53-56 Find x and y in terms of a and b. { a x+b y=1 b x+a y=1 (a^2-b^2 ≠0).

Name: Problem 55, Chapter 11: Algebra and Trigonometry
Uploaded: 2020-07-21T13:59:52-08:00
Duration: 3 min 33 s
Channel: Cullen Miller
Description: 53-56 Find x and y in terms of a and b. { a x+b y=1 b x+a y=1 (a^2-b^2 ≠0).

step 1 Hello, hope you're doing well. So we have our system of equations here, and we're going to try a

53-56 Find x and y in terms of a and b. { a x+b y=1 ...

Key Concepts

System of Linear Equations

A system of linear equations consists of two or more linear equations involving the same set of variables. In this context, the system defines relationships between the unknowns where each equation represents a line in a coordinate system. The solution to the system is the set of variable values that satisfy all equations simultaneously. Understanding the structure of these systems is essential as it leads to strategies for finding unique, infinite, or no solutions.

Determinant and Non-Degeneracy

The determinant of the coefficient matrix in a system of linear equations is a scalar value that indicates whether the system has a unique solution. A non-zero determinant (non-degeneracy condition) ensures that the matrix is invertible and that the system has exactly one solution. In the given problem, the condition a^2 - b^2 ? 0 is essentially the requirement that the determinant is non-zero, ensuring the feasibility of the solution in terms of a and b.

Elimination and Substitution Methods

Elimination and substitution are standard techniques used to solve systems of linear equations. Elimination involves adding or subtracting equations to remove one of the variables, while substitution expresses one variable in terms of another and then replaces it in the other equation. Both methods simplify the process of finding the values of the unknowns, and their effective use is critical in arriving at a solution when dealing with equations that involve parameters.

Transcript

00:01 Hello, hope you're doing well.

00:02 So we have our system of equations here, and we're going to try and solve this using the elimination method.

00:07 So we're going to try and eliminate our variable y.

00:10 So to do that, we're going to multiply this first equation by a, and this whole second equation by minus b.

00:17 So starting with our first equation in distributing the a, if a times a x is a squared x, we have a times b y is plus a b y, and that's equal to a times 1, which is equal to a.

00:32 Moving on to our second equation in distributing the minus v, we have minus b times vx is minus b squared x.

00:39 And we have minus v times a y is minus a b y.

00:42 That's equal to minus b times 1, which is minus b.

00:46 So now adding these two equations together, right here we have ab y minus a b y.

00:51 So these cancel out and go to zero.

00:53 So we're left with a squared x minus v squared x, which is a squared minus v squared x, and that's equal to a minus b.

01:02 Voting both sides of our equation by a squared minus b squared, we end up with, we end up with x is equal to a minus b over a squared minus b squared.

01:17 The a squared minus b squared is the difference of squared.

01:20 So that's equal to a plus b times a minus b.

01:24 And our numerator is still a minus b.

01:27 So these a minus b can cancel and we're left with x is equal to 1 over a plus b.

01:32 All right, so now we want to try and figure out what y value corresponds to this x value...

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