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    how to tell if a system of equations has no solution or infinitely many

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    How to Find If a System of Equations has No Solution or Infinitely Many Solutions

    Learn how to find if a system of equations has no solution or infinitely many solutions. Students can find conditions to check whether the system of equations has no solution or infinitely many solutions

    An equation of the form ax + by + c = 0 where a, b, c ∈ R, a ≠ 0 and b ≠ 0 is a linear equation in two variables. While considering the system of linear equations, we can find the number of solutions by comparing the coefficients of the equations. Also, we can find whether the system of equations has no solution or infinitely many solutions by graphical method. In this article, we will learn how to find if a system of equations has no solution or infinitely many solutions.

    System of Equations has No Solution or Infinitely Many Solutions

    Let us consider the pair of linear equations in two variables x and y.

    a1x + b1y + c1 = 0 a2x + b2y + c2 = 0

    Here a1, b1, c1, a2, b2, c2 are all real numbers.

    Note that a12 + b12 ≠ 0, a22 + b22 ≠ 0.

    Case 1. If (a1/a2) = (b1/b2) = (c1/c2), then there will be infinitely many solutions. This type of equation is called a dependent pair of linear equations in two variables. If we plot the graph of this equation, the lines will coincide.Case 2. If (a1/a2) = (b1/b2) ≠ (c1/c2), then there will be no solution. This type of equation is called an inconsistent pair of linear equations. If we plot the graph, the lines will be parallel. The graph is shown below.

    Solved Examples

    Example 1: How many solutions does the following system have?

    y = -3x + 9 y = -3x – 7 (A) One solution (B) No solution

    (C) Infinitely many solutions

    (D) None of these

    Solution:

    Given equations are y = -3x + 9

    y = -3x – 7

    Here (a1/a2) = (b1/b2) ≠ (c1/c2). So this system of equations has no solution.

    Another method:

    Without graphing them, we can see that both have the same slope -3 which means lines are parallel. Hence the system of equations has no solution.

    So option (B) is the answer.

    Example 2:

    Determine whether the following system of equations have no solution, infinitely many solution or unique solutions. x+2y = 3, 2x+4y = 15

    Solution:

    Given equations are x+2y = 3

    2x+4y = 15

    a1 = 1, b1 = 2, c1 = -3

    a2 = 2, b2 = 4, c2 = -15

    a1/a2 = ½ b1/b2 = ½ c1/c2  = 1/5 a1/a2 = b1/b2≠c1/c2

    So, the system of equations has no solution.

    Related Links:

    Linear equations

    System of linear equations using determinants

    Techniques To Solve System Of Equations

    Solving Linear Equations using Matrix

    Frequently Asked Questions

    Define a Linear equation.

    A Linear equation is an equation that has one or more variables having degree one.

    Give an example of a Linear equation in two variables.

    An example of a Linear equation in two variables: 2x + 3y + 4 = 0.

    Give the condition for a system of linear equations that has no solution.

    If a1/a2 = (b1/b2) ≠ (c1/c2), then there will be no solution.

    Give the condition for a system of linear equations that has infinitely many solutions.

    If (a1/a2) = (b1/b2) = (c1/c2), then there will be infinitely many solutions.

    स्रोत : byjus.com

    Number of solutions to system of equations review (article)

    Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere.

    \small{1} 1 \small{2} 2 \small{3} 3 \small{\llap{-}1} - 1 \small{1} 1 \small{2} 2 \small{3} 3 \small{\llap{-}1} - 1 y y x x

    A coordinate plane. The x- and y-axes both scale by one-half. A graph of a line goes through the points negative one-half, three and three, two. A graph of another line goes through the points zero, zero and one, one. These two lines intersect at an x-value between two and three and a y-value between two and three.

    One solution. A system of linear equations has one solution when the graphs intersect at a point.

    \small{1} 1 \small{2} 2 \small{3} 3 \small{\llap{-}1} - 1 \small{1} 1 \small{2} 2 \small{3} 3 \small{\llap{-}1} - 1 y y x x

    A coordinate plane. The x- and y-axes both scale by one-half. A graph of a line goes through the points one, one and a half and three, one. A graph of another line goes through the points one, two and a half and three, two. These two lines never intersect.

    No solution. A system of linear equations has no solution when the graphs are parallel.

    \small{1} 1 \small{2} 2 \small{3} 3 \small{\llap{-}1} - 1 \small{1} 1 \small{2} 2 \small{3} 3 \small{\llap{-}1} - 1 y y x x

    A coordinate plane. The x- and y-axes both scale by one-half. A graph of a line goes through the points zero, one and a half and three, two. A graph of another line goes through the points zero, one and a half and three, two. These lines overlap entirely.

