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

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

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

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### Video Transcript

What is the value of π for which the simultaneous equations π₯ plus two π¦ 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, letβs take both of these equations and put them in the slope-intercept form. Thatβs the form π¦ equals ππ₯ 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 weβre trying to get the form π¦ equals ππ₯ 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 weβre 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. ππ¦ equals five minus five π₯. We want our π₯ term first. So weβ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 π¦ by itself. So we divide everything by π. π¦ 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.

Letβ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, weβre going to look at the coefficients of π₯ and π¦ in both of these equations. And weβ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 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.

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