Video: Finding the Unknown Coefficient in a Given System of Linear Equations without a Unique Solution

What is the value of π‘˜ for which the simultaneous equations π‘₯ + 2𝑦 = 1 and 5π‘₯ + π‘˜π‘¦ = 5 do not have a unique solution.


Video Transcript

What is the value of π‘˜ for which the simultaneous equations π‘₯ plus two 𝑦 equals one and five π‘₯ plus π‘˜π‘¦ 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, let’s take both of these equations and put them in the slope-intercept form. That’s the form 𝑦 equals π‘šπ‘₯ plus 𝑏. And in this form, π‘š represents the slope.

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

From there, we divide everything by two. 𝑦 equals negative π‘₯ over two plus one-half. But remember we’re looking for that π‘š value. So we wanted to say 𝑦 equals negative one-half times π‘₯ plus one-half. π‘š 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. π‘˜π‘¦ equals five minus five π‘₯. We want our π‘₯ term first. So we’ll flip them around, being careful to keep that negative with our π‘₯. π‘˜π‘¦ equals negative five π‘₯ plus five.

After that, we want 𝑦 by itself. So we divide everything by π‘˜. 𝑦 equals negative five over π‘˜π‘₯ 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‘˜.

Let’s 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, we’re going to look at the coefficients of π‘₯ and 𝑦 in both of these equations. And we’re 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 𝑦 is two. They have a ratio of one to two. In the second equation, the coefficient of π‘₯ is five and the coefficient of 𝑦 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.

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