    Infinite solutions. A system of linear equations has infinite solutions when the graphs are the exact same line.

    Want to learn more about the number of solutions to systems of equations? Check out this video.

    Example system with one solution

    We're asked to find the number of solutions to this system of equations:

    \begin{aligned} y&=-6x+8\\\\ 3x+y&=-4 \end{aligned}

    y 3x+y ​ =−6x+8 =−4 ​

    Let's put them in slope-intercept form:

    \begin{aligned} y&=-6x+8\\\\ y&=-3x-4 \end{aligned}

    y y ​ =−6x+8 =−3x−4 ​

    Since the slopes are different, the lines must intersect. Here are the graphs:

    \small{2} 2 \small{4} 4 \small{6} 6 \small{8} 8 \small{\llap{-}4} - 4 \small{\llap{-}6} - 6 \small{\llap{-}8} - 8 \small{2} 2 \small{4} 4 \small{6} 6 \small{8} 8 \small{\llap{-}4} - 4 \small{\llap{-}6} - 6 \small{\llap{-}8} - 8 y y x x

    A coordinate plane. The x- and y-axes both scale by one-half. The equation y equals negative six x plus eight is graphed going through the points zero, eight and one, two. The equation three x plus y equals negative four is graphed going through the points zero, negative four and one, negative seven. These lines intersect at a value that is below the graph.

    Because the lines intersect at a point, there is one solution to the system of equations the lines represent.

    Example system with no solution

    We're asked to find the number of solutions to this system of equations:

    \begin{aligned} y &= -3x+9\\\\ y &= -3x-7 \end{aligned}

    y y ​ =−3x+9 =−3x−7 ​

    Without graphing these equations, we can observe that they both have a slope of

    -3 −3 minus, 3

    . This means that the lines must be parallel. And since the

    y y y

    -intercepts are different, we know the lines are not on top of each other.

    There is no solution to this system of equations.

    Example system with infinite solutions

    We're asked to find the number of solutions to this system of equations:

    \begin{aligned} -6x+4y &= 2\\\\ 3x-2y &= -1 \end{aligned}

    −6x+4y 3x−2y ​ =2 =−1 ​

    Interestingly, if we multiply the second equation by

    -2 −2 minus, 2

    , we get the first equation:

    \begin{aligned} 3x-2y &= -1\\\\ \blueD{-2}(3x-2y)&=\blueD{-2}(-1)\\\\ -6x+4y &= 2 \end{aligned}

    3x−2y −2(3x−2y) −6x+4y ​ =−1 =−2(−1) =2 ​

    In other words, the equations are equivalent and share the same graph. Any solution that works for one equation will also work for the other equation, so there are infinite solutions to the system.

    Practice

    PROBLEM 1

    How many solutions does the system of linear equations have?

    \begin{aligned} y &= -2x+4\\\\ 7y &= -14x+28 \end{aligned}

    y 7y ​ =−2x+4 =−14x+28 ​ Choose 1 answer: Choose 1 answer:

    Want more practice? Check out these exercises:

    Number of solutions from graph

    Number of solutions from equations

    स्रोत : www.khanacademy.org

    determinant

    For the system $$ \left\{ \begin{array}{rcrcrcr} x &+ &3y &- &z &= &-4 \\ 4x &- &y &+ &2z &= &3 \\ 2x &- &y &- &3z &= &1...

    Set of Linear equation has no solution or unique solution or infinite solution?

    Ask Question

    Asked 8 years, 11 months ago

    Modified 3 years ago

    Viewed 162k times 16 For the system ⎧ ⎩ ⎨ ⎪ ⎪ x 4x 2x + − − 3y y y − + − z 2z 3z = = = −4 3 1

    {x+3y−z=−44x−y+2z=32x−y−3z=1

    what is the condition to determine if there is no solution or unique solution or infinite solution?

