now be written as
As it turns out, the slopes and the y-intercepts of the lines are the same. Thus, the lines are coincidental, which means the system has infinitely many solutions. Graphing the lines, we can see that the equations describe the same line.

2x+3y=12

Solve for y

SubEqn

LHS−2x=RHS−2x

3y=-2x+12

DivEqn

$LHS/3=RHS/3$

$y=3-2x+12 $

WriteSumFrac

Write as a sum of fractions

$y=3-2x +312 $

CalcQuot

Calculate quotient

$y=3-2x +4$

MoveNegNumToFrac

Put minus sign in front of fraction

$y=-32x +4$

MovePartNumRight

$ca⋅b =ca ⋅b$

$y=-32 x+4$

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The given point will be a solution to the system only if it satisfies both equations simultaneously. To do this, we can substitute x=6 and y=3 into both equations and simplify.

${y=-21 x+6y=2x−4 $

(I), (II): x=6, y=3

${3=?-21 ⋅6+63=?2⋅3−4 $

(I), (II): Multiply

${3=?-3+63=?6−4 $

(I), (II): Add and subtract terms

${3=33 =2 $

Show Solution *expand_more*

To algebraically find the number of solutions to the system, either the substitution method or the elimination method can be used.
This resulted in a contradiction, so the system has **no solution**.

${y=2x+7y=2x−3 (I)(II) $

Since there are equally many x- and y-terms in both equations, the elimination method is most convenient. Therefore, the second equation should be subtracted from the first.
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