# This system has solution or not?

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

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

$$-21x-6y=-3$$

$$21x+6y=3$$

$$-21x-6y=-3$$

0x+0y=0

0=0

This seems to be a uncompatible/no solutions system, but graphically the system is
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
so it has infinite solutions.

• The graphs are not correct. You may want to check again. Commented Aug 18, 2021 at 4:51
• Im checking it... Commented Aug 18, 2021 at 4:52
• 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 $7x+2y=1\implies y=\frac{1-7x}{2}$ Commented Aug 18, 2021 at 4:55
• These are coincident lines because the ratios of corresponding coefficients are equal.
– sato
Commented Aug 18, 2021 at 5:14
• A system of equations will have infinitely many solutions if one equation is a scalar multiple of another. Commented Aug 18, 2021 at 5:32

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.
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.
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.