# How do you determine whether a linear system has one solution, many solutions, or no solution when given 2x + 3y= -6 and 3x- 4y= 25?

Oct 19, 2015

One solution only

#### Explanation:

They are both straight lines that are not parallel. As there is no defined limit to the range they are infinitely long. Thus they will cross at some as yet uncalculated point. They would only not cross if they were parallel.

To happen more than once at least one of them has to have at least one curve in it. For example you could have standard form equations of $y = a {x}^{2} + b x + c$ (quadratic which is a horse shoe shape) and the straight line of $y = m x + c$.

It is possible that the straight line curve (yes, it correct to call it a curve) could miss the quadratic entirely, touch it tangentially (just once) or cross it twice. It all depends on the gradients and positioning within the coordinate system.

In the context of your question, if you needed to actually find the point of intersection you would start as follows:

You would treat them as simultaneous equations and solve for each of the unknowns.

Write as:

$3 y = - 2 x - 6$.................... (1)
$4 y = 3 x - 25$.........................(2)

Giving:
$y = - \frac{2}{3} x - 2$.......................(1.1)
$y = \frac{3}{4} x - \frac{25}{4}$.........................(2.1)

You would then proceed to eliminate either y or x to find the value of the other. Once obtained simple substitution would yield the value of the other one.

However, your question does not request that you find this. Only askes how many solutions there are,so I am not going to continue.