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I am working the next set of lineal equations

$\begin{cases} 7x+2y=1 & 1\\ 21x+6y=3 & 2 \end{cases}$
So multiplying (1) by -3

$7x+2y=1(-3)$

$-21x-6y=-3$

Adding both equations

$21x+6y=3$

$-21x-6y=-3$

0x+0y=0

0=0

This seems to be a uncompatible/no solutions system, but graphically the system is enter image description here
So there is a common point in (0,1/2),then it would have ,at least, one solution. And this solution solve the equations:
$\begin{cases} 7(0)+2\frac{1}{2}=1 & \\ 21(0)+6\frac{1}{2}=3 & \end{cases}$
but I have tested the system using the usual methods and gives the $\emptyset$. Then this system has solution or not?

UPDATE
The graphs are wrong the right one is enter image description here
so it has infinite solutions.

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    $\begingroup$ The graphs are not correct. You may want to check again. $\endgroup$ Commented Aug 18, 2021 at 4:51
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    $\begingroup$ Im checking it... $\endgroup$
    – avelardo
    Commented Aug 18, 2021 at 4:52
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    $\begingroup$ You say that "This seems to be a uncompatible/no solutions system". Nope. Because, $$7x+2y=1\iff 21x+6y=3$$ This means, we have infinitely many solutions and this is enough to solve $7x+2y=1\implies y=\frac{1-7x}{2}$ $\endgroup$ Commented Aug 18, 2021 at 4:55
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    $\begingroup$ These are coincident lines because the ratios of corresponding coefficients are equal. $\endgroup$
    – sato
    Commented Aug 18, 2021 at 5:14
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    $\begingroup$ A system of equations will have infinitely many solutions if one equation is a scalar multiple of another. $\endgroup$ Commented Aug 18, 2021 at 5:32

3 Answers 3

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The matrix $\begin{bmatrix} 7 & 2 \\ 21 & 6 \end{bmatrix}$ is a rank 1 matrix i.e. it's not full-ranked. So it's not invertible.

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You have either graphed $-7x+2y=1$ or $-21x+6y=3$ by mistake instead of the equations given in your system. In your original system, equation 2 is simply $3$ times equation 1. As such, you have two unknowns but only one relation between them, so your system has infinite solutions.

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The system has infinitely many solutions because $$7x+2y=1\qquad\Leftrightarrow\qquad 21x+6y=3,$$ so for any $x$ you can take $y=\frac{1-2y}{7}$ to get a solution.

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