Is there a way to solve for $x$ and $y$ in this simultaneous equation.

Question

Is there a way to solve for x and y in this simultaneous equation?

$$2x - 3y = 4 $$

$$4x - 6y = 5 $$

Attempt:

I tried solving it but $x$ and $y$ keeps eliminating.

Answer 1

Notice, re-write the given equations as follows $$2x-3y=4\implies y=\frac{2}{3}x-\frac{4}{3}\tag 1$$ & $$4x-6y=5\implies y=\frac{2}{3}x-\frac{5}{6}\tag 2$$ both (1) & (2) represent the straight lines which are parallel to each other having slope $\frac{2}{3}$ i.e. the pair of lines are not intersecting hence the given system of linear equations does not have any solution.

Answer 2

There are no solutions to this system since you will always get $5 = 8$

Answer 3

There is no solution. The two equations represent parallel lines which will never intersect.

Rewriting each in terms of $y$ should make this clear:

$$y=\frac{2x}{3}-\frac{4}{3}$$

$$y=\frac{2x}{3}-\frac{5}{6}$$

Answer 4

Hint Double the first equation. What does this tell you?

Answer 5

Your simultaneous equations have no solutions in $x$ and $y$.

In general, for simultaneous equations, $$a_1x+b_1y+c_1=0$$ and $$a_2x+b_2y+c_2=0$$ there will exist unique solutions iff $$\frac{a_1}{a_2} \not =\frac{b_1}{b_2}$$ But it is not so in your case.
There will be no solutions iff $$\frac{a_1}{a_2} = \frac{b_1}{b_2} \not = \frac{c_1}{c_2}$$ and infinite solutions iff $$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$$

Answer 6

There are no solutions as these lines are parallel, as one equation is a multiple of the other. Hence, simultaneous equations won't work.

Answer 7

there is no solution because given set of lines are parallel to each other . For confirmation multiply equation number 1 by 2 .

Answer 8

$$ \begin{cases} 2x - 3y = 4 \\ 4x - 6y = 5 \\ \end{cases}\Longleftrightarrow $$

$$ \begin{cases} x = 2- \frac{3y}{2} \\ 4x - 6y = 5 \\ \end{cases}\Longleftrightarrow $$

$$ \begin{cases} x = 2- \frac{3y}{2} \\ 4\left(2- \frac{3y}{2}\right) - 6y = 5 \\ \end{cases}\Longleftrightarrow $$

$$ \begin{cases} x = 2- \frac{3y}{2} \\ 8 = 5 \\ \end{cases} $$

And $8\neq 5$ so no solutions!