Solve the system of equations by using substitution. { x^2+y^2=25 x-y=1 .

Name: Problem 210, Chapter 11: Intermediate Algebra
Uploaded: 2022-09-30T04:29:44-08:00
Duration: 5 min 37 s
Channel: Robin R
Description: Solve the system of equations by using substitution. { x^2+y^2=25 x-y=1 .

Solve the system of equations by using substitution. { ...

Key Concepts

Geometric Interpretation

When a linear equation is combined with a quadratic equation representing a circle, the solutions to the system correspond to the intersection points of the line with the circle. This interpretation aids in visualizing the problem, understanding the number of possible solutions, and verifying the results.

System of Equations

A system of equations consists of two or more equations with multiple variables that are solved simultaneously to find a set of values satisfying all equations at once. This often requires methods that can handle the interdependence of the variables involved.

Substitution Method

The substitution method is a technique used to solve a system of equations where one equation is solved for one variable in terms of the others. This expression is then substituted into another equation, which reduces the system to an equation with a single variable, making it easier to solve.

Quadratic Equations

Quadratic equations are polynomial equations of degree two. They often arise in systems that involve both linear and non-linear equations. Once substitution has reduced the problem to a quadratic equation, methods like factoring, completing the square, or using the quadratic formula are utilized to find the variable values.

Transcript

00:01 And this problem, we are supposed to solve the system of equations by using substitution.

00:08 So one thing i'm going to do is i'm going to take and solve that second one for y, just to make it clear that it's a straight line.

00:15 See if you add y to both sides, you're going to get x equals 1 plus y, subtract one from both sides.

00:26 So this was adding y, adding y, and then subtract one, subtract one, and i'm going to switch sides of the equation on you.

00:36 I'm going to get y equals x minus 1.

00:39 So that's a straight line.

00:41 And so with this setup, the first equation is the equation of a circle, and then you have a straight line.

00:49 So you could have two possible solutions.

00:51 You can have where it crosses at only one point there, or you can have no solutions where they don't intersect at all.

01:00 And so if we take the y equals x minus 1 and we plug it into the first equation, we'll have x squared plus y which is x minus 1 squared equals 25 and so then we're going to foil this would be x minus 1 times x minus 1 so first 2 would be x squared outside 2 would be minus x inside 2 would be minus x and last 2 would be plus 1 equals 25 so we'd have 2x squared minus 2x squared minus 2x and i'm going to take in sub subtract the 25 from both sides.

01:45 So i'm going to get minus 24 is minus 25 plus 1 and minus 24.

01:51 And that'll equal zero.

01:54 And then i can factor out of two.

02:05 And so what we want to look at is factoring x squared plus x minus 12.

02:09 I mean, x squared minus x minus 12.

02:13 So possible factors for 12 would be one in 12, two and six, and three and four.

02:21 And since the last, since this one right here is a negative, then that means we want to look at a difference of negative 1 or a difference of 1.

02:31 So we'll be looking at 3 and 4.

02:36 So that should factor 2.

02:40 The biggest one needs to be negative.

02:42 So the middle term will be negative.

02:43 So that should factor in x minus 4 and x plus 3.

02:47 And then if you foil that real quick, you'd have x squared plus 3x minus 4 x would be minus x and then minus 12.

02:56 So that does, x squared minus x minus 12 does factor 2.

02:59 That...