In Exercises 1–18, solve each system by the substitution method. { x+y=2 y=x^2-4 x+4 .

Name: Problem 3, Chapter 7: Precalculus
Uploaded: 2020-06-10T22:07:41-08:00
Duration: 3 min 41 s
Channel: Cullen Miller
Description: In Exercises 1–18, solve each system by the substitution method. { x+y=2 y=x^2-4 x+4 .

step 1 Hello, hope you're doing well. So we've got our two equations here for the system of equations.

In Exercises 1–18, solve each system by the substitution...

Key Concepts

System of Equations

A system of equations involves finding values for variables that satisfy all equations simultaneously. This is a fundamental concept in algebra where multiple relationships are given, and the goal is to determine the common solution set that makes each equation true.

Substitution Method

The substitution method is a technique used to solve systems of equations by isolating one variable in one equation and then substituting its expression into the other equations. This method reduces the number of variables in an equation step by step, eventually leading to a single equation in one variable, which simplifies the problem-solving process.

Quadratic Equations

Quadratic equations are second-degree polynomial equations, typically taking the form ax² + bx + c = 0. They emerge naturally when using the substitution method on systems where one equation is nonlinear, and solving them is a crucial skill in algebra.

Solving Quadratic Equations

Solving quadratic equations involves determining the values of the variable that satisfy the equation. This can be done using methods such as factoring, completing the square, or applying the quadratic formula. The choice of method depends on the form of the quadratic and can affect the ease and clarity of the solution process.

Transcript

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

00:02 So we've got our two equations here for the system of equations.

00:07 So in order to solve this system by substitution, we're going to take this second equation and plug it into the y value in the first equation.

00:17 So we've got our first equation written here.

00:19 Then when we plug our second equation into y, get x plus x squared minus 4x plus 4x is equal to 2.

00:29 So if we subtract two from both sides, and then we combine this x and minus 4x, get x squared minus 3x plus 2 is equal to 0.

00:43 So now in order to solve for x, we want to see if this expression can be factored.

00:48 So we've got 2 times 1, it's going to give us 2.

00:54 We've got a minus 3 here.

00:57 So we want to see what two numbers multiply to give us 2 and add to give us minus 3.

01:02 It's just going to be minus 2 and minus 1.

01:05 So that means that this expression's factored is going to be x minus 2 times x minus 1 is equal to 0.

01:15 So in order to solve this equation, we're going to set each of these factors equal to 0 and solve for x in order to see what x values will make this left -hand side of the equation equal to 0.

01:27 So first we've got this first factor that we're going to set to 0.

01:30 Get x minus 2 is equal to 0.

01:33 Adding 2 to both sides, we get x is equal to 2.

01:38 This is going to be x1, because this is our first x value that's part of our solution to this system of equations.

01:45 So going on to our second factor, we've got x minus 1 is equal to 0.

01:49 Adding 1 to both sides, get x is equal to 1.

01:54 This is going to be x2.

01:55 So these are our 2x values that are part of our solution to this system of the equation.

02:00 So now what we're going to do is we're going to plug these two x values into this second equation right here to see what y values correspond to these x values.

02:10 We've got our equation for y there...

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