    Thank you!

    linear-algebradeterminantsystems-of-equations

    Share

    edited Jan 16, 2014 at 15:25

    Sammy Black 19.4k3 3 gold badges 30 30 silver badges 48 48 bronze badges

    asked Jan 16, 2014 at 15:11

    Terminal 3431 1 gold badge 2 2 silver badges 6 6 bronze badges 3

    Did you look at the determinant? Did you perform Gaussian ELimination? This system has a unique solution. Also see: math.stackexchange.com/questions/249919/… and math.stackexchange.com/questions/104824/… egards –

    Amzoti

    Jan 16, 2014 at 15:38

    Add a comment

    3 Answers

    29

    Probably the most straightforward method (to fully distinguish between the various possibilities) that I've seen is transforming the corresponding augmented matrix into row-reduced echelon form. In this case, you would start with:

    ⎡ ⎣ ⎢ ⎢ 1 4 2 3 −1 −1 −1 2 −3 −4 3 1 ⎤ ⎦ ⎥ ⎥ [13−1−44−1232−1−31] Subtracting 4 4

    times the first row from the second, and

    2 2

    times the first row from the third, we have:

    ⎡ ⎣ ⎢ ⎢ 1 0 0 3 −13 −7 −1 6 −1 −4 19 9 ⎤ ⎦ ⎥ ⎥

    [13−1−40−136190−7−19]

    Subtracting 2 2

    times the third row from the second, we have:

    ⎡ ⎣ ⎢ ⎢ 1 0 0 3 1 −7 −1 8 −1 −4 1 9 ⎤ ⎦ ⎥ ⎥ [13−1−401810−7−19] Adding 7 7

    times the second row to the third, we have:

    ⎡ ⎣ ⎢ ⎢ 1 0 0 3 1 0 −1 12 55 −4 1 16 ⎤ ⎦ ⎥ ⎥ [13−1−401121005516]

    At this point, we have only zeroes below the main diagonal, but no zeroes on the diagonal, so a unique solution exists. Continuing to reduce until the

    3×3 3×3

    portion of the augmented matrix is just the

    3×3 3×3

    identity matrix, we have

    ⎡ ⎣ ⎢ ⎢ 1 0 0 0 1 0 0 0 1 3/11 −73/55 16/55 ⎤ ⎦ ⎥ ⎥

    [1003/11010−73/5500116/55]

    This tells us that x=3/11, x=3/11, y=−73/55, y=−73/55, z=16/55 z=16/55

    is the unique solution to the system.

    Let's consider another system:

    ⎧ ⎩ ⎨ x+3y−z=4 4x−y+2z=8 2x−7y+4z=−3,

    {x+3y−z=44x−y+2z=82x−7y+4z=−3,

    which has corresponding matrix

    ⎡ ⎣ ⎢ ⎢ 1 4 2 3 −1 −7 −1 2 4 4 8 −3 ⎤ ⎦ ⎥ ⎥ . [13−144−1282−74−3].

    Starting out the same way gets us

    ⎡ ⎣ ⎢ ⎢ 1 0 0 3 −13 −13 −1 6 6 4 −8 −11 ⎤ ⎦ ⎥ ⎥ ,

    [13−140−136−80−136−11],

    and subtracting the second row from the third gives us

    ⎡ ⎣ ⎢ ⎢ 1 0 0 3 −13 0 −1 6 0 4 −8 −3 ⎤ ⎦ ⎥ ⎥ .

    [13−140−136−8000−3].

    Now we have only zeroes below the main diagonal, but we have a zero on the main diagonal, too. This tells us that either there are no solutions or there are infinitely-many. Translated back into terms of

    x,y,z x,y,z

    this is the equivalent system

    ⎧ ⎩ ⎨ x+3y+−z=4 0x−13y+6z=−8 0x+0y+0z=−3,

    {x+3y+−z=40x−13y+6z=−80x+0y+0z=−3,

    or alternatively ⎧ ⎩ ⎨ ⎪ ⎪ x=− 5 13 z+ 28 13 y= 6 13 z+ 8 13 0=3,

    {x=−513z+2813y=613z+8130=3,

    but there is no solution to the last equation, so no solution to the system.

    Upshot: We will have no solutions whenever we end up with one or more rows of all

    0 0

    s except in the last column as we reduce the augmented matrix.

    By contrast, if we'd started with the system

    ⎧ ⎩ ⎨ x+3y−z=4 4x−y+2z=8 2x−7y+4z=0,

    {x+3y−z=44x−y+2z=82x−7y+4z=0,

    which has corresponding matrix

    ⎡ ⎣ ⎢ ⎢ 1 4 2 3 −1 −7 −1 2 4 4 8 0 ⎤ ⎦ ⎥ ⎥ , [13−144−1282−740],

    then our reduction process will get us

    ⎡ ⎣ ⎢ ⎢ 1 0 0 3 −13 0 −1

    स्रोत : math.stackexchange.com

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    Mohammed 1 month ago
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