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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.

If a system has no solution, its graph might look similar to the graph shown.

Recall that the solution to a system is the point where the lines intersect. If a system has no solution, it must be then that the lines never intersect. In fact, the lines must be parallel, meaning that they have the same slope and different $y$-intercepts. An example of one such system is ${y=3x+2y=3x−5. $

The graph of a system that has one solution might look similar to the graph shown. Specifically, it will show that the lines intersect exactly once. The point of intersection is the solution to the system.

In contrast to parallel lines, lines that intersect once must have unequal slopes. For example, the system shown must have exactly one solutin as the two lines have different slopes. ${y=-1x+5y=3x−2 $

For a system to have infinitely many solutions, it must mean that the lines intersect at infinitely many points. In fact, it means the lines lie on top of each other.

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

${y=3x+12y=6x+2. $Find the value of $m$ that ensures the system has infinitely many solutions. ${y=2x+5y=mx+2 $

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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 ${y=2x+5y=2x+2. $ 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.

Determine the number of solutions the system has. ${y=-32 x+42x+3y=12 $

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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 $m=-32 .$
To determine the slope of the second line, let's isolate $y.$
The system can now be written as ${y=-32 x+4y=-32 x+4. $
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 $

WriteSumFracWrite as a sum of fractions

$y=3-2x +312 $

CalcQuotCalculate quotient

$y=3-2x +4$

MoveNegNumToFracPut minus sign in front of fraction

$y=-32x +4$

MovePartNumRight$ca⋅b =ca ⋅b$

$y=-32 x+4$

Determine if the point $(6,3)$ is a solution to the system ${y=-21 x+6y=2x−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 $

Determine the number of solutions the system has algebraically. ${y=2x+7y=2x−3 $

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To algebraically find the number of solutions to the system, either the substitution method or the elimination method can be used.
${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.
$− yy0 == = 2x+72x−30+10 $
This resulted in a contradiction, so the system has **no solution**.

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