### Video Transcript

What is the value of π for which
the simultaneous equations π₯ plus two π¦ equals one and five π₯ plus ππ¦ equals
five do not have a unique solution?

Equations that do not have a unique
solution do not intersect, which means they are parallel lines and they have equals
slopes. To find the slope, letβs take both
of these equations and put them in the slope-intercept form. Thatβs the form π¦ equals ππ₯ plus
π. And in this form, π represents the
slope.

To get equation one in the
slope-intercept form, we need to isolate π¦. We can do that by subtracting π₯
from both sides of the equation. We then have two π¦ equals one
minus π₯. But because weβre trying to get the
form π¦ equals ππ₯ plus π, we can flip the π₯ and the π. It will then say two π¦ equals
negative π₯ plus one.

From there, we divide everything by
two. π¦ equals negative π₯ over two plus
one-half. But remember weβre looking for that
π value. So we wanted to say π¦ equals
negative one-half times π₯ plus one-half. π equals the slope and it equals
negative one-half.

We want to get our second equation
in the same form. So we subtract five π₯ from both
sides. ππ¦ equals five minus five π₯. We want our π₯ term first. So weβll flip them around, being
careful to keep that negative with our π₯. ππ¦ equals negative five π₯ plus
five.

After that, we want π¦ by
itself. So we divide everything by π. π¦ equals negative five over ππ₯
plus five over π. The slope of our second equation is
negative five over π. And we want to know what π is. These two lines do not have a
unique solution. And that means they have equal
slopes. Negative one-half equals negative
five over π.

Letβs get the π out of the
denominator, multiply by π on both sides. π times negative one equals
negative π. The denominator stays negative
two. And that equals negative five. To get two out of the denominator,
we multiply by two on both sides. Negative π equals negative 10. And that means positive π equals
10.

When π equals 10, these two lines
have the same slope. And that means they have no unique
solution. But before we leave this problem, I
want to show you one other way to think about it.

This time, weβre going to look at
the coefficients of π₯ and π¦ in both of these equations. And weβre going to make a ratio of
these coefficients. The coefficient of π₯ in the first
equation is one. We could say one π₯. And the coefficient of π¦ is
two. They have a ratio of one to
two. In the second equation, the
coefficient of π₯ is five and the coefficient of π¦ is π.

Notice that to get from one to two,
we multiply by two. And if we want these two ratios to
be equal, then to go from five to π, we will also multiply by two. Five times two is 10. So in order for these equations to
be parallel, π must be equal to 10.

Both methods are equally valid. One just requires a little bit more
algebra than the other.